Gravitational Potential Energy of the earth

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To determine how high above Earth's surface a mass must be for its gravitational potential energy to be 17% of its surface value, the equation PEg = (-GMm)/r is used. The user initially set r1 to 1 meter, leading to confusion as it did not yield a sensible height. Clarification was provided that the radius of Earth is significantly larger than 1 meter, which is crucial for accurate calculations. By correctly applying the Earth's radius, the user was able to arrive at the right answer. Understanding that gravitational potential energy approaches zero at infinity is also an important aspect of the problem.
kt102188
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Homework Statement



How high above the surface of the Earth must a mass be located so its gravitational potential energy is only 17.00% of its value on the surface?


Homework Equations



PEg=(-GMm)/r, PElocal= mgy where y is height the object can fall

The Attempt at a Solution



I've tried a couple of different things but all my variables cancel out.

(-GMm)/r1=0.17((-GMm)/r2

r1/0.17=r2 If I randomly set r1 to be 1 r2= 5.88235 but that doesn't make sense because if you raise an object by 5m it would have a greater PEg

Any suggestions on where to go?
Thanks!
 
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Welcome to PF!

Hi kt102188! Welcome to PF! :smile:
kt102188 said:
r1/0.17=r2 If I randomly set r1 to be 1 r2= 5.88235 but that doesn't make sense because if you raise an object by 5m it would have a greater PEg

But the radius of the Earth is a lot larger than 1m. :confused:

(and remember, the question assumes that the PE at ∞ is zero :wink:)
 
Thank you! I was totally thinking of this wrong. Actually using the radius of Earth got the right answer.
 
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