Can an object lose weight just by moving higher above the Earth?

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An object can lose weight by moving higher above the Earth due to the gravitational force decreasing with distance. To determine how high it must be to lose 30% of its weight, the gravitational force equation is applied, considering the mass of the Earth. The calculation involves rearranging the formula to solve for the radius, factoring in the necessary conversions for time. It is emphasized that the problem should focus on the decrease in gravitational force rather than the mass of the Earth itself. Ultimately, the radius must be greater than the Earth's radius to achieve the specified weight loss.
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Homework Statement


How far above the Earth must an object be in order to lose 30% of its weight?


Homework Equations


ma=Gm(Mearth)/r^2
a=v^2/r
v=2pir/T


The Attempt at a Solution


I think the 30% lost weight would be the same as the Earth's weight losing 30%. Therefore, I can put Mearth=(0.30)(6x10^24)

We have to solve for r to find the radius and then subtract it from the radius of the Earth to find where this object is.

So rearranging the equation we get:
4pi^2r^3/T^2=0.30GMearth/r^2, where T=1 day and needs to be converted to seconds.

Finally, we rearrange to solve for r and get:
r=0.30GMearth/4pi^2
 
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physics(L)10 said:

Homework Statement


How far above the Earth must an object be in order to lose 30% of its weight?


Homework Equations


ma=Gm(Mearth)/r^2
a=v^2/r
v=2pir/T


The Attempt at a Solution


I think the 30% lost weight would be the same as the Earth's weight losing 30%. Therefore, I can put Mearth=(0.30)(6x10^24)

We have to solve for r to find the radius and then subtract it from the radius of the Earth to find where this object is.

So rearranging the equation we get:
4pi^2r^3/T^2=0.30GMearth/r^2, where T=1 day and needs to be converted to seconds.

Finally, we rearrange to solve for r and get:
r=0.30GMearth/4pi^2

Time has no place in this calculation.

Don't try to think of it in terms of reducing the mass of the Earth. That's not good intuition.

Use the main equation you list, and calculate the radius that's larger than the Earth's radius that results in the listed decrease in *force*.
 
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