If Earth is tilted, why is Polaris always above the same spot?

Why do money exchangers give different rates to different bills?

Reconstruct a matrix from its traces

I caught several of my students plagiarizing. Could it be my fault as a teacher?

To customize a predefined symbol with different colors

What is the unit of the area when geometry attributes are calculated in QGIS?

Is it cheaper to drop cargo than to land it?

Was there ever a Kickstart that took advantage of 68020+ instructions that would work on an A2000?

What is the minimal installation possible in order to run a .jar Java file?

How would adding a darkvision racial trait to Dragonborn affect balance?

Does this article imply that Turing-Computability is not the same as "effectively computable"?

Would glacier 'trees' be plausible?

In Avengers 1, why does Thanos need Loki?

Besides the up and down quark, what other quarks are present in daily matter around us?

Why wasn't the Night King naked in S08E03?

CRT Oscilloscope - part of the plot is missing

Theorem won't go to multiple lines and is causing text to run off the page

I may be a fraud

Is Cola "probably the best-known" Latin word in the world? If not, which might it be?

Upside-Down Pyramid Addition...REVERSED!

What are the spoon bit of a spoon and fork bit of a fork called?

What are the differences between credential stuffing and password spraying?

Is there a legal ground for stripping the UK of its UN Veto if Scotland and/or N.Ireland split from the UK?

Why is Arya visibly scared in the library in S8E3?

How to calculate density of unknown planet?

How long does one orbit around the Moon take in a typical parking orbit for a stage a lander is going to meet?Which satellites can hear emergency signals from Scott Kelley's watch?How does orbital eccentricity affect positions of Lagrange points $L_4$ and $L_5$?Through what process does MESSENGER undergo orbital decay?How can you predict an orbit, based on an initial position and an initial direction vector?Which of the planets could we detect today from just the movements of the sun?Could twin planets exist and share the same orbit?Calculate Low/High/Geo orbits for other planets? - math lite - gamedevStationary Inter-planetary Satellite?Orbits that allows observation of a polar regionAre images of exoplanets' surfaces technically possible?How does lightsail 2 plan to raise its apogee

9

$begingroup$

Having approached an unknown planet, a
spaceship went in a low circular orbit. Would
astronauts be able to determine the average
density of the planets using only a watch

orbit exoplanet

share|improve this question

asked Apr 23 at 3:11

Danish Danish

511

$endgroup$

• 
  
  
  

  
  6
  

  
  
  
  
  
  

  
  $begingroup$
  Why would the astronauts only have a watch?
  $endgroup$
  – Ingolifs
  Apr 23 at 3:20
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  It is a question i came across.....even i cant figure that out
  $endgroup$
  – Danish
  Apr 23 at 3:28
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  This is a perfectly clear question. *Welcome to Space! @Danish ;-)
  $endgroup$
  – uhoh
  Apr 23 at 4:14
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It raises a second question: how would you be able to calculate an orbital trajectory without knowing the density of the planet? Or are we assuming the spaceship crew lucked out? :)
  $endgroup$
  – Robin Whittleton
  Apr 23 at 10:33
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @RobinWhittleton Assuming the planet had a basically normal density (i.e. somewhere in the 1.5 (dirty-ice) to 6 (iron core) g/cc range) it wouldn't be hard to pick a trajectory that would lead to a safe orbit, circular if the planet turns out to be dense, elliptical with a very high apoapsis if it's light.
  $endgroup$
  – Russell Borogove
  Apr 23 at 16:01
  

  
  
  
  

add a comment | 

9

$begingroup$

Having approached an unknown planet, a
spaceship went in a low circular orbit. Would
astronauts be able to determine the average
density of the planets using only a watch

orbit exoplanet

share|improve this question

asked Apr 23 at 3:11

Danish Danish

511

$endgroup$

• 
  
  
  

  
  6
  

  
  
  
  
  
  

  
  $begingroup$
  Why would the astronauts only have a watch?
  $endgroup$
  – Ingolifs
  Apr 23 at 3:20
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  It is a question i came across.....even i cant figure that out
  $endgroup$
  – Danish
  Apr 23 at 3:28
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  This is a perfectly clear question. *Welcome to Space! @Danish ;-)
  $endgroup$
  – uhoh
  Apr 23 at 4:14
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It raises a second question: how would you be able to calculate an orbital trajectory without knowing the density of the planet? Or are we assuming the spaceship crew lucked out? :)
  $endgroup$
  – Robin Whittleton
  Apr 23 at 10:33
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @RobinWhittleton Assuming the planet had a basically normal density (i.e. somewhere in the 1.5 (dirty-ice) to 6 (iron core) g/cc range) it wouldn't be hard to pick a trajectory that would lead to a safe orbit, circular if the planet turns out to be dense, elliptical with a very high apoapsis if it's light.
  $endgroup$
  – Russell Borogove
  Apr 23 at 16:01
  

  
  
  
  

add a comment | 

$begingroup$

Having approached an unknown planet, a
spaceship went in a low circular orbit. Would
astronauts be able to determine the average
density of the planets using only a watch

orbit exoplanet

share|improve this question

asked Apr 23 at 3:11

Danish Danish

511

$endgroup$

share|improve this question

asked Apr 23 at 3:11

Danish Danish

511

share|improve this question

asked Apr 23 at 3:11

Danish Danish

511

asked Apr 23 at 3:11

Danish Danish

511

asked Apr 23 at 3:11

Danish Danish

511

1 Answer
1

active

oldest

votes

10

$begingroup$

Planets are usually rotating, and so you can't time two passes over a single geographical feature to time an orbit. But you can time successive sunrises to get a good approximation of the orbital period, or some other astronomical coordination between the planet's limb and the celestial sphere.

From this answer:

Starting with

$$T = 2pisqrtfraca^3mu = 2pisqrtfraca^3Gm$$

and

$$rho = fracmfrac43 pi R^3$$

$$frac1m = frac1frac43 rho pi R^3$$

where $rho$ is the density and $R$ is the body's radius. Then:

$$T = 2pisqrtfraca^3Gfrac43 rho pi R^3,$$

and if you set $a=R$, you get:

$$T = 2pisqrtfrac34Gpi rho,$$

and finally

$$T = sqrtfrac3 piGrho.$$

And finally, flipping that around, we get:

$$rho = frac3 pi G T^2.$$

The catch is the step where you set the radius of the circular orbit equal to the radius of the planet; $a=R$.

On an airless world you can get pretty close for a little while, but bodies are usually nonuniform (lumpy) and so an orbit that skims near the surface will quickly get perturbed and crash.

So to do the experiment safely you need to orbit a safe distance above the surface, and this makes $a=R$ a source of error.

One solution is to use a hand-held laser altimeter (if such a thing exists) but probably that won't be found in an astronaut's watch, not even a fancy Scott Kelly watch.

With a conventional watch, one way to estimate altitude would be to measure the time of more celestial events, like a sunrise plus a sunset, as long as you can work out the 3D geometry to convert that to an altitude. The higher the altitude, the longer the day and shorter the night is.

Once you have altitude $h$, you can solve for both the radii of the planet $R$ and of the orbit $a$.

share|improve this answer

edited Apr 23 at 4:17

answered Apr 23 at 4:06

uhohuhoh

41.7k19160525

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  If $a/R$ is fixed then a similar calculation gives$rho$ In terms of T. That ratio can be measured by measuring the angle the planet subtends in your field of view. Fur instance of it is 120 degrees (from 4 to 8 on the face of your watch, then the ratio is $sqrt3/2$
  $endgroup$
  – Steve Linton
  Apr 23 at 6:32
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  If we have only clock - we can measure durations of shadow passes and starshine passes during orbit. Make much enough orbits, take maximum t_shadow/t_light, and if we are sure our orbit is circular than we can estimate a/R ratio.
  $endgroup$
  – Heopps
  Apr 23 at 6:40
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  In other words, no. You cannot determine the density of the planet with just a watch.
  $endgroup$
  – GdD
  Apr 23 at 8:13
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @GdD Not to unlimited precision, no, but you can't do that no matter what instruments you have. You can get a pretty decent approximation (enough to distinguish Venus density 5.24 from Earth density 5.5) by entering the lowest stable orbit that you can, and then using the watch to time the interval between successive rises of a suitable star over the horizon.
  $endgroup$
  – Steve Linton
  Apr 23 at 8:45
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @GdD No, I think yes! I believe you can by measuring complementary intervals to get both. Just for a simple illustration, $t_1, t_2, t_3$ = sunrise, sunset, sunrise. Two equations, two unknowns. Two time intervals ($t_2-t_1$ and $t_3-t_1$) and two radii ($r$ and $R$). Of course it would be more complicated due to (at least) two inclinations in the problem, but I think the only instrument you need (besides your eye and a window) is a watch.
  $endgroup$
  – uhoh
  Apr 23 at 15:56
  
  

  
  
  
  

add a comment | 

Your Answer

StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "508"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);

else
createEditor();

);

function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);



);

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name

Email

Required, but never shown

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fspace.stackexchange.com%2fquestions%2f35722%2fhow-to-calculate-density-of-unknown-planet%23new-answer', 'question_page');

);

Post as a guest

Name

Email

Required, but never shown

10

$begingroup$

Planets are usually rotating, and so you can't time two passes over a single geographical feature to time an orbit. But you can time successive sunrises to get a good approximation of the orbital period, or some other astronomical coordination between the planet's limb and the celestial sphere.

From this answer:

Starting with

$$T = 2pisqrtfraca^3mu = 2pisqrtfraca^3Gm$$

and

$$rho = fracmfrac43 pi R^3$$

$$frac1m = frac1frac43 rho pi R^3$$

where $rho$ is the density and $R$ is the body's radius. Then:

$$T = 2pisqrtfraca^3Gfrac43 rho pi R^3,$$

and if you set $a=R$, you get:

$$T = 2pisqrtfrac34Gpi rho,$$

and finally

$$T = sqrtfrac3 piGrho.$$

And finally, flipping that around, we get:

$$rho = frac3 pi G T^2.$$

The catch is the step where you set the radius of the circular orbit equal to the radius of the planet; $a=R$.

On an airless world you can get pretty close for a little while, but bodies are usually nonuniform (lumpy) and so an orbit that skims near the surface will quickly get perturbed and crash.

So to do the experiment safely you need to orbit a safe distance above the surface, and this makes $a=R$ a source of error.

One solution is to use a hand-held laser altimeter (if such a thing exists) but probably that won't be found in an astronaut's watch, not even a fancy Scott Kelly watch.

With a conventional watch, one way to estimate altitude would be to measure the time of more celestial events, like a sunrise plus a sunset, as long as you can work out the 3D geometry to convert that to an altitude. The higher the altitude, the longer the day and shorter the night is.

Once you have altitude $h$, you can solve for both the radii of the planet $R$ and of the orbit $a$.

share|improve this answer

edited Apr 23 at 4:17

answered Apr 23 at 4:06

uhohuhoh

41.7k19160525

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  If $a/R$ is fixed then a similar calculation gives$rho$ In terms of T. That ratio can be measured by measuring the angle the planet subtends in your field of view. Fur instance of it is 120 degrees (from 4 to 8 on the face of your watch, then the ratio is $sqrt3/2$
  $endgroup$
  – Steve Linton
  Apr 23 at 6:32
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  If we have only clock - we can measure durations of shadow passes and starshine passes during orbit. Make much enough orbits, take maximum t_shadow/t_light, and if we are sure our orbit is circular than we can estimate a/R ratio.
  $endgroup$
  – Heopps
  Apr 23 at 6:40
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  In other words, no. You cannot determine the density of the planet with just a watch.
  $endgroup$
  – GdD
  Apr 23 at 8:13
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @GdD Not to unlimited precision, no, but you can't do that no matter what instruments you have. You can get a pretty decent approximation (enough to distinguish Venus density 5.24 from Earth density 5.5) by entering the lowest stable orbit that you can, and then using the watch to time the interval between successive rises of a suitable star over the horizon.
  $endgroup$
  – Steve Linton
  Apr 23 at 8:45
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @GdD No, I think yes! I believe you can by measuring complementary intervals to get both. Just for a simple illustration, $t_1, t_2, t_3$ = sunrise, sunset, sunrise. Two equations, two unknowns. Two time intervals ($t_2-t_1$ and $t_3-t_1$) and two radii ($r$ and $R$). Of course it would be more complicated due to (at least) two inclinations in the problem, but I think the only instrument you need (besides your eye and a window) is a watch.
  $endgroup$
  – uhoh
  Apr 23 at 15:56
  
  

  
  
  
  

add a comment | 

10

$begingroup$

Planets are usually rotating, and so you can't time two passes over a single geographical feature to time an orbit. But you can time successive sunrises to get a good approximation of the orbital period, or some other astronomical coordination between the planet's limb and the celestial sphere.

From this answer:

Starting with

$$T = 2pisqrtfraca^3mu = 2pisqrtfraca^3Gm$$

and

$$rho = fracmfrac43 pi R^3$$

$$frac1m = frac1frac43 rho pi R^3$$

where $rho$ is the density and $R$ is the body's radius. Then:

$$T = 2pisqrtfraca^3Gfrac43 rho pi R^3,$$

and if you set $a=R$, you get:

$$T = 2pisqrtfrac34Gpi rho,$$

and finally

$$T = sqrtfrac3 piGrho.$$

And finally, flipping that around, we get:

$$rho = frac3 pi G T^2.$$

The catch is the step where you set the radius of the circular orbit equal to the radius of the planet; $a=R$.

On an airless world you can get pretty close for a little while, but bodies are usually nonuniform (lumpy) and so an orbit that skims near the surface will quickly get perturbed and crash.

So to do the experiment safely you need to orbit a safe distance above the surface, and this makes $a=R$ a source of error.

One solution is to use a hand-held laser altimeter (if such a thing exists) but probably that won't be found in an astronaut's watch, not even a fancy Scott Kelly watch.

With a conventional watch, one way to estimate altitude would be to measure the time of more celestial events, like a sunrise plus a sunset, as long as you can work out the 3D geometry to convert that to an altitude. The higher the altitude, the longer the day and shorter the night is.

Once you have altitude $h$, you can solve for both the radii of the planet $R$ and of the orbit $a$.

share|improve this answer

edited Apr 23 at 4:17

answered Apr 23 at 4:06

uhohuhoh

41.7k19160525

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  If $a/R$ is fixed then a similar calculation gives$rho$ In terms of T. That ratio can be measured by measuring the angle the planet subtends in your field of view. Fur instance of it is 120 degrees (from 4 to 8 on the face of your watch, then the ratio is $sqrt3/2$
  $endgroup$
  – Steve Linton
  Apr 23 at 6:32
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  If we have only clock - we can measure durations of shadow passes and starshine passes during orbit. Make much enough orbits, take maximum t_shadow/t_light, and if we are sure our orbit is circular than we can estimate a/R ratio.
  $endgroup$
  – Heopps
  Apr 23 at 6:40
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  In other words, no. You cannot determine the density of the planet with just a watch.
  $endgroup$
  – GdD
  Apr 23 at 8:13
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @GdD Not to unlimited precision, no, but you can't do that no matter what instruments you have. You can get a pretty decent approximation (enough to distinguish Venus density 5.24 from Earth density 5.5) by entering the lowest stable orbit that you can, and then using the watch to time the interval between successive rises of a suitable star over the horizon.
  $endgroup$
  – Steve Linton
  Apr 23 at 8:45
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @GdD No, I think yes! I believe you can by measuring complementary intervals to get both. Just for a simple illustration, $t_1, t_2, t_3$ = sunrise, sunset, sunrise. Two equations, two unknowns. Two time intervals ($t_2-t_1$ and $t_3-t_1$) and two radii ($r$ and $R$). Of course it would be more complicated due to (at least) two inclinations in the problem, but I think the only instrument you need (besides your eye and a window) is a watch.
  $endgroup$
  – uhoh
  Apr 23 at 15:56
  
  

  
  
  
  

add a comment | 

$begingroup$

Planets are usually rotating, and so you can't time two passes over a single geographical feature to time an orbit. But you can time successive sunrises to get a good approximation of the orbital period, or some other astronomical coordination between the planet's limb and the celestial sphere.

From this answer:

Starting with

$$T = 2pisqrtfraca^3mu = 2pisqrtfraca^3Gm$$

and

$$rho = fracmfrac43 pi R^3$$

$$frac1m = frac1frac43 rho pi R^3$$

where $rho$ is the density and $R$ is the body's radius. Then:

$$T = 2pisqrtfraca^3Gfrac43 rho pi R^3,$$

and if you set $a=R$, you get:

$$T = 2pisqrtfrac34Gpi rho,$$

and finally

$$T = sqrtfrac3 piGrho.$$

And finally, flipping that around, we get:

$$rho = frac3 pi G T^2.$$

The catch is the step where you set the radius of the circular orbit equal to the radius of the planet; $a=R$.

On an airless world you can get pretty close for a little while, but bodies are usually nonuniform (lumpy) and so an orbit that skims near the surface will quickly get perturbed and crash.

So to do the experiment safely you need to orbit a safe distance above the surface, and this makes $a=R$ a source of error.

One solution is to use a hand-held laser altimeter (if such a thing exists) but probably that won't be found in an astronaut's watch, not even a fancy Scott Kelly watch.

With a conventional watch, one way to estimate altitude would be to measure the time of more celestial events, like a sunrise plus a sunset, as long as you can work out the 3D geometry to convert that to an altitude. The higher the altitude, the longer the day and shorter the night is.

Once you have altitude $h$, you can solve for both the radii of the planet $R$ and of the orbit $a$.

share|improve this answer

edited Apr 23 at 4:17

answered Apr 23 at 4:06

uhohuhoh

41.7k19160525

$endgroup$

share|improve this answer

edited Apr 23 at 4:17

answered Apr 23 at 4:06

uhohuhoh

41.7k19160525

answered Apr 23 at 4:06

uhohuhoh

41.7k19160525

answered Apr 23 at 4:06

uhohuhoh

41.7k19160525