# Thread: Gravity from a mountain observation.

1. ## Gravity from a mountain observation.

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?

2. ## Re: Gravity from a mountain observation.

Use more significant figures.

Gravity should be taken as standard gravity, 9.81 m/s^2

3. ## Re: Gravity from a mountain observation.

Does the problem state the gravity at sea level is 9.83?
If so, use that to find r. Don't use the average earth radius.

Don't have paper or a calculator near me but it seems like the numbers would work. 9.83 means stronger gravity, so smaller radius than the given average.

Do the problem using that equation twice - once to find r_sea, and use that value for r in the equation you evaluated in the OP.

4. ## Re: Gravity from a mountain observation.

The problem states that gravity at sea level is 9.83.
Solving for r_sea gets me 6.37*10^6, which is the radius I plugged into the problem, but then I noticed that I rounded it. If I didn't round it and do all the math, it gets me 2,594.

5. ## Re: Gravity from a mountain observation.

When you're doing a precise calculation like this one, don't round anything until the final answer.
It takes some practice but you have to recognize what the problem is asking for and think about what would be acceptable for rounding.

For this scenario, you're trying to find a distance in the low thousands of feet, so decimals aren't going to matter 99% of the time.

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