Free fall acceleration astronauts problem

  • Thread starter bcjochim07
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  • #1
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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.


Homework Equations


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)
 

Answers and Replies

  • #2
Chi Meson
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Bang on!

That is, your are right.
 
  • #3
tiny-tim
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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! :smile:

(Why were you worried about it? :confused:)

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). :smile:

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). :smile:
 

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