# Universal Law of Gravitation problem

Discussion in 'MCAT Study Question Q&A' started by kmcgrath, Jan 7, 2009.

1. ### kmcgrath 5+ Year Member

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How do I get to this answer step by step? Any help would be greatly appreciated. Thanks!

Q: At what height above the earth's surface is acceleration due to gravity 10% of that at sea level?

A: 1.36x10^7 m

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2. ### TooMuchResearch i'm goin' to Kathmandu... Lifetime DonorVerified Account 7+ Year Member

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I think we would just set up the normal gravitational force equation:

____Gm1m2
F = -------
_____r^2

Next, recall that F=ma. In this case, that m probably equals the mass of an object (that isn't the earth)...lets call it m1.

Now, m1 will cancel from each side, leaving us with:

____Gm2
a = ------
_____r^2

Since we are given acceleration due to gravity is 10% of that at sea level (recall: g=9.8 m/s^2 or ~10 m/s^2), we know that a=g=1.0 m/s^2.

Just solve for r and should be okay...you might have to subtract the radius of the earth from the value you determine for r since the question asks for the height above earth's surface.

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3. OP

### kmcgrath 5+ Year Member

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That's funny b/c I did exactly that and got the wrong answer...I am trying it again with a fresh mind. At least I am going the right direction with this one. Probably a math error I was repeating from exhaustion. Thank you for your help.

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4. ### Vandau New Member 15+ Year Member

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Don't plug radius of earth into the equation, since g is calculated at earth's surface already. 10% of g would be the distance starting from earth's surface so just solve for r, the distance above earth's surface.

g=GM/r^2
1=GM/r^2
r^2=GM
r=sq rt of GM

5. ### Mantis Toboggin 2+ Year Member

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h= r'-r

To find r, use the value of a=g=10.0 m/s^2

10= GM/(r^2)

10r^2= GM

(we know the square root of 10 will fall between the squares of 3 and 4, closer to 3... just guestimate it to 3)

So r= .67 * 10^7 m

r'= 2 * 10^7 m

h= 1.3 * 10^7 m or 1.3 * 10^4 km

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