Simple kinematics problem — falling from a geostationary satellite

[AlternativePhysics]

Homework Statement

Given that gravitational acceleration at geostationary altitude is 0.221 m/s², and neglecting air resistance, how long does it take for a marble released from a geostationary satellite in orbit to reach the ground?

Relevant Equations

F = mg, where g varies with altitude: g(r) = GM/r²
F = mg, where g varies with altitude: g(r) = GM/r²
Orbital velocity: v = √(GM/r)

I know g at that altitude is 0.221 m/s² but I'm not sure how to integrate the increasing gravitational acceleration as the object approaches Earth. I attempted using kinematic equations but g is not constant during the fall.

 

[anuttarasammyak]

How about integrating
$$dt=\frac{dr}{v}$$
where
$$\frac{v^2}{2}-\frac{GM}{r}=const.$$?

 

[berkeman]

Welcome to PF.

how long does it take for a marble released from a geostationary satellite in orbit to reach the ground?

That's a trick question. If you release a marble from a geostationary satellite, it just continues in the same orbit as the satellite. If you mean "a marble is released from rest as a geostationary satellite goes by...", that is a different question. Did you quote the schoolwork question word-for-word?

 

[Andrew Mason]

Homework Statement: Given that gravitational acceleration at geostationary altitude is 0.221 m/s², and neglecting air resistance, how long does it take for a marble released from a geostationary satellite in orbit to reach the ground?
Relevant Equations: F = mg, where g varies with altitude: g(r) = GM/r²
F = mg, where g varies with altitude: g(r) = GM/r²
Orbital velocity: v = √(GM/r)

I know g at that altitude is 0.221 m/s² but I'm not sure how to integrate the increasing gravitational acceleration as the object approaches Earth. I attempted using kinematic equations but g is not constant during the fall.

As Berkeman has pointed out you would need to cancel the marble’s orbital speed first in order for the marble to drop vertically to the Earth surface.

The orbital speed is about 3066 m/s given by $v=\sqrt{\frac{GM}{r}}$ where r=42,164 km. So it has to lose a lot of velocity make it drop.

Assuming you could do that, the easiest way to calculate its speed would be to equate kinetic energy per unit mass to change of potential.

$\frac{1}{2}v^2=-\Delta U=GM\left(\frac{1}{r_f}-\frac{1}{r_i}\right)$

Also, as a practical matter, there will be effects after entering the atmosphere. The upper atmosphere 100 km above the surface which would be 6,478 km from Earth centre.

AM

 

[sophiecentaur]

Homework Statement: Given that gravitational acceleration at geostationary altitude is 0.221 m/s², and neglecting air resistance, how long does it take for a marble released from a geostationary satellite in orbit to reach the ground?
Relevant Equations: F = mg, where g varies with altitude: g(r) = GM/r²
F = mg, where g varies with altitude: g(r) = GM/r²
Orbital velocity: v = √(GM/r)

I know g at that altitude is 0.221 m/s² but I'm not sure how to integrate the increasing gravitational acceleration as the object approaches Earth. I attempted using kinematic equations but g is not constant during the fall.

A quick check on your answer, achieved 'the hard', accurate way:

The period of a geostationary satellite would be (of course) 24hours. How long would the object take to fall to Earth? Treating the Earth as a point mass with no atmosphere, the fall time would be the same as the time as one quarter of the orbit time. A circular orbit can be analysed as the sum of two linear SHMs at right angles. so ignore one of the axes. The majority of the fall time would not involve atmospheric resistance or getting through the few thousand km for the final phase. So an arm waving value would be about one quarter of a day length.

Something to hang on to and to help you spot any gross errors in your initial calculation. (Haha - as if . . . .)

 

[AlternativePhysics]

Berkeman, your answer is very accurate — it indeed DEPENDS, and you have perfectly identified the two distinct situations.

Now I'll add a detail I didn't mention in the original problem. Imagine three marbles released simultaneously:

— Marble 1: released from the geostationary satellite at a given point
— Marble 2: released from the top of a rigid tower anchored to the Earth — whose upper end permanently and indefinitely shares its position with the satellite, not momentarily
— Marble 3: launched by a sharpshooter from the base of the tower, aiming directly at the satellite, with exactly the velocity needed to reach that altitude with zero velocity relative to the Earth

All three marbles coincide at the same point in space. Marble 1 does not fall. Marbles 2 and 3 do fall.

Why?"




 

[sophiecentaur]

Berkeman, your answer is very accurate — it indeed DEPENDS, and you have perfectly identified the two distinct situations.

Now I'll add a detail I didn't mention in the original problem. Imagine three marbles released simultaneously:

— Marble 1: released from the geostationary satellite at a given point
— Marble 2: released from the top of a rigid tower anchored to the Earth — whose upper end permanently and indefinitely shares its position with the satellite, not momentarily
— Marble 3: launched by a sharpshooter from the base of the tower, aiming directly at the satellite, with exactly the velocity needed to reach that altitude with zero velocity relative to the Earth

All three marbles coincide at the same point in space. Marble 1 does not fall. Marbles 2 and 3 do fall.

Why?"




The answer to that involves considering Energy in the three cases. (GPE and KE)

 

[nasu]

The top of the tower moves with the same velocity as the satellite.

 

[Ibix]

@nasu has given a good hint.

I would add that I don't think what Marble 3 is doing is particularly clear (possibly a translation issue). Is it meant to mean that Marble 3 comes to rest relative to the tip of the tower and the satellite? If so, note that it's a trick question.

 

[berkeman]

Marbles 2 and 3 do fall.

Why does marble #2 fall?

 

[sophiecentaur]

Why does marble #2 fall?

My suggestion to consider the Energy situations makes it easy to appreciate what goes on without bothering with all that pesky calculus. Marble 2 has the same KE as the one on board the satellite. They both have the same GPE. because they are at the same height. So the situations for 2 and 3 are actually the same.[ Edit: sorry; 1 and 2 are the same!]

If marble 3 started out with enough KE to get it to the height of the tip of the tower (minimum energy to take it to that height but it will travel in a straight line so the tip of the tower will leave it behind as it follows the orbiting satellite. By that time the marble will have had all its KE converted to GPE and will be stationary. It will fall back to ground and arrive at a speed somewhat lower than its original launch speed. It's lost energy on the journey out and the journey back to ground.

So the basic answer to the original question is NEVER.

 

[sophiecentaur]

All three marbles coincide at the same point in space.

No they don't, for the reasons above.

 

[anuttarasammyak]

— Marble 2: released from the top of a rigid tower anchored to the Earth — whose upper end permanently and indefinitely shares its position with the satellite, not momentarily

Such a tower is considered as the space elevator

 

[PeterDonis]

— Marble 1: released from the geostationary satellite at a given point

Meaning, with the same velocity as the satellite? Then, as has already been said, it will have the same orbit as the satellite.

— Marble 2: released from the top of a rigid tower anchored to the Earth — whose upper end permanently and indefinitely shares its position with the satellite, not momentarily

Which means marble #2 will also "indefinitely share its position with the satellite", just as marble #1 does.

— Marble 3: launched by a sharpshooter from the base of the tower, aiming directly at the satellite, with exactly the velocity needed to reach that altitude with zero velocity relative to the Earth

In other words, not sharing the same velocity as the satellite, correct?

Marble 1 does not fall

Yes.

Marbles 2 and 3 do fall.

Not quite. Marble 3 falls, but marble 2 does not. See above.

 

[PeterDonis]

Why does marble #2 fall?

It doesn't. See post #14.

 

[PeterDonis]

If marble 3 started out with enough KE to get it to the height of the tip of the tower (minimum energy to take it to that height but it will travel in a straight line

In an inertial frame centered on the Earth, in general, its trajectory will be an ellipse (assuming it never achieves escape velocity). But which ellipse? We can control that by controlling how the sharpshooter aims.

The OP has the sharpshooter at the base of the tower, aiming straight up. You are correct that, for this case, the marble will fall behind the tower. In an Earth-centered inertial frame, this is because the marble is moving eastward at only about 450 m/s (the equatorial velocity of the surface of the Earth), whereas the tip of the tower and the satellite have a much higher eastward velocity (because they have the same angular velocity and are much higher up). In the rotating frame that the OP appears to be implicitly using to imagine the scenario, this is because of the Coriolis effect.

But we can compensate for this by having the sharpshooter aim forward and launch the marble with a higher velocity. In the inertial frame, the marble will then have an elliptical orbit whose apogee is placed just right to meet the satellite and the tower as they go by. In the rotating frame, the marble curves backwards due to the Coriolis effect, so that it still meets the tip of the tower and the satellite.

None of the above changes the key point, which is that, unlike the other two marbles, this one falls back to Earth.

 

[Dale]

unlike the other two marbles, this one falls back to Earth.

Yes. And at the point where they all meet the first two have the same velocity while the third does not.

a detail I didn't mention in the original problem

In the future please fill out the template completely.

 

[berkeman]

Thread is closed temporarily for Moderation...

 

[berkeman]

Thread is reopened provisionally so that the OP can respond to my DM in-thread. Short leash.

 

[sophiecentaur]

in general, its trajectory will be an ellipse

My sloppy reply. Of course an ellipse, in general. If, as the question says, it will (just) reach GSO height then it would have been given minimum KE then its path will be radial so there will be no tangential component. I think that part of the question was not clear enough in describing where the marble is aimed.

[EDIT: The GPE for a circular orbit will be half the KE so that would mean the marble launch (without losses, of course) would be 1/√2 of the orbital velocity - another bit of info)

 

[PeterDonis]

If, as the question says, it will (just) reach GSO height then it would have been given minimum KE then its path will be radial

If you want the path to be radial in an Earth-centered inertial frame, launching from the rotating Earth, you need to fire it backwards from the standpoint of the rotating Earth--and from a point far enough forward of the tower that the marble at the top of its radial path will just meet the tip of the tower (and the satellite) as they rotate. You're correct that this would be the path of minimum kinetic energy in the inertial frame--but it doesn't have the marble being launched from the base of the tower.

There is no way to have a free-fall path be purely radial in the rotating frame in which the tower and the GSO satellite are at rest. The Coriolis effect prohibits it. This is the part that I don't think the OP understood: you can't launch a free-fall projectile from the base of the tower and have it just rise up the tower and come to rest at the top--not on the rotating Earth.

 

[Ibix]

Obvs the general path would be an ellipse to but there has to be a launch site or time which will give a radial path to coincide with the satellite.

Sure. You calculate the velocity from the change in GPE, then adjust the launch angle so that $v\sin(\theta)$ (or $v\cos(\theta)$ depending on your angle convention) is equal to the linear velocity of the Earth at the equator. In the non-rotating Earth-centered frame the shot is then purely radial.

Then calculate the flight time, $T$, and position your gun $\omega T$ ahead of the tower, where $\omega=2\pi/86400\mathrm{s^{-1}}$ is the angular velocity of the Earth.

 

[PeterDonis]

Thread closed for moderation.

 

[berkeman]

After a Mentor discussion and some thread cleanup, the thread will remain closed. Thanks to all who contributed trying to help the OP with their misunderstandings of gravity.