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Gravity from a mountain observation.
#1
A sensitive gravimeter at a mountain observatory finds that the free-fall acceleration is 8.00×10−3m/s^2 less than that at sea level.
What is the observatory's altitude?

Given gravity at sea level: 9.83 m/s^2

I used:
G = 6.673 * 10 ^ -11 m^3 kg^-1 s^-2
r = 6.37*10^6 m
mass of the earth = 5.98 *10^24 kg

The equation I used was: g = Gm/(r^2)

How I did the problem:
9.822 = Gm/(r+h)^2
(r+h)^2 = Gm/9.822
h = sqrt(Gm/9.822) - r
The answer I got was 3987 m, while the correct answer is 2590 m. Plugging in their answer I get 9.826 m/s^2, while plugging in my answer gets me 9.822 m/s^2.

Is their answer wrong, or am I using the wrong equation?
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Messages In This Thread
Gravity from a mountain observation. - by Imagine - 2014-01-13, 02:39 AM
Gravity from a mountain observation. - by JoeTang - 2014-01-13, 09:05 AM
Gravity from a mountain observation. - by Imagine - 2014-01-13, 04:45 PM

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