A question involcing gravitational acceleration (1 Viewer)

[independantz]

calculate, how far an astronaut would need to be away above the Earth in order for his weight to be 0.01 his weigth on the Earth's surface?

 

[SpinCobra]

Err.. Lemme think..

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Weight = mass x gravitational acceleration

Let m = 100kg

Therefore..

980N = 100kg x 9.8

For weight to be 0.1 of normal, acceleration due to gravity would need to be 0.98ms²

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g = Gm/r²

0.98 = (6.67x10^-11 x 5.94x10²⁴) / r²

r² = (6.67x10^-11 x 5.94x10²⁴) / 0.98

I think...

 

[SpinCobra]

Oops. Edited above.

g = Gm/r²

Which then gives the answer.. 2.01x10⁷ metres.

 

[benji6667]

you would then need to subtract the radius of the earth to find the altitude above the earths surface

my answer is 6.373 x 10^7 m above the earths surface... is this value anywhere near the correct answer (im prone to calculating errors)

 

[benji6667]

oops ignore that last post, i just realised a conversion error

my final answer is 5.737 x 10^7 m or 57372km above the earths surface

 

[namburger]

W = mg
g = GM/r^2

Solving simultaneously gives you W = GMm/r^2
N.B. M is mass of planet and m is mass of object

W = 9.8m is the weight of the object on surface
0.01 of its W = 0.098m

Therfore
0.098m = [6.67 x 10^-11 x 6 x 10^24 x m] / (6380000 + h)^2
0.098 = [6.67 x 10^-11 x 6 x 10^24] / (6380000 + h)^2
(6380000 + h)^2 = [6.67 x 10^-11 x 6 x 10^24] / 0.098
(6380000 + h)^2 = 4.08 x 10^15
6380000 + h = 63903626.42
h = 57523626 m
= 57523.63 km

 

[SpinCobra]

Oh right. I did 0.1 of normal instead of 0.01. My bad.