Why aren't satellites disintegrated even though they orbit earth within earth's Roche Limits?Tidal force when planets and smaller objects collideWould a considerably big asteroid be disintegrated by the Earth's Roche limit?How many celestial bodies could be in stable orbit at roughly the same distance from a star?Why aren't there more natural satellites orbiting the Earth than just the Moon?How long can a satellite stay in orbit?Is the lay explanation of the equivalence principle wrong?Why aren't satellites increasing speed because of gravitational acceleration?Tidal Forces MisunderstandingWhy aren't Roche limit and the difference in gravitational acceleration the same?Debris ring formed from breakup of extreme-high rotation rate of rigid ring
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Why aren't satellites disintegrated even though they orbit earth within earth's Roche Limits?
Tidal force when planets and smaller objects collideWould a considerably big asteroid be disintegrated by the Earth's Roche limit?How many celestial bodies could be in stable orbit at roughly the same distance from a star?Why aren't there more natural satellites orbiting the Earth than just the Moon?How long can a satellite stay in orbit?Is the lay explanation of the equivalence principle wrong?Why aren't satellites increasing speed because of gravitational acceleration?Tidal Forces MisunderstandingWhy aren't Roche limit and the difference in gravitational acceleration the same?Debris ring formed from breakup of extreme-high rotation rate of rigid ring
$begingroup$
I was wondering about the Roche limit and its effects on satellites.
Why aren't artificial satellites ripped apart by gravitational tidal forces of the earth?
I think it's due to the satellites being stronger than rocks?
Is this true?
Also, is the Roche limit just a line (very narrow band) around the planet or is it a range (broad cross sectional area) of distance around the planet?
newtonian-gravity orbital-motion tidal-effect satellites
asked May 15 at 10:07
Rajkumar Maurya Rajkumar Maurya
43129
$endgroup$
add a comment |
$begingroup$
I was wondering about the Roche limit and its effects on satellites.
Why aren't artificial satellites ripped apart by gravitational tidal forces of the earth?
I think it's due to the satellites being stronger than rocks?
Is this true?
Also, is the Roche limit just a line (very narrow band) around the planet or is it a range (broad cross sectional area) of distance around the planet?
newtonian-gravity orbital-motion tidal-effect satellites
asked May 15 at 10:07
Rajkumar Maurya Rajkumar Maurya
43129
$endgroup$
63
$begingroup$
Basically, it's because artificial satellites aren't piles of loose gravel.
$endgroup$
– Ilmari Karonen
May 15 at 13:51
add a comment |
$begingroup$
I was wondering about the Roche limit and its effects on satellites.
Why aren't artificial satellites ripped apart by gravitational tidal forces of the earth?
I think it's due to the satellites being stronger than rocks?
Is this true?
Also, is the Roche limit just a line (very narrow band) around the planet or is it a range (broad cross sectional area) of distance around the planet?
newtonian-gravity orbital-motion tidal-effect satellites
asked May 15 at 10:07
Rajkumar Maurya Rajkumar Maurya
43129
$endgroup$
I was wondering about the Roche limit and its effects on satellites.
Why aren't artificial satellites ripped apart by gravitational tidal forces of the earth?
I think it's due to the satellites being stronger than rocks?
Is this true?
Also, is the Roche limit just a line (very narrow band) around the planet or is it a range (broad cross sectional area) of distance around the planet?
newtonian-gravity orbital-motion tidal-effect satellites
newtonian-gravity orbital-motion tidal-effect satellites
asked May 15 at 10:07
Rajkumar Maurya Rajkumar Maurya
43129
asked May 15 at 10:07
Rajkumar Maurya Rajkumar Maurya
43129
edited May 16 at 7:08
Rajkumar Maurya
asked May 15 at 10:07
Rajkumar Maurya Rajkumar Maurya
43129
asked May 15 at 10:07
Rajkumar Maurya Rajkumar Maurya
43129
asked May 15 at 10:07
Rajkumar Maurya Rajkumar Maurya
43129
43129
63
$begingroup$
Basically, it's because artificial satellites aren't piles of loose gravel.
$endgroup$
– Ilmari Karonen
May 15 at 13:51
add a comment |
63
$begingroup$
Basically, it's because artificial satellites aren't piles of loose gravel.
$endgroup$
– Ilmari Karonen
May 15 at 13:51
63
63
$begingroup$
Basically, it's because artificial satellites aren't piles of loose gravel.
$endgroup$
– Ilmari Karonen
May 15 at 13:51
$begingroup$
Basically, it's because artificial satellites aren't piles of loose gravel.
$endgroup$
– Ilmari Karonen
May 15 at 13:51
add a comment |
6 Answers
6
active
oldest
votes
$begingroup$
The Roche limit denotes how close a body held together by its own gravity can come. Since gravity tends to be the only thing holding moon-sized objects together, you won't find natural moons closer than the Roche limit. [Strictly speaking, the Roche Limit is a function of both the primary (in the case of this question, Earth) and the secondary (satellites) bodies; there is a different Roche limit for objects with different densities, but for simplicity I'll be treating the Roche Limit as being a function just of the primary.] For instance, Saturn's rings lie inside its Roche limit, and may be the debris from a satellite that was ripped apart. The rings are made up of small particles, and each particle is held together by molecular bonds. Since they have something other than gravity holding them together, they are not ripped apart any further. Similarly, an artificial satellite is also held together by molecular bonds, not internal gravity.
The molecular-bonds-will-be-ripped-apart-by-tidal-forces limit is obviously much smaller than a satellite's orbit, as we, on the surface of the Earth, are even closer, and we are not ripped apart. You would have to have an extremely dense object, such as a neutron star or black hole, for that limit to exist. Being inside the Roche limit does mean that if an astronaut were to go on a space walk without a tether, tidal forces would pull them away from the larger satellite. Outside the Roche limit, the gravity of the larger satellite would pull the astronaut back (although not before the astronaut runs out of air).
If you look at the influence of the Moon's tides on Earth, you can see that the oceans are pulled towards the Moon, but the land is (relatively) stationary. The fact that tides are only a few meters shows that the Earth is well outside the Moon's Roche limit (and of course, the Earth's Roche limit is further out than the Moon's, so the Moon would reach the Earth's Roche limit long before the Earth reached the Moon's). If the Moon were to move towards the Earth, the tides would get higher and higher. The Moon's Roche limit is the point at which the tides would get so high that the water is ripped away from the Earth. The land would still survive slightly past that point, because the crust has some rigidity beyond mere gravitational attraction.
Regarding your second question: there is a region in which the tidal forces would be larger than internal gravitational attraction, and a region in which internal gravitational attraction would be larger than tidal forces. The Roche limit is the boundary between those two regions. Everything inside the Roche limit constitutes the former region, while everything outside the Roche limit constitutes the latter.
answered May 15 at 15:05
AcccumulationAcccumulation
3,694715
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14
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@Ed999 Yes. "Were" is the subjunctive form of "be", which indicates things contrary to fact.
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– Acccumulation
May 15 at 19:24
1
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And the innermost muns of Saturn, which orbit in gaps in the main rings, also lie inside the planet's Roche limit, and are (like artificial satellites) held together by their own tensile strength.
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– Sean
May 15 at 20:37
1
$begingroup$
Some of the questions posted here (not this one!) are asked by people whose knowledge of the most basic concepts of physics is only at a schoolboy level. It is therefore incumbent on us not to muddy the waters even further by posting scenarios which are (to us) obviously impossible, but which might be misunderstood by a casual reader as implying that, for example, the Moon could fall out of the sky (if you were to wait a million years). Where something is definitely impossible it is incumbent on us not to present it as being possible in the event of some circumstance that can't happen.
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– Ed999
May 15 at 23:25
8
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@Ed999 I don't think your nitpicking is in place here. How would you better propose an artificial scenario to depict the effect other than quite literally saying "if you somehow placed the moon close to Earth"? Nothing in the answer suggest it's going to happen. Writing answers in such defensive way against people misunderstanding hypotetical scenario for a fact in "If moon were moving towards Earth" is madness. IMHO it's also an insult to intelligence of general audience. You can't teach physics with out considering idealized or impossible scenarios, and some of best teachers did it
$endgroup$
– luk32
May 16 at 15:08
1
$begingroup$
This answer could be improved by pointing out explicitly that there's no definite "Earth's Roche limit". The limit depends on the density of the orbiting body -- concentrating the same mass in a smaller volume means that the self-gravitation on the surface is larger whereas the tidal acceleration on the surface is weaker, so a denser object withstands tides better and the Roche limit that applies to it is smaller.
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– Henning Makholm
May 17 at 15:14
|
show 2 more comments
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The Roche limit applies only to bodies which are held together purely by internal gravitational attraction. Compact objects such as artificial satellites are held together by the much stronger inter-molecular electromagnetic forces (this is another demonstration of just how weak gravity is compared to electromagnetism).
As for your second question: the Roche limit usually defined as the radius away from a body at which magnitude of the tidal forces exactly equal that of the internal gravitational attraction of the smaller body. Of course, the magnitude of the tidal forces becomes significant at further radii, and so the distance at which tidal forces become significant is a much broader area/range.
answered May 15 at 10:19
Bob KnightonBob Knighton
5,97931233
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10
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@DarrelHoffman: No, there isn't. A satellite being pulled apart by tidal forces either breaks up or it doesn't. The tidal forces don't get any stronger over time, unless of course the satellite itself is slowly spiraling towards the planet (like Phobos is, for example). Of course, if a solid satellite is orbiting within the Roche limit, an impact or some other source of momentary stress might break off a chunk that then goes flying away, because the satellite's own gravity isn't enough to hold it down. But while impacts do tend to occur gradually over time, they're not caused by time alone.
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– Ilmari Karonen
May 15 at 13:49
2
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When a man-made satellite in low earth orbit runs out of fuel for its reaction jets, it can no longer make orbital corrections to counteract the (very thin) atmospheric friction, so over a long period will be decelerated by it, and pulled down to Earth, as occurred with Skylab in the 1970s. But one in a high orbit (a prograde orbit about a rotating mass) is accelerated by the Earth's gravity (i.e. by the rotation-dragging effect), so will gradually gain increased orbital altitude (so does not "de-orbit"). This is the effect which causes the Moon to increase its distance from the Earth.
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– Ed999
May 15 at 18:17
1
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@DarrelHoffman The satellites will mostely de-orbit. Some might break apart for one reason or another. None of the satellites in orbit is going to break apart due to the tidal forces from the earth pulling the satellite to pieces - because the tidal forces are not sufficient to overcome the friction forces that are keeping the satellite together.
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– Taemyr
May 16 at 10:26
1
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@J... Bolts, tape and the like all still work through electromagnetic interaction.
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– Bob Knighton
May 16 at 17:37
1
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@BobKnighton Naturally, but as a macroscopic system it is geometrically and physically distinct from a contiguous piece of material held together by purely microscopic interactions.
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– J...
May 16 at 18:30
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show 3 more comments
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The Roche limit is a limit on objects being held together by their own gravity. Satellites are held together by much stronger forces. Different parts of the satellite are ultimately connected by chemical bonds, which are electromagnetic.
answered May 15 at 10:18
knzhouknzhou
49.9k12137245
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add a comment |
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To add to other answers, also consider that artificial satellites are much smaller than natural satellites. This means that the difference between the gravitational force at the point the closest to the planet and at the point the furthest from the planet is much smaller in artificial satellites.
answered May 15 at 20:25
stackzebrastackzebra
1312
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... and therefore tidal factors are negligible is what you're saying, right? Just making sure I understand.
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– user45266
May 16 at 3:18
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@user45266 Yes.
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– stackzebra
May 16 at 11:03
2
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The size of the bodies does go into the Roche formula--the Roche "limit" isn't a precise number, when it is presented as one number that's what it would be for an ordinary moon. (Note that Titan would start coming apart earlier.)
$endgroup$
– Loren Pechtel
May 17 at 13:40
add a comment |
$begingroup$
When I was a kid I also wondered why artificial satellites within the Roche Limit were not pulled apart by tidal forces.
When I was a kid I also wondered, if any body within the Roche Limit would be pulled apart by tidal forces, and since the surface of the Earth is deep within the Roche limit, why aren't all objects on the surface of the Earth - including Human bodies - pulled apart by tidal forces.
Since my body was not being pulled apart by tidal forces the statement that all bodies within the Roche Limit were pulled apart by tidal forces must not be correct. Therefore the simple statement that all bodies within the Roche Limit are pulled apart by tidal forces must be an oversimplification as stated.
But since such statements were made in non fiction sources it seemed probable that they were not totally false. Therefore I expected that sometime in the future I would read a fuller and more complex account of the Roche limit that would explain the seeming paradoxes.
And I did. Eventually I learned that the Roche limit was not a single absolute distance but varied with the sizes, masses, and densities of the larger and the smaller objects. I also learned that the Roche Limit only applied to objects that were held together only by their internal gravitational attraction and not to objects like artificial satellites or Human bodies.
Wondering why the Roche Limit didn't apply to my Human body was an example of using reductio ad absurdum to show that a statement was an oversimplification of a more complex situation.
answered May 17 at 17:26
M.A. GoldingM.A. Golding
111
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add a comment |
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Why aren't artificial satellites ripped apart by gravitational tidal forces of the earth?
In short, the pieces of a satellite are held together by chemical, molecular bonds, which are strong compared to the tidal forces that are weak for objects the size of artificial satellites and gravity sources the strength of the earth.
A tidal force is the difference in attraction between two distances from a gravitational source. For a small object such as a natural satellite, that is just a few meters, gravity simply won't vary much between the nearest and farthest points unless you're talking about a orbiting a neutron star or black hole extremely closely.
For instance the ISS is 100m along its longest axis. Say that axis is perpendicular to the earth's surface. If its nearest point to earth is at 400km from the surface, the farthest point is therefore 400.1km. But earth is 5371km radius, so the distance from the center (where the gravity "seems" to be coming from) is 5771km and 5771.1km.
Gravity falls off at 1/(distance squared), so the difference in gravity is (1/(5771^2)) / (1/(5771.1^2)). That's a difference of 3*10^-8, or .000003% difference.
Gravity at earth's surface is 1G, giving 1kg mass 9.8N (newton) force (what you feel when you hold up 1kg on earth's surface). At 400km altitude it'd be 9.8*5371/(5371+400))^2=8.4N . So if you had even 1,000,000kg at each end of a 100m rope at that altitude, 1,000,000 * 8.4N * .000003% = .25N, about what you feel were you to pick up 25g of coins right now. (About 5 nickels.)
Even a human hair is strong enough to hold 25g at the earth's surface, and further, satellites aren't that heavy, and are far stronger than a hair.
answered May 18 at 1:31
Swiss FrankSwiss Frank
1112
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6 Answers
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6 Answers
6
active
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$begingroup$
The Roche limit denotes how close a body held together by its own gravity can come. Since gravity tends to be the only thing holding moon-sized objects together, you won't find natural moons closer than the Roche limit. [Strictly speaking, the Roche Limit is a function of both the primary (in the case of this question, Earth) and the secondary (satellites) bodies; there is a different Roche limit for objects with different densities, but for simplicity I'll be treating the Roche Limit as being a function just of the primary.] For instance, Saturn's rings lie inside its Roche limit, and may be the debris from a satellite that was ripped apart. The rings are made up of small particles, and each particle is held together by molecular bonds. Since they have something other than gravity holding them together, they are not ripped apart any further. Similarly, an artificial satellite is also held together by molecular bonds, not internal gravity.
The molecular-bonds-will-be-ripped-apart-by-tidal-forces limit is obviously much smaller than a satellite's orbit, as we, on the surface of the Earth, are even closer, and we are not ripped apart. You would have to have an extremely dense object, such as a neutron star or black hole, for that limit to exist. Being inside the Roche limit does mean that if an astronaut were to go on a space walk without a tether, tidal forces would pull them away from the larger satellite. Outside the Roche limit, the gravity of the larger satellite would pull the astronaut back (although not before the astronaut runs out of air).
If you look at the influence of the Moon's tides on Earth, you can see that the oceans are pulled towards the Moon, but the land is (relatively) stationary. The fact that tides are only a few meters shows that the Earth is well outside the Moon's Roche limit (and of course, the Earth's Roche limit is further out than the Moon's, so the Moon would reach the Earth's Roche limit long before the Earth reached the Moon's). If the Moon were to move towards the Earth, the tides would get higher and higher. The Moon's Roche limit is the point at which the tides would get so high that the water is ripped away from the Earth. The land would still survive slightly past that point, because the crust has some rigidity beyond mere gravitational attraction.
Regarding your second question: there is a region in which the tidal forces would be larger than internal gravitational attraction, and a region in which internal gravitational attraction would be larger than tidal forces. The Roche limit is the boundary between those two regions. Everything inside the Roche limit constitutes the former region, while everything outside the Roche limit constitutes the latter.
answered May 15 at 15:05
AcccumulationAcccumulation
3,694715
$endgroup$
14
$begingroup$
@Ed999 Yes. "Were" is the subjunctive form of "be", which indicates things contrary to fact.
$endgroup$
– Acccumulation
May 15 at 19:24
1
$begingroup$
And the innermost muns of Saturn, which orbit in gaps in the main rings, also lie inside the planet's Roche limit, and are (like artificial satellites) held together by their own tensile strength.
$endgroup$
– Sean
May 15 at 20:37
1
$begingroup$
Some of the questions posted here (not this one!) are asked by people whose knowledge of the most basic concepts of physics is only at a schoolboy level. It is therefore incumbent on us not to muddy the waters even further by posting scenarios which are (to us) obviously impossible, but which might be misunderstood by a casual reader as implying that, for example, the Moon could fall out of the sky (if you were to wait a million years). Where something is definitely impossible it is incumbent on us not to present it as being possible in the event of some circumstance that can't happen.
$endgroup$
– Ed999
May 15 at 23:25
8
$begingroup$
@Ed999 I don't think your nitpicking is in place here. How would you better propose an artificial scenario to depict the effect other than quite literally saying "if you somehow placed the moon close to Earth"? Nothing in the answer suggest it's going to happen. Writing answers in such defensive way against people misunderstanding hypotetical scenario for a fact in "If moon were moving towards Earth" is madness. IMHO it's also an insult to intelligence of general audience. You can't teach physics with out considering idealized or impossible scenarios, and some of best teachers did it
$endgroup$
– luk32
May 16 at 15:08
1
$begingroup$
This answer could be improved by pointing out explicitly that there's no definite "Earth's Roche limit". The limit depends on the density of the orbiting body -- concentrating the same mass in a smaller volume means that the self-gravitation on the surface is larger whereas the tidal acceleration on the surface is weaker, so a denser object withstands tides better and the Roche limit that applies to it is smaller.
$endgroup$
– Henning Makholm
May 17 at 15:14
|
show 2 more comments
$begingroup$
The Roche limit denotes how close a body held together by its own gravity can come. Since gravity tends to be the only thing holding moon-sized objects together, you won't find natural moons closer than the Roche limit. [Strictly speaking, the Roche Limit is a function of both the primary (in the case of this question, Earth) and the secondary (satellites) bodies; there is a different Roche limit for objects with different densities, but for simplicity I'll be treating the Roche Limit as being a function just of the primary.] For instance, Saturn's rings lie inside its Roche limit, and may be the debris from a satellite that was ripped apart. The rings are made up of small particles, and each particle is held together by molecular bonds. Since they have something other than gravity holding them together, they are not ripped apart any further. Similarly, an artificial satellite is also held together by molecular bonds, not internal gravity.
The molecular-bonds-will-be-ripped-apart-by-tidal-forces limit is obviously much smaller than a satellite's orbit, as we, on the surface of the Earth, are even closer, and we are not ripped apart. You would have to have an extremely dense object, such as a neutron star or black hole, for that limit to exist. Being inside the Roche limit does mean that if an astronaut were to go on a space walk without a tether, tidal forces would pull them away from the larger satellite. Outside the Roche limit, the gravity of the larger satellite would pull the astronaut back (although not before the astronaut runs out of air).
If you look at the influence of the Moon's tides on Earth, you can see that the oceans are pulled towards the Moon, but the land is (relatively) stationary. The fact that tides are only a few meters shows that the Earth is well outside the Moon's Roche limit (and of course, the Earth's Roche limit is further out than the Moon's, so the Moon would reach the Earth's Roche limit long before the Earth reached the Moon's). If the Moon were to move towards the Earth, the tides would get higher and higher. The Moon's Roche limit is the point at which the tides would get so high that the water is ripped away from the Earth. The land would still survive slightly past that point, because the crust has some rigidity beyond mere gravitational attraction.
Regarding your second question: there is a region in which the tidal forces would be larger than internal gravitational attraction, and a region in which internal gravitational attraction would be larger than tidal forces. The Roche limit is the boundary between those two regions. Everything inside the Roche limit constitutes the former region, while everything outside the Roche limit constitutes the latter.
answered May 15 at 15:05
AcccumulationAcccumulation
3,694715
$endgroup$
14
$begingroup$
@Ed999 Yes. "Were" is the subjunctive form of "be", which indicates things contrary to fact.
$endgroup$
– Acccumulation
May 15 at 19:24
1
$begingroup$
And the innermost muns of Saturn, which orbit in gaps in the main rings, also lie inside the planet's Roche limit, and are (like artificial satellites) held together by their own tensile strength.
$endgroup$
– Sean
May 15 at 20:37
1
$begingroup$
Some of the questions posted here (not this one!) are asked by people whose knowledge of the most basic concepts of physics is only at a schoolboy level. It is therefore incumbent on us not to muddy the waters even further by posting scenarios which are (to us) obviously impossible, but which might be misunderstood by a casual reader as implying that, for example, the Moon could fall out of the sky (if you were to wait a million years). Where something is definitely impossible it is incumbent on us not to present it as being possible in the event of some circumstance that can't happen.
$endgroup$
– Ed999
May 15 at 23:25
8
$begingroup$
@Ed999 I don't think your nitpicking is in place here. How would you better propose an artificial scenario to depict the effect other than quite literally saying "if you somehow placed the moon close to Earth"? Nothing in the answer suggest it's going to happen. Writing answers in such defensive way against people misunderstanding hypotetical scenario for a fact in "If moon were moving towards Earth" is madness. IMHO it's also an insult to intelligence of general audience. You can't teach physics with out considering idealized or impossible scenarios, and some of best teachers did it
$endgroup$
– luk32
May 16 at 15:08
1
$begingroup$
This answer could be improved by pointing out explicitly that there's no definite "Earth's Roche limit". The limit depends on the density of the orbiting body -- concentrating the same mass in a smaller volume means that the self-gravitation on the surface is larger whereas the tidal acceleration on the surface is weaker, so a denser object withstands tides better and the Roche limit that applies to it is smaller.
$endgroup$
– Henning Makholm
May 17 at 15:14
|
show 2 more comments
$begingroup$
The Roche limit denotes how close a body held together by its own gravity can come. Since gravity tends to be the only thing holding moon-sized objects together, you won't find natural moons closer than the Roche limit. [Strictly speaking, the Roche Limit is a function of both the primary (in the case of this question, Earth) and the secondary (satellites) bodies; there is a different Roche limit for objects with different densities, but for simplicity I'll be treating the Roche Limit as being a function just of the primary.] For instance, Saturn's rings lie inside its Roche limit, and may be the debris from a satellite that was ripped apart. The rings are made up of small particles, and each particle is held together by molecular bonds. Since they have something other than gravity holding them together, they are not ripped apart any further. Similarly, an artificial satellite is also held together by molecular bonds, not internal gravity.
The molecular-bonds-will-be-ripped-apart-by-tidal-forces limit is obviously much smaller than a satellite's orbit, as we, on the surface of the Earth, are even closer, and we are not ripped apart. You would have to have an extremely dense object, such as a neutron star or black hole, for that limit to exist. Being inside the Roche limit does mean that if an astronaut were to go on a space walk without a tether, tidal forces would pull them away from the larger satellite. Outside the Roche limit, the gravity of the larger satellite would pull the astronaut back (although not before the astronaut runs out of air).
If you look at the influence of the Moon's tides on Earth, you can see that the oceans are pulled towards the Moon, but the land is (relatively) stationary. The fact that tides are only a few meters shows that the Earth is well outside the Moon's Roche limit (and of course, the Earth's Roche limit is further out than the Moon's, so the Moon would reach the Earth's Roche limit long before the Earth reached the Moon's). If the Moon were to move towards the Earth, the tides would get higher and higher. The Moon's Roche limit is the point at which the tides would get so high that the water is ripped away from the Earth. The land would still survive slightly past that point, because the crust has some rigidity beyond mere gravitational attraction.
Regarding your second question: there is a region in which the tidal forces would be larger than internal gravitational attraction, and a region in which internal gravitational attraction would be larger than tidal forces. The Roche limit is the boundary between those two regions. Everything inside the Roche limit constitutes the former region, while everything outside the Roche limit constitutes the latter.
answered May 15 at 15:05
AcccumulationAcccumulation
3,694715
$endgroup$
The Roche limit denotes how close a body held together by its own gravity can come. Since gravity tends to be the only thing holding moon-sized objects together, you won't find natural moons closer than the Roche limit. [Strictly speaking, the Roche Limit is a function of both the primary (in the case of this question, Earth) and the secondary (satellites) bodies; there is a different Roche limit for objects with different densities, but for simplicity I'll be treating the Roche Limit as being a function just of the primary.] For instance, Saturn's rings lie inside its Roche limit, and may be the debris from a satellite that was ripped apart. The rings are made up of small particles, and each particle is held together by molecular bonds. Since they have something other than gravity holding them together, they are not ripped apart any further. Similarly, an artificial satellite is also held together by molecular bonds, not internal gravity.
The molecular-bonds-will-be-ripped-apart-by-tidal-forces limit is obviously much smaller than a satellite's orbit, as we, on the surface of the Earth, are even closer, and we are not ripped apart. You would have to have an extremely dense object, such as a neutron star or black hole, for that limit to exist. Being inside the Roche limit does mean that if an astronaut were to go on a space walk without a tether, tidal forces would pull them away from the larger satellite. Outside the Roche limit, the gravity of the larger satellite would pull the astronaut back (although not before the astronaut runs out of air).
If you look at the influence of the Moon's tides on Earth, you can see that the oceans are pulled towards the Moon, but the land is (relatively) stationary. The fact that tides are only a few meters shows that the Earth is well outside the Moon's Roche limit (and of course, the Earth's Roche limit is further out than the Moon's, so the Moon would reach the Earth's Roche limit long before the Earth reached the Moon's). If the Moon were to move towards the Earth, the tides would get higher and higher. The Moon's Roche limit is the point at which the tides would get so high that the water is ripped away from the Earth. The land would still survive slightly past that point, because the crust has some rigidity beyond mere gravitational attraction.
Regarding your second question: there is a region in which the tidal forces would be larger than internal gravitational attraction, and a region in which internal gravitational attraction would be larger than tidal forces. The Roche limit is the boundary between those two regions. Everything inside the Roche limit constitutes the former region, while everything outside the Roche limit constitutes the latter.
answered May 15 at 15:05
AcccumulationAcccumulation
3,694715
edited May 17 at 15:23
answered May 15 at 15:05
AcccumulationAcccumulation
3,694715
answered May 15 at 15:05
AcccumulationAcccumulation
3,694715
answered May 15 at 15:05
AcccumulationAcccumulation
3,694715
3,694715
14
$begingroup$
@Ed999 Yes. "Were" is the subjunctive form of "be", which indicates things contrary to fact.
$endgroup$
– Acccumulation
May 15 at 19:24
1
$begingroup$
And the innermost muns of Saturn, which orbit in gaps in the main rings, also lie inside the planet's Roche limit, and are (like artificial satellites) held together by their own tensile strength.
$endgroup$
– Sean
May 15 at 20:37
1
$begingroup$
Some of the questions posted here (not this one!) are asked by people whose knowledge of the most basic concepts of physics is only at a schoolboy level. It is therefore incumbent on us not to muddy the waters even further by posting scenarios which are (to us) obviously impossible, but which might be misunderstood by a casual reader as implying that, for example, the Moon could fall out of the sky (if you were to wait a million years). Where something is definitely impossible it is incumbent on us not to present it as being possible in the event of some circumstance that can't happen.
$endgroup$
– Ed999
May 15 at 23:25
8
$begingroup$
@Ed999 I don't think your nitpicking is in place here. How would you better propose an artificial scenario to depict the effect other than quite literally saying "if you somehow placed the moon close to Earth"? Nothing in the answer suggest it's going to happen. Writing answers in such defensive way against people misunderstanding hypotetical scenario for a fact in "If moon were moving towards Earth" is madness. IMHO it's also an insult to intelligence of general audience. You can't teach physics with out considering idealized or impossible scenarios, and some of best teachers did it
$endgroup$
– luk32
May 16 at 15:08
1
$begingroup$
This answer could be improved by pointing out explicitly that there's no definite "Earth's Roche limit". The limit depends on the density of the orbiting body -- concentrating the same mass in a smaller volume means that the self-gravitation on the surface is larger whereas the tidal acceleration on the surface is weaker, so a denser object withstands tides better and the Roche limit that applies to it is smaller.
$endgroup$
– Henning Makholm
May 17 at 15:14
|
show 2 more comments
14
$begingroup$
@Ed999 Yes. "Were" is the subjunctive form of "be", which indicates things contrary to fact.
$endgroup$
– Acccumulation
May 15 at 19:24
1
$begingroup$
And the innermost muns of Saturn, which orbit in gaps in the main rings, also lie inside the planet's Roche limit, and are (like artificial satellites) held together by their own tensile strength.
$endgroup$
– Sean
May 15 at 20:37
1
$begingroup$
Some of the questions posted here (not this one!) are asked by people whose knowledge of the most basic concepts of physics is only at a schoolboy level. It is therefore incumbent on us not to muddy the waters even further by posting scenarios which are (to us) obviously impossible, but which might be misunderstood by a casual reader as implying that, for example, the Moon could fall out of the sky (if you were to wait a million years). Where something is definitely impossible it is incumbent on us not to present it as being possible in the event of some circumstance that can't happen.
$endgroup$
– Ed999
May 15 at 23:25
8
$begingroup$
@Ed999 I don't think your nitpicking is in place here. How would you better propose an artificial scenario to depict the effect other than quite literally saying "if you somehow placed the moon close to Earth"? Nothing in the answer suggest it's going to happen. Writing answers in such defensive way against people misunderstanding hypotetical scenario for a fact in "If moon were moving towards Earth" is madness. IMHO it's also an insult to intelligence of general audience. You can't teach physics with out considering idealized or impossible scenarios, and some of best teachers did it
$endgroup$
– luk32
May 16 at 15:08
1
$begingroup$
This answer could be improved by pointing out explicitly that there's no definite "Earth's Roche limit". The limit depends on the density of the orbiting body -- concentrating the same mass in a smaller volume means that the self-gravitation on the surface is larger whereas the tidal acceleration on the surface is weaker, so a denser object withstands tides better and the Roche limit that applies to it is smaller.
$endgroup$
– Henning Makholm
May 17 at 15:14
14
14
$begingroup$
@Ed999 Yes. "Were" is the subjunctive form of "be", which indicates things contrary to fact.
$endgroup$
– Acccumulation
May 15 at 19:24
$begingroup$
@Ed999 Yes. "Were" is the subjunctive form of "be", which indicates things contrary to fact.
$endgroup$
– Acccumulation
May 15 at 19:24
1
1
$begingroup$
And the innermost muns of Saturn, which orbit in gaps in the main rings, also lie inside the planet's Roche limit, and are (like artificial satellites) held together by their own tensile strength.
$endgroup$
– Sean
May 15 at 20:37
$begingroup$
And the innermost muns of Saturn, which orbit in gaps in the main rings, also lie inside the planet's Roche limit, and are (like artificial satellites) held together by their own tensile strength.
$endgroup$
– Sean
May 15 at 20:37
1
1
$begingroup$
Some of the questions posted here (not this one!) are asked by people whose knowledge of the most basic concepts of physics is only at a schoolboy level. It is therefore incumbent on us not to muddy the waters even further by posting scenarios which are (to us) obviously impossible, but which might be misunderstood by a casual reader as implying that, for example, the Moon could fall out of the sky (if you were to wait a million years). Where something is definitely impossible it is incumbent on us not to present it as being possible in the event of some circumstance that can't happen.
$endgroup$
– Ed999
May 15 at 23:25
$begingroup$
Some of the questions posted here (not this one!) are asked by people whose knowledge of the most basic concepts of physics is only at a schoolboy level. It is therefore incumbent on us not to muddy the waters even further by posting scenarios which are (to us) obviously impossible, but which might be misunderstood by a casual reader as implying that, for example, the Moon could fall out of the sky (if you were to wait a million years). Where something is definitely impossible it is incumbent on us not to present it as being possible in the event of some circumstance that can't happen.
$endgroup$
– Ed999
May 15 at 23:25
8
8
$begingroup$
@Ed999 I don't think your nitpicking is in place here. How would you better propose an artificial scenario to depict the effect other than quite literally saying "if you somehow placed the moon close to Earth"? Nothing in the answer suggest it's going to happen. Writing answers in such defensive way against people misunderstanding hypotetical scenario for a fact in "If moon were moving towards Earth" is madness. IMHO it's also an insult to intelligence of general audience. You can't teach physics with out considering idealized or impossible scenarios, and some of best teachers did it
$endgroup$
– luk32
May 16 at 15:08
$begingroup$
@Ed999 I don't think your nitpicking is in place here. How would you better propose an artificial scenario to depict the effect other than quite literally saying "if you somehow placed the moon close to Earth"? Nothing in the answer suggest it's going to happen. Writing answers in such defensive way against people misunderstanding hypotetical scenario for a fact in "If moon were moving towards Earth" is madness. IMHO it's also an insult to intelligence of general audience. You can't teach physics with out considering idealized or impossible scenarios, and some of best teachers did it
$endgroup$
– luk32
May 16 at 15:08
1
1
$begingroup$
This answer could be improved by pointing out explicitly that there's no definite "Earth's Roche limit". The limit depends on the density of the orbiting body -- concentrating the same mass in a smaller volume means that the self-gravitation on the surface is larger whereas the tidal acceleration on the surface is weaker, so a denser object withstands tides better and the Roche limit that applies to it is smaller.
$endgroup$
– Henning Makholm
May 17 at 15:14
$begingroup$
This answer could be improved by pointing out explicitly that there's no definite "Earth's Roche limit". The limit depends on the density of the orbiting body -- concentrating the same mass in a smaller volume means that the self-gravitation on the surface is larger whereas the tidal acceleration on the surface is weaker, so a denser object withstands tides better and the Roche limit that applies to it is smaller.
$endgroup$
– Henning Makholm
May 17 at 15:14
|
show 2 more comments
$begingroup$
The Roche limit applies only to bodies which are held together purely by internal gravitational attraction. Compact objects such as artificial satellites are held together by the much stronger inter-molecular electromagnetic forces (this is another demonstration of just how weak gravity is compared to electromagnetism).
As for your second question: the Roche limit usually defined as the radius away from a body at which magnitude of the tidal forces exactly equal that of the internal gravitational attraction of the smaller body. Of course, the magnitude of the tidal forces becomes significant at further radii, and so the distance at which tidal forces become significant is a much broader area/range.
answered May 15 at 10:19
Bob KnightonBob Knighton
5,97931233
$endgroup$
10
$begingroup$
@DarrelHoffman: No, there isn't. A satellite being pulled apart by tidal forces either breaks up or it doesn't. The tidal forces don't get any stronger over time, unless of course the satellite itself is slowly spiraling towards the planet (like Phobos is, for example). Of course, if a solid satellite is orbiting within the Roche limit, an impact or some other source of momentary stress might break off a chunk that then goes flying away, because the satellite's own gravity isn't enough to hold it down. But while impacts do tend to occur gradually over time, they're not caused by time alone.
$endgroup$
– Ilmari Karonen
May 15 at 13:49
2
$begingroup$
When a man-made satellite in low earth orbit runs out of fuel for its reaction jets, it can no longer make orbital corrections to counteract the (very thin) atmospheric friction, so over a long period will be decelerated by it, and pulled down to Earth, as occurred with Skylab in the 1970s. But one in a high orbit (a prograde orbit about a rotating mass) is accelerated by the Earth's gravity (i.e. by the rotation-dragging effect), so will gradually gain increased orbital altitude (so does not "de-orbit"). This is the effect which causes the Moon to increase its distance from the Earth.
$endgroup$
– Ed999
May 15 at 18:17
1
$begingroup$
@DarrelHoffman The satellites will mostely de-orbit. Some might break apart for one reason or another. None of the satellites in orbit is going to break apart due to the tidal forces from the earth pulling the satellite to pieces - because the tidal forces are not sufficient to overcome the friction forces that are keeping the satellite together.
$endgroup$
– Taemyr
May 16 at 10:26
1
$begingroup$
@J... Bolts, tape and the like all still work through electromagnetic interaction.
$endgroup$
– Bob Knighton
May 16 at 17:37
1
$begingroup$
@BobKnighton Naturally, but as a macroscopic system it is geometrically and physically distinct from a contiguous piece of material held together by purely microscopic interactions.
$endgroup$
– J...
May 16 at 18:30
|
show 3 more comments
$begingroup$
The Roche limit applies only to bodies which are held together purely by internal gravitational attraction. Compact objects such as artificial satellites are held together by the much stronger inter-molecular electromagnetic forces (this is another demonstration of just how weak gravity is compared to electromagnetism).
As for your second question: the Roche limit usually defined as the radius away from a body at which magnitude of the tidal forces exactly equal that of the internal gravitational attraction of the smaller body. Of course, the magnitude of the tidal forces becomes significant at further radii, and so the distance at which tidal forces become significant is a much broader area/range.
answered May 15 at 10:19
Bob KnightonBob Knighton
5,97931233
$endgroup$
10
$begingroup$
@DarrelHoffman: No, there isn't. A satellite being pulled apart by tidal forces either breaks up or it doesn't. The tidal forces don't get any stronger over time, unless of course the satellite itself is slowly spiraling towards the planet (like Phobos is, for example). Of course, if a solid satellite is orbiting within the Roche limit, an impact or some other source of momentary stress might break off a chunk that then goes flying away, because the satellite's own gravity isn't enough to hold it down. But while impacts do tend to occur gradually over time, they're not caused by time alone.
$endgroup$
– Ilmari Karonen
May 15 at 13:49
2
$begingroup$
When a man-made satellite in low earth orbit runs out of fuel for its reaction jets, it can no longer make orbital corrections to counteract the (very thin) atmospheric friction, so over a long period will be decelerated by it, and pulled down to Earth, as occurred with Skylab in the 1970s. But one in a high orbit (a prograde orbit about a rotating mass) is accelerated by the Earth's gravity (i.e. by the rotation-dragging effect), so will gradually gain increased orbital altitude (so does not "de-orbit"). This is the effect which causes the Moon to increase its distance from the Earth.
$endgroup$
– Ed999
May 15 at 18:17
1
$begingroup$
@DarrelHoffman The satellites will mostely de-orbit. Some might break apart for one reason or another. None of the satellites in orbit is going to break apart due to the tidal forces from the earth pulling the satellite to pieces - because the tidal forces are not sufficient to overcome the friction forces that are keeping the satellite together.
$endgroup$
– Taemyr
May 16 at 10:26
1
$begingroup$
@J... Bolts, tape and the like all still work through electromagnetic interaction.
$endgroup$
– Bob Knighton
May 16 at 17:37
1
$begingroup$
@BobKnighton Naturally, but as a macroscopic system it is geometrically and physically distinct from a contiguous piece of material held together by purely microscopic interactions.
$endgroup$
– J...
May 16 at 18:30
|
show 3 more comments
$begingroup$
The Roche limit applies only to bodies which are held together purely by internal gravitational attraction. Compact objects such as artificial satellites are held together by the much stronger inter-molecular electromagnetic forces (this is another demonstration of just how weak gravity is compared to electromagnetism).
As for your second question: the Roche limit usually defined as the radius away from a body at which magnitude of the tidal forces exactly equal that of the internal gravitational attraction of the smaller body. Of course, the magnitude of the tidal forces becomes significant at further radii, and so the distance at which tidal forces become significant is a much broader area/range.
answered May 15 at 10:19
Bob KnightonBob Knighton
5,97931233
$endgroup$
The Roche limit applies only to bodies which are held together purely by internal gravitational attraction. Compact objects such as artificial satellites are held together by the much stronger inter-molecular electromagnetic forces (this is another demonstration of just how weak gravity is compared to electromagnetism).
As for your second question: the Roche limit usually defined as the radius away from a body at which magnitude of the tidal forces exactly equal that of the internal gravitational attraction of the smaller body. Of course, the magnitude of the tidal forces becomes significant at further radii, and so the distance at which tidal forces become significant is a much broader area/range.
answered May 15 at 10:19
Bob KnightonBob Knighton
5,97931233
answered May 15 at 10:19
Bob KnightonBob Knighton
5,97931233
answered May 15 at 10:19
Bob KnightonBob Knighton
5,97931233
answered May 15 at 10:19
Bob KnightonBob Knighton
5,97931233
5,97931233
10
$begingroup$
@DarrelHoffman: No, there isn't. A satellite being pulled apart by tidal forces either breaks up or it doesn't. The tidal forces don't get any stronger over time, unless of course the satellite itself is slowly spiraling towards the planet (like Phobos is, for example). Of course, if a solid satellite is orbiting within the Roche limit, an impact or some other source of momentary stress might break off a chunk that then goes flying away, because the satellite's own gravity isn't enough to hold it down. But while impacts do tend to occur gradually over time, they're not caused by time alone.
$endgroup$
– Ilmari Karonen
May 15 at 13:49
2
$begingroup$
When a man-made satellite in low earth orbit runs out of fuel for its reaction jets, it can no longer make orbital corrections to counteract the (very thin) atmospheric friction, so over a long period will be decelerated by it, and pulled down to Earth, as occurred with Skylab in the 1970s. But one in a high orbit (a prograde orbit about a rotating mass) is accelerated by the Earth's gravity (i.e. by the rotation-dragging effect), so will gradually gain increased orbital altitude (so does not "de-orbit"). This is the effect which causes the Moon to increase its distance from the Earth.
$endgroup$
– Ed999
May 15 at 18:17
1
$begingroup$
@DarrelHoffman The satellites will mostely de-orbit. Some might break apart for one reason or another. None of the satellites in orbit is going to break apart due to the tidal forces from the earth pulling the satellite to pieces - because the tidal forces are not sufficient to overcome the friction forces that are keeping the satellite together.
$endgroup$
– Taemyr
May 16 at 10:26
1
$begingroup$
@J... Bolts, tape and the like all still work through electromagnetic interaction.
$endgroup$
– Bob Knighton
May 16 at 17:37
1
$begingroup$
@BobKnighton Naturally, but as a macroscopic system it is geometrically and physically distinct from a contiguous piece of material held together by purely microscopic interactions.
$endgroup$
– J...
May 16 at 18:30
|
show 3 more comments
10
$begingroup$
@DarrelHoffman: No, there isn't. A satellite being pulled apart by tidal forces either breaks up or it doesn't. The tidal forces don't get any stronger over time, unless of course the satellite itself is slowly spiraling towards the planet (like Phobos is, for example). Of course, if a solid satellite is orbiting within the Roche limit, an impact or some other source of momentary stress might break off a chunk that then goes flying away, because the satellite's own gravity isn't enough to hold it down. But while impacts do tend to occur gradually over time, they're not caused by time alone.
$endgroup$
– Ilmari Karonen
May 15 at 13:49
2
$begingroup$
When a man-made satellite in low earth orbit runs out of fuel for its reaction jets, it can no longer make orbital corrections to counteract the (very thin) atmospheric friction, so over a long period will be decelerated by it, and pulled down to Earth, as occurred with Skylab in the 1970s. But one in a high orbit (a prograde orbit about a rotating mass) is accelerated by the Earth's gravity (i.e. by the rotation-dragging effect), so will gradually gain increased orbital altitude (so does not "de-orbit"). This is the effect which causes the Moon to increase its distance from the Earth.
$endgroup$
– Ed999
May 15 at 18:17
1
$begingroup$
@DarrelHoffman The satellites will mostely de-orbit. Some might break apart for one reason or another. None of the satellites in orbit is going to break apart due to the tidal forces from the earth pulling the satellite to pieces - because the tidal forces are not sufficient to overcome the friction forces that are keeping the satellite together.
$endgroup$
– Taemyr
May 16 at 10:26
1
$begingroup$
@J... Bolts, tape and the like all still work through electromagnetic interaction.
$endgroup$
– Bob Knighton
May 16 at 17:37
1
$begingroup$
@BobKnighton Naturally, but as a macroscopic system it is geometrically and physically distinct from a contiguous piece of material held together by purely microscopic interactions.
$endgroup$
– J...
May 16 at 18:30
10
10
$begingroup$
@DarrelHoffman: No, there isn't. A satellite being pulled apart by tidal forces either breaks up or it doesn't. The tidal forces don't get any stronger over time, unless of course the satellite itself is slowly spiraling towards the planet (like Phobos is, for example). Of course, if a solid satellite is orbiting within the Roche limit, an impact or some other source of momentary stress might break off a chunk that then goes flying away, because the satellite's own gravity isn't enough to hold it down. But while impacts do tend to occur gradually over time, they're not caused by time alone.
$endgroup$
– Ilmari Karonen
May 15 at 13:49
$begingroup$
@DarrelHoffman: No, there isn't. A satellite being pulled apart by tidal forces either breaks up or it doesn't. The tidal forces don't get any stronger over time, unless of course the satellite itself is slowly spiraling towards the planet (like Phobos is, for example). Of course, if a solid satellite is orbiting within the Roche limit, an impact or some other source of momentary stress might break off a chunk that then goes flying away, because the satellite's own gravity isn't enough to hold it down. But while impacts do tend to occur gradually over time, they're not caused by time alone.
$endgroup$
– Ilmari Karonen
May 15 at 13:49
2
2
$begingroup$
When a man-made satellite in low earth orbit runs out of fuel for its reaction jets, it can no longer make orbital corrections to counteract the (very thin) atmospheric friction, so over a long period will be decelerated by it, and pulled down to Earth, as occurred with Skylab in the 1970s. But one in a high orbit (a prograde orbit about a rotating mass) is accelerated by the Earth's gravity (i.e. by the rotation-dragging effect), so will gradually gain increased orbital altitude (so does not "de-orbit"). This is the effect which causes the Moon to increase its distance from the Earth.
$endgroup$
– Ed999
May 15 at 18:17
$begingroup$
When a man-made satellite in low earth orbit runs out of fuel for its reaction jets, it can no longer make orbital corrections to counteract the (very thin) atmospheric friction, so over a long period will be decelerated by it, and pulled down to Earth, as occurred with Skylab in the 1970s. But one in a high orbit (a prograde orbit about a rotating mass) is accelerated by the Earth's gravity (i.e. by the rotation-dragging effect), so will gradually gain increased orbital altitude (so does not "de-orbit"). This is the effect which causes the Moon to increase its distance from the Earth.
$endgroup$
– Ed999
May 15 at 18:17
1
1
$begingroup$
@DarrelHoffman The satellites will mostely de-orbit. Some might break apart for one reason or another. None of the satellites in orbit is going to break apart due to the tidal forces from the earth pulling the satellite to pieces - because the tidal forces are not sufficient to overcome the friction forces that are keeping the satellite together.
$endgroup$
– Taemyr
May 16 at 10:26
$begingroup$
@DarrelHoffman The satellites will mostely de-orbit. Some might break apart for one reason or another. None of the satellites in orbit is going to break apart due to the tidal forces from the earth pulling the satellite to pieces - because the tidal forces are not sufficient to overcome the friction forces that are keeping the satellite together.
$endgroup$
– Taemyr
May 16 at 10:26
1
1
$begingroup$
@J... Bolts, tape and the like all still work through electromagnetic interaction.
$endgroup$
– Bob Knighton
May 16 at 17:37
$begingroup$
@J... Bolts, tape and the like all still work through electromagnetic interaction.
$endgroup$
– Bob Knighton
May 16 at 17:37
1
1
$begingroup$
@BobKnighton Naturally, but as a macroscopic system it is geometrically and physically distinct from a contiguous piece of material held together by purely microscopic interactions.
$endgroup$
– J...
May 16 at 18:30
$begingroup$
@BobKnighton Naturally, but as a macroscopic system it is geometrically and physically distinct from a contiguous piece of material held together by purely microscopic interactions.
$endgroup$
– J...
May 16 at 18:30
|
show 3 more comments
$begingroup$
The Roche limit is a limit on objects being held together by their own gravity. Satellites are held together by much stronger forces. Different parts of the satellite are ultimately connected by chemical bonds, which are electromagnetic.
answered May 15 at 10:18
knzhouknzhou
49.9k12137245
$endgroup$
add a comment |
$begingroup$
The Roche limit is a limit on objects being held together by their own gravity. Satellites are held together by much stronger forces. Different parts of the satellite are ultimately connected by chemical bonds, which are electromagnetic.
answered May 15 at 10:18
knzhouknzhou
49.9k12137245
$endgroup$
add a comment |
$begingroup$
The Roche limit is a limit on objects being held together by their own gravity. Satellites are held together by much stronger forces. Different parts of the satellite are ultimately connected by chemical bonds, which are electromagnetic.
answered May 15 at 10:18
knzhouknzhou
49.9k12137245
$endgroup$
The Roche limit is a limit on objects being held together by their own gravity. Satellites are held together by much stronger forces. Different parts of the satellite are ultimately connected by chemical bonds, which are electromagnetic.
answered May 15 at 10:18
knzhouknzhou
49.9k12137245
answered May 15 at 10:18
knzhouknzhou
49.9k12137245
answered May 15 at 10:18
knzhouknzhou
49.9k12137245
answered May 15 at 10:18
knzhouknzhou
49.9k12137245
49.9k12137245
add a comment |
add a comment |
$begingroup$
To add to other answers, also consider that artificial satellites are much smaller than natural satellites. This means that the difference between the gravitational force at the point the closest to the planet and at the point the furthest from the planet is much smaller in artificial satellites.
answered May 15 at 20:25
stackzebrastackzebra
1312
$endgroup$
$begingroup$
... and therefore tidal factors are negligible is what you're saying, right? Just making sure I understand.
$endgroup$
– user45266
May 16 at 3:18
$begingroup$
@user45266 Yes.
$endgroup$
– stackzebra
May 16 at 11:03
2
$begingroup$
The size of the bodies does go into the Roche formula--the Roche "limit" isn't a precise number, when it is presented as one number that's what it would be for an ordinary moon. (Note that Titan would start coming apart earlier.)
$endgroup$
– Loren Pechtel
May 17 at 13:40
add a comment |
$begingroup$
To add to other answers, also consider that artificial satellites are much smaller than natural satellites. This means that the difference between the gravitational force at the point the closest to the planet and at the point the furthest from the planet is much smaller in artificial satellites.
answered May 15 at 20:25
stackzebrastackzebra
1312
$endgroup$
$begingroup$
... and therefore tidal factors are negligible is what you're saying, right? Just making sure I understand.
$endgroup$
– user45266
May 16 at 3:18
$begingroup$
@user45266 Yes.
$endgroup$
– stackzebra
May 16 at 11:03
2
$begingroup$
The size of the bodies does go into the Roche formula--the Roche "limit" isn't a precise number, when it is presented as one number that's what it would be for an ordinary moon. (Note that Titan would start coming apart earlier.)
$endgroup$
– Loren Pechtel
May 17 at 13:40
add a comment |
$begingroup$
To add to other answers, also consider that artificial satellites are much smaller than natural satellites. This means that the difference between the gravitational force at the point the closest to the planet and at the point the furthest from the planet is much smaller in artificial satellites.
answered May 15 at 20:25
stackzebrastackzebra
1312
$endgroup$
To add to other answers, also consider that artificial satellites are much smaller than natural satellites. This means that the difference between the gravitational force at the point the closest to the planet and at the point the furthest from the planet is much smaller in artificial satellites.
answered May 15 at 20:25
stackzebrastackzebra
1312
answered May 15 at 20:25
stackzebrastackzebra
1312
answered May 15 at 20:25
stackzebrastackzebra
1312
answered May 15 at 20:25
stackzebrastackzebra
1312
1312
$begingroup$
... and therefore tidal factors are negligible is what you're saying, right? Just making sure I understand.
$endgroup$
– user45266
May 16 at 3:18
$begingroup$
@user45266 Yes.
$endgroup$
– stackzebra
May 16 at 11:03
2
$begingroup$
The size of the bodies does go into the Roche formula--the Roche "limit" isn't a precise number, when it is presented as one number that's what it would be for an ordinary moon. (Note that Titan would start coming apart earlier.)
$endgroup$
– Loren Pechtel
May 17 at 13:40
add a comment |
$begingroup$
... and therefore tidal factors are negligible is what you're saying, right? Just making sure I understand.
$endgroup$
– user45266
May 16 at 3:18
$begingroup$
@user45266 Yes.
$endgroup$
– stackzebra
May 16 at 11:03
2
$begingroup$
The size of the bodies does go into the Roche formula--the Roche "limit" isn't a precise number, when it is presented as one number that's what it would be for an ordinary moon. (Note that Titan would start coming apart earlier.)
$endgroup$
– Loren Pechtel
May 17 at 13:40
$begingroup$
... and therefore tidal factors are negligible is what you're saying, right? Just making sure I understand.
$endgroup$
– user45266
May 16 at 3:18
$begingroup$
... and therefore tidal factors are negligible is what you're saying, right? Just making sure I understand.
$endgroup$
– user45266
May 16 at 3:18
$begingroup$
@user45266 Yes.
$endgroup$
– stackzebra
May 16 at 11:03
$begingroup$
@user45266 Yes.
$endgroup$
– stackzebra
May 16 at 11:03
2
2
$begingroup$
The size of the bodies does go into the Roche formula--the Roche "limit" isn't a precise number, when it is presented as one number that's what it would be for an ordinary moon. (Note that Titan would start coming apart earlier.)
$endgroup$
– Loren Pechtel
May 17 at 13:40
$begingroup$
The size of the bodies does go into the Roche formula--the Roche "limit" isn't a precise number, when it is presented as one number that's what it would be for an ordinary moon. (Note that Titan would start coming apart earlier.)
$endgroup$
– Loren Pechtel
May 17 at 13:40
add a comment |
$begingroup$
When I was a kid I also wondered why artificial satellites within the Roche Limit were not pulled apart by tidal forces.
When I was a kid I also wondered, if any body within the Roche Limit would be pulled apart by tidal forces, and since the surface of the Earth is deep within the Roche limit, why aren't all objects on the surface of the Earth - including Human bodies - pulled apart by tidal forces.
Since my body was not being pulled apart by tidal forces the statement that all bodies within the Roche Limit were pulled apart by tidal forces must not be correct. Therefore the simple statement that all bodies within the Roche Limit are pulled apart by tidal forces must be an oversimplification as stated.
But since such statements were made in non fiction sources it seemed probable that they were not totally false. Therefore I expected that sometime in the future I would read a fuller and more complex account of the Roche limit that would explain the seeming paradoxes.
And I did. Eventually I learned that the Roche limit was not a single absolute distance but varied with the sizes, masses, and densities of the larger and the smaller objects. I also learned that the Roche Limit only applied to objects that were held together only by their internal gravitational attraction and not to objects like artificial satellites or Human bodies.
Wondering why the Roche Limit didn't apply to my Human body was an example of using reductio ad absurdum to show that a statement was an oversimplification of a more complex situation.
answered May 17 at 17:26
M.A. GoldingM.A. Golding
111
$endgroup$
add a comment |
$begingroup$
When I was a kid I also wondered why artificial satellites within the Roche Limit were not pulled apart by tidal forces.
When I was a kid I also wondered, if any body within the Roche Limit would be pulled apart by tidal forces, and since the surface of the Earth is deep within the Roche limit, why aren't all objects on the surface of the Earth - including Human bodies - pulled apart by tidal forces.
Since my body was not being pulled apart by tidal forces the statement that all bodies within the Roche Limit were pulled apart by tidal forces must not be correct. Therefore the simple statement that all bodies within the Roche Limit are pulled apart by tidal forces must be an oversimplification as stated.
But since such statements were made in non fiction sources it seemed probable that they were not totally false. Therefore I expected that sometime in the future I would read a fuller and more complex account of the Roche limit that would explain the seeming paradoxes.
And I did. Eventually I learned that the Roche limit was not a single absolute distance but varied with the sizes, masses, and densities of the larger and the smaller objects. I also learned that the Roche Limit only applied to objects that were held together only by their internal gravitational attraction and not to objects like artificial satellites or Human bodies.
Wondering why the Roche Limit didn't apply to my Human body was an example of using reductio ad absurdum to show that a statement was an oversimplification of a more complex situation.
answered May 17 at 17:26
M.A. GoldingM.A. Golding
111
$endgroup$
add a comment |
$begingroup$
When I was a kid I also wondered why artificial satellites within the Roche Limit were not pulled apart by tidal forces.
When I was a kid I also wondered, if any body within the Roche Limit would be pulled apart by tidal forces, and since the surface of the Earth is deep within the Roche limit, why aren't all objects on the surface of the Earth - including Human bodies - pulled apart by tidal forces.
Since my body was not being pulled apart by tidal forces the statement that all bodies within the Roche Limit were pulled apart by tidal forces must not be correct. Therefore the simple statement that all bodies within the Roche Limit are pulled apart by tidal forces must be an oversimplification as stated.
But since such statements were made in non fiction sources it seemed probable that they were not totally false. Therefore I expected that sometime in the future I would read a fuller and more complex account of the Roche limit that would explain the seeming paradoxes.
And I did. Eventually I learned that the Roche limit was not a single absolute distance but varied with the sizes, masses, and densities of the larger and the smaller objects. I also learned that the Roche Limit only applied to objects that were held together only by their internal gravitational attraction and not to objects like artificial satellites or Human bodies.
Wondering why the Roche Limit didn't apply to my Human body was an example of using reductio ad absurdum to show that a statement was an oversimplification of a more complex situation.
answered May 17 at 17:26
M.A. GoldingM.A. Golding
111
$endgroup$
When I was a kid I also wondered why artificial satellites within the Roche Limit were not pulled apart by tidal forces.
When I was a kid I also wondered, if any body within the Roche Limit would be pulled apart by tidal forces, and since the surface of the Earth is deep within the Roche limit, why aren't all objects on the surface of the Earth - including Human bodies - pulled apart by tidal forces.
Since my body was not being pulled apart by tidal forces the statement that all bodies within the Roche Limit were pulled apart by tidal forces must not be correct. Therefore the simple statement that all bodies within the Roche Limit are pulled apart by tidal forces must be an oversimplification as stated.
But since such statements were made in non fiction sources it seemed probable that they were not totally false. Therefore I expected that sometime in the future I would read a fuller and more complex account of the Roche limit that would explain the seeming paradoxes.
And I did. Eventually I learned that the Roche limit was not a single absolute distance but varied with the sizes, masses, and densities of the larger and the smaller objects. I also learned that the Roche Limit only applied to objects that were held together only by their internal gravitational attraction and not to objects like artificial satellites or Human bodies.
Wondering why the Roche Limit didn't apply to my Human body was an example of using reductio ad absurdum to show that a statement was an oversimplification of a more complex situation.
answered May 17 at 17:26
M.A. GoldingM.A. Golding
111
answered May 17 at 17:26
M.A. GoldingM.A. Golding
111
answered May 17 at 17:26
M.A. GoldingM.A. Golding
111
answered May 17 at 17:26
M.A. GoldingM.A. Golding
111
111
add a comment |
add a comment |
$begingroup$
Why aren't artificial satellites ripped apart by gravitational tidal forces of the earth?
In short, the pieces of a satellite are held together by chemical, molecular bonds, which are strong compared to the tidal forces that are weak for objects the size of artificial satellites and gravity sources the strength of the earth.
A tidal force is the difference in attraction between two distances from a gravitational source. For a small object such as a natural satellite, that is just a few meters, gravity simply won't vary much between the nearest and farthest points unless you're talking about a orbiting a neutron star or black hole extremely closely.
For instance the ISS is 100m along its longest axis. Say that axis is perpendicular to the earth's surface. If its nearest point to earth is at 400km from the surface, the farthest point is therefore 400.1km. But earth is 5371km radius, so the distance from the center (where the gravity "seems" to be coming from) is 5771km and 5771.1km.
Gravity falls off at 1/(distance squared), so the difference in gravity is (1/(5771^2)) / (1/(5771.1^2)). That's a difference of 3*10^-8, or .000003% difference.
Gravity at earth's surface is 1G, giving 1kg mass 9.8N (newton) force (what you feel when you hold up 1kg on earth's surface). At 400km altitude it'd be 9.8*5371/(5371+400))^2=8.4N . So if you had even 1,000,000kg at each end of a 100m rope at that altitude, 1,000,000 * 8.4N * .000003% = .25N, about what you feel were you to pick up 25g of coins right now. (About 5 nickels.)
Even a human hair is strong enough to hold 25g at the earth's surface, and further, satellites aren't that heavy, and are far stronger than a hair.
answered May 18 at 1:31
Swiss FrankSwiss Frank
1112
$endgroup$
add a comment |
$begingroup$
Why aren't artificial satellites ripped apart by gravitational tidal forces of the earth?
In short, the pieces of a satellite are held together by chemical, molecular bonds, which are strong compared to the tidal forces that are weak for objects the size of artificial satellites and gravity sources the strength of the earth.
A tidal force is the difference in attraction between two distances from a gravitational source. For a small object such as a natural satellite, that is just a few meters, gravity simply won't vary much between the nearest and farthest points unless you're talking about a orbiting a neutron star or black hole extremely closely.
For instance the ISS is 100m along its longest axis. Say that axis is perpendicular to the earth's surface. If its nearest point to earth is at 400km from the surface, the farthest point is therefore 400.1km. But earth is 5371km radius, so the distance from the center (where the gravity "seems" to be coming from) is 5771km and 5771.1km.
Gravity falls off at 1/(distance squared), so the difference in gravity is (1/(5771^2)) / (1/(5771.1^2)). That's a difference of 3*10^-8, or .000003% difference.
Gravity at earth's surface is 1G, giving 1kg mass 9.8N (newton) force (what you feel when you hold up 1kg on earth's surface). At 400km altitude it'd be 9.8*5371/(5371+400))^2=8.4N . So if you had even 1,000,000kg at each end of a 100m rope at that altitude, 1,000,000 * 8.4N * .000003% = .25N, about what you feel were you to pick up 25g of coins right now. (About 5 nickels.)
Even a human hair is strong enough to hold 25g at the earth's surface, and further, satellites aren't that heavy, and are far stronger than a hair.
answered May 18 at 1:31
Swiss FrankSwiss Frank
1112
$endgroup$
add a comment |
$begingroup$
Why aren't artificial satellites ripped apart by gravitational tidal forces of the earth?
In short, the pieces of a satellite are held together by chemical, molecular bonds, which are strong compared to the tidal forces that are weak for objects the size of artificial satellites and gravity sources the strength of the earth.
A tidal force is the difference in attraction between two distances from a gravitational source. For a small object such as a natural satellite, that is just a few meters, gravity simply won't vary much between the nearest and farthest points unless you're talking about a orbiting a neutron star or black hole extremely closely.
For instance the ISS is 100m along its longest axis. Say that axis is perpendicular to the earth's surface. If its nearest point to earth is at 400km from the surface, the farthest point is therefore 400.1km. But earth is 5371km radius, so the distance from the center (where the gravity "seems" to be coming from) is 5771km and 5771.1km.
Gravity falls off at 1/(distance squared), so the difference in gravity is (1/(5771^2)) / (1/(5771.1^2)). That's a difference of 3*10^-8, or .000003% difference.
Gravity at earth's surface is 1G, giving 1kg mass 9.8N (newton) force (what you feel when you hold up 1kg on earth's surface). At 400km altitude it'd be 9.8*5371/(5371+400))^2=8.4N . So if you had even 1,000,000kg at each end of a 100m rope at that altitude, 1,000,000 * 8.4N * .000003% = .25N, about what you feel were you to pick up 25g of coins right now. (About 5 nickels.)
Even a human hair is strong enough to hold 25g at the earth's surface, and further, satellites aren't that heavy, and are far stronger than a hair.
answered May 18 at 1:31
Swiss FrankSwiss Frank
1112
$endgroup$
Why aren't artificial satellites ripped apart by gravitational tidal forces of the earth?
In short, the pieces of a satellite are held together by chemical, molecular bonds, which are strong compared to the tidal forces that are weak for objects the size of artificial satellites and gravity sources the strength of the earth.
A tidal force is the difference in attraction between two distances from a gravitational source. For a small object such as a natural satellite, that is just a few meters, gravity simply won't vary much between the nearest and farthest points unless you're talking about a orbiting a neutron star or black hole extremely closely.
For instance the ISS is 100m along its longest axis. Say that axis is perpendicular to the earth's surface. If its nearest point to earth is at 400km from the surface, the farthest point is therefore 400.1km. But earth is 5371km radius, so the distance from the center (where the gravity "seems" to be coming from) is 5771km and 5771.1km.
Gravity falls off at 1/(distance squared), so the difference in gravity is (1/(5771^2)) / (1/(5771.1^2)). That's a difference of 3*10^-8, or .000003% difference.
Gravity at earth's surface is 1G, giving 1kg mass 9.8N (newton) force (what you feel when you hold up 1kg on earth's surface). At 400km altitude it'd be 9.8*5371/(5371+400))^2=8.4N . So if you had even 1,000,000kg at each end of a 100m rope at that altitude, 1,000,000 * 8.4N * .000003% = .25N, about what you feel were you to pick up 25g of coins right now. (About 5 nickels.)
Even a human hair is strong enough to hold 25g at the earth's surface, and further, satellites aren't that heavy, and are far stronger than a hair.
answered May 18 at 1:31
Swiss FrankSwiss Frank
1112
edited May 20 at 4:34
answered May 18 at 1:31
Swiss FrankSwiss Frank
1112
answered May 18 at 1:31
Swiss FrankSwiss Frank
1112
answered May 18 at 1:31
Swiss FrankSwiss Frank
1112
1112
add a comment |
add a comment |
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$begingroup$
Basically, it's because artificial satellites aren't piles of loose gravel.
$endgroup$
– Ilmari Karonen
May 15 at 13:51