# Object to lose weight

## 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

berkeman
Mentor

## 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*.