Potential Energy problem

  • #1

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.
 

Answers and Replies

  • #2
CompuChip
Science Advisor
Homework Helper
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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?
 
  • #3
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 ...?
 
  • #4
CompuChip
Science Advisor
Homework Helper
4,306
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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,
[tex]- \frac{G M m}{r} = m g r[/tex]
where r is the distance from the center of the earth, you can solve this for g:
[tex]g = - \frac{G M}{r^2}[/tex].
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).
 

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