A gravatational homework question

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The discussion centers on a gravitational homework problem involving the weight of an object decreasing to 0.99mg at a certain height above Earth's surface. The acceleration due to gravity at sea level is given as 9.81 m/s^2, with Earth's radius at 6370 km. The user applied the formula .99mg = (Gm(me)/(6370 + x)^2) to find the height x, arriving at an answer of approximately 1000 km. There is uncertainty regarding the correctness of this solution, prompting requests for clarification on the calculation process. The conversation highlights the need for verification of the derived height in relation to gravitational principles.
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Homework Statement


If an object is near the surface of the earth, the variation of its weight with distance from the center of the Earth can often be neglected. The accelertaion due to gravity at sea level is 9.81 m/s^2. The radius of the Earth is 6370 km. The weight of an object at sea level is mg, where m is its mass. At what height above the surface of the Earth does the weight of the object decrease to .99mg?


Homework Equations


I used .99mg= (Gm(me)/(6370 + x)^2) ... where G= 6.67 x 10^-11 and me= mass of the earth

The Attempt at a Solution



I used the formula above and solved for x getting an answer around 1000km, there is no way for me to check my solution... Any hints?
Thanks
 
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Can you show exactly how you got 'x'? Because x=1000km is a bit odd.
 

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