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What happens when the drag force exceeds the weight of an object falling into earth?
How does potential energy work in the context of objects in space?Calculating wind force and drag force on a falling objectWhy aren't Roche limit and the difference in gravitational acceleration the same?How to calculate the distance an accelerating object will be pulled by another object in a given amount of time?Motion of dropped object relative to an accelerating observerWhy isn't the apparent weight of a body in a fluid equal to the buoyant force? Why does buoyancy reduce it instead?Why arrow stays tangent to its trajectory?Drag force equals weightHow would a massless object behave on Earth?How does orbiting work exactly?
$begingroup$
Let's say a meteor is coming towards earth. It's not accelerating, but it does have an initial velocity. This meteor is shaped so it has an insane amount of drag, enough to even exceed its weight (not mass) as it gets closer. What happens? Why?
At first I thought it would just stop, but that doesn't really make sense. The forces cancel but that doesn't mean the body slows down. Would it keep going in another direction?
newtonian-mechanics newtonian-gravity acceleration projectile drag
edited May 7 at 6:07
Qmechanic♦
109k122051271
asked May 7 at 2:15
Laura IglesiasLaura Iglesias
8614
$endgroup$
add a comment |
$begingroup$
Let's say a meteor is coming towards earth. It's not accelerating, but it does have an initial velocity. This meteor is shaped so it has an insane amount of drag, enough to even exceed its weight (not mass) as it gets closer. What happens? Why?
At first I thought it would just stop, but that doesn't really make sense. The forces cancel but that doesn't mean the body slows down. Would it keep going in another direction?
newtonian-mechanics newtonian-gravity acceleration projectile drag
edited May 7 at 6:07
Qmechanic♦
109k122051271
asked May 7 at 2:15
Laura IglesiasLaura Iglesias
8614
$endgroup$
add a comment |
$begingroup$
Let's say a meteor is coming towards earth. It's not accelerating, but it does have an initial velocity. This meteor is shaped so it has an insane amount of drag, enough to even exceed its weight (not mass) as it gets closer. What happens? Why?
At first I thought it would just stop, but that doesn't really make sense. The forces cancel but that doesn't mean the body slows down. Would it keep going in another direction?
newtonian-mechanics newtonian-gravity acceleration projectile drag
edited May 7 at 6:07
Qmechanic♦
109k122051271
asked May 7 at 2:15
Laura IglesiasLaura Iglesias
8614
$endgroup$
Let's say a meteor is coming towards earth. It's not accelerating, but it does have an initial velocity. This meteor is shaped so it has an insane amount of drag, enough to even exceed its weight (not mass) as it gets closer. What happens? Why?
At first I thought it would just stop, but that doesn't really make sense. The forces cancel but that doesn't mean the body slows down. Would it keep going in another direction?
newtonian-mechanics newtonian-gravity acceleration projectile drag
newtonian-mechanics newtonian-gravity acceleration projectile drag
edited May 7 at 6:07
Qmechanic♦
109k122051271
asked May 7 at 2:15
Laura IglesiasLaura Iglesias
8614
edited May 7 at 6:07
Qmechanic♦
109k122051271
asked May 7 at 2:15
Laura IglesiasLaura Iglesias
8614
edited May 7 at 6:07
Qmechanic♦
109k122051271
edited May 7 at 6:07
Qmechanic♦
109k122051271
edited May 7 at 6:07
Qmechanic♦
109k122051271
109k122051271
asked May 7 at 2:15
Laura IglesiasLaura Iglesias
8614
asked May 7 at 2:15
Laura IglesiasLaura Iglesias
8614
asked May 7 at 2:15
Laura IglesiasLaura Iglesias
8614
8614
add a comment |
add a comment |
5 Answers
5
active
oldest
votes
$begingroup$
The acceleration (or deceleration) of an object is $a$: $$a = f/m.$$
Acceleration (or deceleration) is the rate of change of velocity. So, to find the rate of change of velocity of an object you divide all the forces acting on it, by the object's mass.
The forces acting on your hypothetical object are the object's weight and the atmospheric drag. Someone being very picky might also say that there is a relatively very small atmospheric buoyancy at work too, but it can be ignored in most cases.
Let's say the object is falling straight down. Atmospheric drag is velocity-dependent, with low drag at low speeds and much higher drag at higher speeds. Drag will cause the object to decelerate, until drag equals the object's weight. At that point the object is said to be moving at terminal velocity, and it just keeps falling at terminal velocity.
The whole scenario gets more complicated when altitude-dependent atmospheric density is taken into account, but what's described above captures the essence of an answer to your question.
answered May 7 at 2:33
S. McGrewS. McGrew
10.4k21342
$endgroup$
23
$begingroup$
Someone being very picky... On StackExchange? Never!...
$endgroup$
– Oscar Bravo
May 7 at 6:29
18
$begingroup$
To give a different perspective, rather than a meteor one can consider a parachutist. When they jump out of a plane their speed increases until they reach terminal velocity. When they open a parachute, the drag increases and now our parachutist becomes an object from OPs question. Their speed will slow down until reaching the new terminal velocity and then more or less gently float to the ground at a constant speed. I ignore here a bunch of things but unlike meteors that never have occasion to slow down to terminal velocity, parachutist usually do that making them a better example IMO.
$endgroup$
– Ister
May 7 at 10:49
2
$begingroup$
Do not forget the gravitational force from the Sun and the Moon. (For an object with a mass of several tons it's several Newton, if I'm not mistaken.)
$endgroup$
– Peter A. Schneider
May 7 at 12:28
3
$begingroup$
@PeterA.Schneider Is it day or is it night?
$endgroup$
– Spitemaster
May 7 at 15:56
2
$begingroup$
I would argue that atmospheric buyoancy is accounted for in the object weight.
$endgroup$
– John Dvorak
May 7 at 18:34
|
show 6 more comments
$begingroup$
Drag is proportional to the object's velocity (or velocity squared, depending on regime), so as the object's speed decreases, the drag decreases as well. Therefore, the body won't start "falling upwards" because for that to happen, the velocity must reach zero, and at that point the drag also drops to zero.
In practice, the falling object slows down until the drag equals the weight, at which point it keeps moving at constant speed (so-called "terminal velocity"). In the hypothetical scenario where you have a constant force that's greater than the weight pulling on the object, then the object will indeed start falling upwards.
answered May 7 at 2:22
AllureAllure
2,7471028
$endgroup$
1
$begingroup$
I don't think drag is proportional to velocity. en.wikipedia.org/wiki/Drag_equation is proportional to square of velocity.
$endgroup$
– Taemyr
May 7 at 5:19
3
$begingroup$
@Taemyr indeed, I was simplifying things. Edited.
$endgroup$
– Allure
May 7 at 5:27
7
$begingroup$
@Taemyr That depends on the regime. en.wikipedia.org/wiki/Stokes%27_law
$endgroup$
– Vladimir F
May 7 at 6:40
2
$begingroup$
@Taemyr They are proportional, but not directly proportional. The area of a circle is proportional to its radius, but directly proportional to the square of the radius ($A=πr²$)
$endgroup$
– Chronocidal
May 8 at 13:23
add a comment |
$begingroup$
It would accelerate upward. This is exactly what happens when skydivers open their parachutes, for instance.
The underlying confusion behind this question is likely due to mixing up velocity and acceleration. You can accelerate in one direction without also having a velocity in that direction: you can accelerate up while moving down. Hitting the brakes on a car is not the same as putting it in reverse.
answered May 8 at 13:17
knzhouknzhou
49.5k12136242
$endgroup$
add a comment |
$begingroup$
You've probably already seen what happens yourself with extremely small particles, such as those that comprise smoke. Lots of meteors are this small.
At this scale, things still fall, of course, but it takes a long time.
If things get small enough, eventually Brownian motion takes over.
answered May 7 at 16:50
RogerRoger
1192
$endgroup$
add a comment |
$begingroup$
It will decelerate, yes. If the vector force exceed the vector force of the weight.
Note that both will increase as altitude decreases.
If the drag exceeds the weight and speed becomes 0, then the velocity will become negative. (Think parachute in a thermal or strong updraught). It may be a pumice-like meteor or quite small.
If it is lighter than water then the weight will become negative when it hits water, so there is that to consider.
answered May 8 at 4:07
mckenzmmckenzm
1112
$endgroup$
add a comment |
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The acceleration (or deceleration) of an object is $a$: $$a = f/m.$$
Acceleration (or deceleration) is the rate of change of velocity. So, to find the rate of change of velocity of an object you divide all the forces acting on it, by the object's mass.
The forces acting on your hypothetical object are the object's weight and the atmospheric drag. Someone being very picky might also say that there is a relatively very small atmospheric buoyancy at work too, but it can be ignored in most cases.
Let's say the object is falling straight down. Atmospheric drag is velocity-dependent, with low drag at low speeds and much higher drag at higher speeds. Drag will cause the object to decelerate, until drag equals the object's weight. At that point the object is said to be moving at terminal velocity, and it just keeps falling at terminal velocity.
The whole scenario gets more complicated when altitude-dependent atmospheric density is taken into account, but what's described above captures the essence of an answer to your question.
answered May 7 at 2:33
S. McGrewS. McGrew
10.4k21342
$endgroup$
23
$begingroup$
Someone being very picky... On StackExchange? Never!...
$endgroup$
– Oscar Bravo
May 7 at 6:29
18
$begingroup$
To give a different perspective, rather than a meteor one can consider a parachutist. When they jump out of a plane their speed increases until they reach terminal velocity. When they open a parachute, the drag increases and now our parachutist becomes an object from OPs question. Their speed will slow down until reaching the new terminal velocity and then more or less gently float to the ground at a constant speed. I ignore here a bunch of things but unlike meteors that never have occasion to slow down to terminal velocity, parachutist usually do that making them a better example IMO.
$endgroup$
– Ister
May 7 at 10:49
2
$begingroup$
Do not forget the gravitational force from the Sun and the Moon. (For an object with a mass of several tons it's several Newton, if I'm not mistaken.)
$endgroup$
– Peter A. Schneider
May 7 at 12:28
3
$begingroup$
@PeterA.Schneider Is it day or is it night?
$endgroup$
– Spitemaster
May 7 at 15:56
2
$begingroup$
I would argue that atmospheric buyoancy is accounted for in the object weight.
$endgroup$
– John Dvorak
May 7 at 18:34
|
show 6 more comments
$begingroup$
The acceleration (or deceleration) of an object is $a$: $$a = f/m.$$
Acceleration (or deceleration) is the rate of change of velocity. So, to find the rate of change of velocity of an object you divide all the forces acting on it, by the object's mass.
The forces acting on your hypothetical object are the object's weight and the atmospheric drag. Someone being very picky might also say that there is a relatively very small atmospheric buoyancy at work too, but it can be ignored in most cases.
Let's say the object is falling straight down. Atmospheric drag is velocity-dependent, with low drag at low speeds and much higher drag at higher speeds. Drag will cause the object to decelerate, until drag equals the object's weight. At that point the object is said to be moving at terminal velocity, and it just keeps falling at terminal velocity.
The whole scenario gets more complicated when altitude-dependent atmospheric density is taken into account, but what's described above captures the essence of an answer to your question.
answered May 7 at 2:33
S. McGrewS. McGrew
10.4k21342
$endgroup$
23
$begingroup$
Someone being very picky... On StackExchange? Never!...
$endgroup$
– Oscar Bravo
May 7 at 6:29
18
$begingroup$
To give a different perspective, rather than a meteor one can consider a parachutist. When they jump out of a plane their speed increases until they reach terminal velocity. When they open a parachute, the drag increases and now our parachutist becomes an object from OPs question. Their speed will slow down until reaching the new terminal velocity and then more or less gently float to the ground at a constant speed. I ignore here a bunch of things but unlike meteors that never have occasion to slow down to terminal velocity, parachutist usually do that making them a better example IMO.
$endgroup$
– Ister
May 7 at 10:49
2
$begingroup$
Do not forget the gravitational force from the Sun and the Moon. (For an object with a mass of several tons it's several Newton, if I'm not mistaken.)
$endgroup$
– Peter A. Schneider
May 7 at 12:28
3
$begingroup$
@PeterA.Schneider Is it day or is it night?
$endgroup$
– Spitemaster
May 7 at 15:56
2
$begingroup$
I would argue that atmospheric buyoancy is accounted for in the object weight.
$endgroup$
– John Dvorak
May 7 at 18:34
|
show 6 more comments
$begingroup$
The acceleration (or deceleration) of an object is $a$: $$a = f/m.$$
Acceleration (or deceleration) is the rate of change of velocity. So, to find the rate of change of velocity of an object you divide all the forces acting on it, by the object's mass.
The forces acting on your hypothetical object are the object's weight and the atmospheric drag. Someone being very picky might also say that there is a relatively very small atmospheric buoyancy at work too, but it can be ignored in most cases.
Let's say the object is falling straight down. Atmospheric drag is velocity-dependent, with low drag at low speeds and much higher drag at higher speeds. Drag will cause the object to decelerate, until drag equals the object's weight. At that point the object is said to be moving at terminal velocity, and it just keeps falling at terminal velocity.
The whole scenario gets more complicated when altitude-dependent atmospheric density is taken into account, but what's described above captures the essence of an answer to your question.
answered May 7 at 2:33
S. McGrewS. McGrew
10.4k21342
$endgroup$
The acceleration (or deceleration) of an object is $a$: $$a = f/m.$$
Acceleration (or deceleration) is the rate of change of velocity. So, to find the rate of change of velocity of an object you divide all the forces acting on it, by the object's mass.
The forces acting on your hypothetical object are the object's weight and the atmospheric drag. Someone being very picky might also say that there is a relatively very small atmospheric buoyancy at work too, but it can be ignored in most cases.
Let's say the object is falling straight down. Atmospheric drag is velocity-dependent, with low drag at low speeds and much higher drag at higher speeds. Drag will cause the object to decelerate, until drag equals the object's weight. At that point the object is said to be moving at terminal velocity, and it just keeps falling at terminal velocity.
The whole scenario gets more complicated when altitude-dependent atmospheric density is taken into account, but what's described above captures the essence of an answer to your question.
answered May 7 at 2:33
S. McGrewS. McGrew
10.4k21342
answered May 7 at 2:33
S. McGrewS. McGrew
10.4k21342
answered May 7 at 2:33
S. McGrewS. McGrew
10.4k21342
answered May 7 at 2:33
S. McGrewS. McGrew
10.4k21342
10.4k21342
23
$begingroup$
Someone being very picky... On StackExchange? Never!...
$endgroup$
– Oscar Bravo
May 7 at 6:29
18
$begingroup$
To give a different perspective, rather than a meteor one can consider a parachutist. When they jump out of a plane their speed increases until they reach terminal velocity. When they open a parachute, the drag increases and now our parachutist becomes an object from OPs question. Their speed will slow down until reaching the new terminal velocity and then more or less gently float to the ground at a constant speed. I ignore here a bunch of things but unlike meteors that never have occasion to slow down to terminal velocity, parachutist usually do that making them a better example IMO.
$endgroup$
– Ister
May 7 at 10:49
2
$begingroup$
Do not forget the gravitational force from the Sun and the Moon. (For an object with a mass of several tons it's several Newton, if I'm not mistaken.)
$endgroup$
– Peter A. Schneider
May 7 at 12:28
3
$begingroup$
@PeterA.Schneider Is it day or is it night?
$endgroup$
– Spitemaster
May 7 at 15:56
2
$begingroup$
I would argue that atmospheric buyoancy is accounted for in the object weight.
$endgroup$
– John Dvorak
May 7 at 18:34
|
show 6 more comments
23
$begingroup$
Someone being very picky... On StackExchange? Never!...
$endgroup$
– Oscar Bravo
May 7 at 6:29
18
$begingroup$
To give a different perspective, rather than a meteor one can consider a parachutist. When they jump out of a plane their speed increases until they reach terminal velocity. When they open a parachute, the drag increases and now our parachutist becomes an object from OPs question. Their speed will slow down until reaching the new terminal velocity and then more or less gently float to the ground at a constant speed. I ignore here a bunch of things but unlike meteors that never have occasion to slow down to terminal velocity, parachutist usually do that making them a better example IMO.
$endgroup$
– Ister
May 7 at 10:49
2
$begingroup$
Do not forget the gravitational force from the Sun and the Moon. (For an object with a mass of several tons it's several Newton, if I'm not mistaken.)
$endgroup$
– Peter A. Schneider
May 7 at 12:28
3
$begingroup$
@PeterA.Schneider Is it day or is it night?
$endgroup$
– Spitemaster
May 7 at 15:56
2
$begingroup$
I would argue that atmospheric buyoancy is accounted for in the object weight.
$endgroup$
– John Dvorak
May 7 at 18:34
23
23
$begingroup$
Someone being very picky... On StackExchange? Never!...
$endgroup$
– Oscar Bravo
May 7 at 6:29
$begingroup$
Someone being very picky... On StackExchange? Never!...
$endgroup$
– Oscar Bravo
May 7 at 6:29
18
18
$begingroup$
To give a different perspective, rather than a meteor one can consider a parachutist. When they jump out of a plane their speed increases until they reach terminal velocity. When they open a parachute, the drag increases and now our parachutist becomes an object from OPs question. Their speed will slow down until reaching the new terminal velocity and then more or less gently float to the ground at a constant speed. I ignore here a bunch of things but unlike meteors that never have occasion to slow down to terminal velocity, parachutist usually do that making them a better example IMO.
$endgroup$
– Ister
May 7 at 10:49
$begingroup$
To give a different perspective, rather than a meteor one can consider a parachutist. When they jump out of a plane their speed increases until they reach terminal velocity. When they open a parachute, the drag increases and now our parachutist becomes an object from OPs question. Their speed will slow down until reaching the new terminal velocity and then more or less gently float to the ground at a constant speed. I ignore here a bunch of things but unlike meteors that never have occasion to slow down to terminal velocity, parachutist usually do that making them a better example IMO.
$endgroup$
– Ister
May 7 at 10:49
2
2
$begingroup$
Do not forget the gravitational force from the Sun and the Moon. (For an object with a mass of several tons it's several Newton, if I'm not mistaken.)
$endgroup$
– Peter A. Schneider
May 7 at 12:28
$begingroup$
Do not forget the gravitational force from the Sun and the Moon. (For an object with a mass of several tons it's several Newton, if I'm not mistaken.)
$endgroup$
– Peter A. Schneider
May 7 at 12:28
3
3
$begingroup$
@PeterA.Schneider Is it day or is it night?
$endgroup$
– Spitemaster
May 7 at 15:56
$begingroup$
@PeterA.Schneider Is it day or is it night?
$endgroup$
– Spitemaster
May 7 at 15:56
2
2
$begingroup$
I would argue that atmospheric buyoancy is accounted for in the object weight.
$endgroup$
– John Dvorak
May 7 at 18:34
$begingroup$
I would argue that atmospheric buyoancy is accounted for in the object weight.
$endgroup$
– John Dvorak
May 7 at 18:34
|
show 6 more comments
$begingroup$
Drag is proportional to the object's velocity (or velocity squared, depending on regime), so as the object's speed decreases, the drag decreases as well. Therefore, the body won't start "falling upwards" because for that to happen, the velocity must reach zero, and at that point the drag also drops to zero.
In practice, the falling object slows down until the drag equals the weight, at which point it keeps moving at constant speed (so-called "terminal velocity"). In the hypothetical scenario where you have a constant force that's greater than the weight pulling on the object, then the object will indeed start falling upwards.
answered May 7 at 2:22
AllureAllure
2,7471028
$endgroup$
1
$begingroup$
I don't think drag is proportional to velocity. en.wikipedia.org/wiki/Drag_equation is proportional to square of velocity.
$endgroup$
– Taemyr
May 7 at 5:19
3
$begingroup$
@Taemyr indeed, I was simplifying things. Edited.
$endgroup$
– Allure
May 7 at 5:27
7
$begingroup$
@Taemyr That depends on the regime. en.wikipedia.org/wiki/Stokes%27_law
$endgroup$
– Vladimir F
May 7 at 6:40
2
$begingroup$
@Taemyr They are proportional, but not directly proportional. The area of a circle is proportional to its radius, but directly proportional to the square of the radius ($A=πr²$)
$endgroup$
– Chronocidal
May 8 at 13:23
add a comment |
$begingroup$
Drag is proportional to the object's velocity (or velocity squared, depending on regime), so as the object's speed decreases, the drag decreases as well. Therefore, the body won't start "falling upwards" because for that to happen, the velocity must reach zero, and at that point the drag also drops to zero.
In practice, the falling object slows down until the drag equals the weight, at which point it keeps moving at constant speed (so-called "terminal velocity"). In the hypothetical scenario where you have a constant force that's greater than the weight pulling on the object, then the object will indeed start falling upwards.
answered May 7 at 2:22
AllureAllure
2,7471028
$endgroup$
1
$begingroup$
I don't think drag is proportional to velocity. en.wikipedia.org/wiki/Drag_equation is proportional to square of velocity.
$endgroup$
– Taemyr
May 7 at 5:19
3
$begingroup$
@Taemyr indeed, I was simplifying things. Edited.
$endgroup$
– Allure
May 7 at 5:27
7
$begingroup$
@Taemyr That depends on the regime. en.wikipedia.org/wiki/Stokes%27_law
$endgroup$
– Vladimir F
May 7 at 6:40
2
$begingroup$
@Taemyr They are proportional, but not directly proportional. The area of a circle is proportional to its radius, but directly proportional to the square of the radius ($A=πr²$)
$endgroup$
– Chronocidal
May 8 at 13:23
add a comment |
$begingroup$
Drag is proportional to the object's velocity (or velocity squared, depending on regime), so as the object's speed decreases, the drag decreases as well. Therefore, the body won't start "falling upwards" because for that to happen, the velocity must reach zero, and at that point the drag also drops to zero.
In practice, the falling object slows down until the drag equals the weight, at which point it keeps moving at constant speed (so-called "terminal velocity"). In the hypothetical scenario where you have a constant force that's greater than the weight pulling on the object, then the object will indeed start falling upwards.
answered May 7 at 2:22
AllureAllure
2,7471028
$endgroup$
Drag is proportional to the object's velocity (or velocity squared, depending on regime), so as the object's speed decreases, the drag decreases as well. Therefore, the body won't start "falling upwards" because for that to happen, the velocity must reach zero, and at that point the drag also drops to zero.
In practice, the falling object slows down until the drag equals the weight, at which point it keeps moving at constant speed (so-called "terminal velocity"). In the hypothetical scenario where you have a constant force that's greater than the weight pulling on the object, then the object will indeed start falling upwards.
answered May 7 at 2:22
AllureAllure
2,7471028
edited May 7 at 6:47
answered May 7 at 2:22
AllureAllure
2,7471028
answered May 7 at 2:22
AllureAllure
2,7471028
answered May 7 at 2:22
AllureAllure
2,7471028
2,7471028
1
$begingroup$
I don't think drag is proportional to velocity. en.wikipedia.org/wiki/Drag_equation is proportional to square of velocity.
$endgroup$
– Taemyr
May 7 at 5:19
3
$begingroup$
@Taemyr indeed, I was simplifying things. Edited.
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– Allure
May 7 at 5:27
7
$begingroup$
@Taemyr That depends on the regime. en.wikipedia.org/wiki/Stokes%27_law
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– Vladimir F
May 7 at 6:40
2
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@Taemyr They are proportional, but not directly proportional. The area of a circle is proportional to its radius, but directly proportional to the square of the radius ($A=πr²$)
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– Chronocidal
May 8 at 13:23
add a comment |
1
$begingroup$
I don't think drag is proportional to velocity. en.wikipedia.org/wiki/Drag_equation is proportional to square of velocity.
$endgroup$
– Taemyr
May 7 at 5:19
3
$begingroup$
@Taemyr indeed, I was simplifying things. Edited.
$endgroup$
– Allure
May 7 at 5:27
7
$begingroup$
@Taemyr That depends on the regime. en.wikipedia.org/wiki/Stokes%27_law
$endgroup$
– Vladimir F
May 7 at 6:40
2
$begingroup$
@Taemyr They are proportional, but not directly proportional. The area of a circle is proportional to its radius, but directly proportional to the square of the radius ($A=πr²$)
$endgroup$
– Chronocidal
May 8 at 13:23
1
1
$begingroup$
I don't think drag is proportional to velocity. en.wikipedia.org/wiki/Drag_equation is proportional to square of velocity.
$endgroup$
– Taemyr
May 7 at 5:19
$begingroup$
I don't think drag is proportional to velocity. en.wikipedia.org/wiki/Drag_equation is proportional to square of velocity.
$endgroup$
– Taemyr
May 7 at 5:19
3
3
$begingroup$
@Taemyr indeed, I was simplifying things. Edited.
$endgroup$
– Allure
May 7 at 5:27
$begingroup$
@Taemyr indeed, I was simplifying things. Edited.
$endgroup$
– Allure
May 7 at 5:27
7
7
$begingroup$
@Taemyr That depends on the regime. en.wikipedia.org/wiki/Stokes%27_law
$endgroup$
– Vladimir F
May 7 at 6:40
$begingroup$
@Taemyr That depends on the regime. en.wikipedia.org/wiki/Stokes%27_law
$endgroup$
– Vladimir F
May 7 at 6:40
2
2
$begingroup$
@Taemyr They are proportional, but not directly proportional. The area of a circle is proportional to its radius, but directly proportional to the square of the radius ($A=πr²$)
$endgroup$
– Chronocidal
May 8 at 13:23
$begingroup$
@Taemyr They are proportional, but not directly proportional. The area of a circle is proportional to its radius, but directly proportional to the square of the radius ($A=πr²$)
$endgroup$
– Chronocidal
May 8 at 13:23
add a comment |
$begingroup$
It would accelerate upward. This is exactly what happens when skydivers open their parachutes, for instance.
The underlying confusion behind this question is likely due to mixing up velocity and acceleration. You can accelerate in one direction without also having a velocity in that direction: you can accelerate up while moving down. Hitting the brakes on a car is not the same as putting it in reverse.
answered May 8 at 13:17
knzhouknzhou
49.5k12136242
$endgroup$
add a comment |
$begingroup$
It would accelerate upward. This is exactly what happens when skydivers open their parachutes, for instance.
The underlying confusion behind this question is likely due to mixing up velocity and acceleration. You can accelerate in one direction without also having a velocity in that direction: you can accelerate up while moving down. Hitting the brakes on a car is not the same as putting it in reverse.
answered May 8 at 13:17
knzhouknzhou
49.5k12136242
$endgroup$
add a comment |
$begingroup$
It would accelerate upward. This is exactly what happens when skydivers open their parachutes, for instance.
The underlying confusion behind this question is likely due to mixing up velocity and acceleration. You can accelerate in one direction without also having a velocity in that direction: you can accelerate up while moving down. Hitting the brakes on a car is not the same as putting it in reverse.
answered May 8 at 13:17
knzhouknzhou
49.5k12136242
$endgroup$
It would accelerate upward. This is exactly what happens when skydivers open their parachutes, for instance.
The underlying confusion behind this question is likely due to mixing up velocity and acceleration. You can accelerate in one direction without also having a velocity in that direction: you can accelerate up while moving down. Hitting the brakes on a car is not the same as putting it in reverse.
answered May 8 at 13:17
knzhouknzhou
49.5k12136242
answered May 8 at 13:17
knzhouknzhou
49.5k12136242
answered May 8 at 13:17
knzhouknzhou
49.5k12136242
answered May 8 at 13:17
knzhouknzhou
49.5k12136242
49.5k12136242
add a comment |
add a comment |
$begingroup$
You've probably already seen what happens yourself with extremely small particles, such as those that comprise smoke. Lots of meteors are this small.
At this scale, things still fall, of course, but it takes a long time.
If things get small enough, eventually Brownian motion takes over.
answered May 7 at 16:50
RogerRoger
1192
$endgroup$
add a comment |
$begingroup$
You've probably already seen what happens yourself with extremely small particles, such as those that comprise smoke. Lots of meteors are this small.
At this scale, things still fall, of course, but it takes a long time.
If things get small enough, eventually Brownian motion takes over.
answered May 7 at 16:50
RogerRoger
1192
$endgroup$
add a comment |
$begingroup$
You've probably already seen what happens yourself with extremely small particles, such as those that comprise smoke. Lots of meteors are this small.
At this scale, things still fall, of course, but it takes a long time.
If things get small enough, eventually Brownian motion takes over.
answered May 7 at 16:50
RogerRoger
1192
$endgroup$
You've probably already seen what happens yourself with extremely small particles, such as those that comprise smoke. Lots of meteors are this small.
At this scale, things still fall, of course, but it takes a long time.
If things get small enough, eventually Brownian motion takes over.
answered May 7 at 16:50
RogerRoger
1192
answered May 7 at 16:50
RogerRoger
1192
answered May 7 at 16:50
RogerRoger
1192
answered May 7 at 16:50
RogerRoger
1192
1192
add a comment |
add a comment |
$begingroup$
It will decelerate, yes. If the vector force exceed the vector force of the weight.
Note that both will increase as altitude decreases.
If the drag exceeds the weight and speed becomes 0, then the velocity will become negative. (Think parachute in a thermal or strong updraught). It may be a pumice-like meteor or quite small.
If it is lighter than water then the weight will become negative when it hits water, so there is that to consider.
answered May 8 at 4:07
mckenzmmckenzm
1112
$endgroup$
add a comment |
$begingroup$
It will decelerate, yes. If the vector force exceed the vector force of the weight.
Note that both will increase as altitude decreases.
If the drag exceeds the weight and speed becomes 0, then the velocity will become negative. (Think parachute in a thermal or strong updraught). It may be a pumice-like meteor or quite small.
If it is lighter than water then the weight will become negative when it hits water, so there is that to consider.
answered May 8 at 4:07
mckenzmmckenzm
1112
$endgroup$
add a comment |
$begingroup$
It will decelerate, yes. If the vector force exceed the vector force of the weight.
Note that both will increase as altitude decreases.
If the drag exceeds the weight and speed becomes 0, then the velocity will become negative. (Think parachute in a thermal or strong updraught). It may be a pumice-like meteor or quite small.
If it is lighter than water then the weight will become negative when it hits water, so there is that to consider.
answered May 8 at 4:07
mckenzmmckenzm
1112
$endgroup$
It will decelerate, yes. If the vector force exceed the vector force of the weight.
Note that both will increase as altitude decreases.
If the drag exceeds the weight and speed becomes 0, then the velocity will become negative. (Think parachute in a thermal or strong updraught). It may be a pumice-like meteor or quite small.
If it is lighter than water then the weight will become negative when it hits water, so there is that to consider.
answered May 8 at 4:07
mckenzmmckenzm
1112
answered May 8 at 4:07
mckenzmmckenzm
1112
answered May 8 at 4:07
mckenzmmckenzm
1112
answered May 8 at 4:07
mckenzmmckenzm
1112
1112
add a comment |
add a comment |
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