[Experst only] How much time does it take to fall from high altitude?

Question

An isolated planet has orbital period T = 2 hours
for low circular orbits.

A body initially located at distance equal to 2 radii of the planet
from the center of the planet is released with zero initial speed.

How much time does it take for the body
to reach the surface of the planet?

Answer 1

From Kepler's 3rd, we get

T=(2π r^(3/2))/(GM)^(1/2)

If we use conservation of energy, we get

v=
Sqrt[2GM(1/r_i-1/r_f)]

(Using r_a to mean r sub a).

If we integrate

∫dx/Sqrt[1/x-1/2] from 2 to 1 we get (using Mathematica)

(2+π)/Sqrt[2]

So t=
(1/Sqrt[2 GM])*
((2+π)/Sqrt[2])*
r^(3/2)

If we combine this with the result you got from Kepler's 3rd law, the rest is pretty easy.

Answer 2

All right, since the other posting of this question is already closed, I will put the exact answer to the other question, not this one. For a free-fall towards the surface of a planet of radius R and gravitational acceleration g at the surface, and letting H be the height from the center of the planet, the equation in time H(t) is:

H(t) = (H^3/2 - ((3R - √g)/√2) t)^2/3

so that the time to reach R from H is:

t = (√2 / (3R√g))(H^3/2 - R^3/2)

Dimensionally, this works out.