# Potential Energy problem

## Homework Statement

A 1kg block rests on earth's surface. How much energy is required to move the block very far from the earth, ending up at rest again?

## Homework Equations

U = -G(m1m2)/r
Uf-Ui= -W (W= work internal)
U=mgy

## The Attempt at a Solution

I know that my final potential energy will be zero, and because the object ends at rest i can say my final kinetic energy will also be zero. Not sure where to go from here, or what equation to use.

CompuChip
Homework Helper
You are on the right track... as you say, the final total energy will be zero.
Can you also say something about the initial energy?

So assuming i use the equation Kf + Uf = Ki + Ui the left side of the equation goes to zero. Im then only left with my initial conditions, and my Ki = 1/2mv^2 and U=mgy..? or does U= -G (Mm)/r ...?

CompuChip
Homework Helper
So assuming i use the equation Kf + Uf = Ki + Ui the left side of the equation goes to zero. Im then only left with my initial conditions,
Exactly.

and my Ki = 1/2mv^2
Where v is the initial velocity (which is ... ?)

and U=mgy..? or does U= -G (Mm)/r ...?
Yes. The whole idea is that they are the same at or very close to the surface of the earth. You can easily show this yourself, if you equate the two,
$$- \frac{G M m}{r} = m g r$$
where r is the distance from the center of the earth, you can solve this for g:
$$g = - \frac{G M}{r^2}$$.
Just plug in the numbers for r at (or negligibly close to) the surface of the earth, at 6400 km from the center, and you will find a value very close to the usually quoted value of g = -9.81 m/s².
(This also explains the reason that g varies between different locations on earth: since the earth is not a perfect sphere, its radius changes a bit from one point to another. It is a bit smaller at the poles, so g will be a bit larger there than at the equator).