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