# Free fall acceleration astronauts problem

bcjochim07

## Homework Statement

A starship is circling a distant planet of radius R. The astronauts find that the free fall acceleration of their altitude is is half the value at the surface of the planet. How far above the surface are they orbiting? The answer should be a multiple of R.

F=GMm/r^2 = ma

a=g=GM/r^2

## The Attempt at a Solution

h is the altitude of orbit

GM/(R+h)^2 = .5* GM/R^2

2GM/(R+h)^2 = GM/R^2

2GM=(GM/R^2)(R+h)^2
2= (R+h)^2 / R^2
2R^2 = (R+h)^2
sqrt 2 * R = R+h
h= R(sqrt2 -1)

Homework Helper
Bang on!

Homework Helper
GM/(R+h)² = .5* GM/R²

2GM/(R+h)² = GM/R²

2GM=(GM/R²)(R+h)²
2= (R+h)² / R²
2R² = (R+h)²
√2 * R = R+h
h= R(√2 - 1)

Hi bcjochim07!

Yes that's right!

(Why were you worried about it? )

But a bit long-winded … you could have cut it down to:

GM/(R+h)² = .5* GM/R²

so 2= (R+h)² / R²

so √2 = (R+h)/R = 1 + h/R

so h= R(√2 - 1).

Alternatively, start by saying let r be the height above the planet's centre.

Then GM/r² = .5* GM/R²,

so r = R√2, so h= R(√2 - 1).