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.