How Much Energy Is Needed to Move a 1kg Block Away from Earth?

[MillerGenuine]

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]

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?

 

[MillerGenuine]

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

 

[CompuChip]

So assuming i use the equation Kf + Uf = Ki + Ui the left side of the equation goes to zero. I am 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).