# Gravitation Potential Energy Help

1. Jan 30, 2005

### krypt0nite

Calculate the ratio of the energy that would be required to put a mass M into orbit near the earth's surface (if there were no friction) and the energy to put the same into distant space.

I'm stuck. How am i suppose to calculate the ratio. I know the that the energy to put M into orbit near the earth's surface would be a small positive energy while putting it into distant space would be a large positive energy.

2. Jan 30, 2005

### dextercioby

The signs are purely conventional.Actually for gravitational potential energy,it is MINUS INSTEAD OF PLUS...
The gravity field of a spherically symmetric celestial body (i.e.Earth) is CONSERVATIVE,which means u can apply the law of conservation of total energy.

Can u compute the energy required to put a mass into orbit near the surface of the Earth??

Daniel.

3. Jan 30, 2005

### krypt0nite

There are no values given to me so i dont know how to compute it

4. Jan 30, 2005

### christinono

I'm not sure how to find the answer, but I can tell you this: Binding energy is the additional energy you have to supply a satellite (or anything in orbit) to escape the Earth's gravitational field. This is the formula for it:
$$E_{binding}=\frac{Gm_1m_2}{2r}$$

When a satellite is in orbit, it's total energy (total mechanical energy) is:
$$E_{mechanical}=-\frac{Gm_1m_2}{2r}$$

So I guess the ratio would be 1:1?

5. Jan 30, 2005

### dextercioby

Not really.The second formula doesn't hold for bodies on the earth (which do not orbit,hence do not have KE),so i guess the OP is correct.The problem is missing some data...

Daniel.

6. Jan 30, 2005

### krypt0nite

Wouldn't the "r" be different values so I dont think they can be 1:1?
I was thinking more like 1:1/r ratio would make sense.

7. Jan 31, 2005

### Andrew Mason

For near earth orbit:

$$mv^2/R = F = GMm/R^2$$

Which is:
(1)$$2KE = GMm/R$$
where KE is the kinetic energy of the orbiting body, which is just -1/2 * its gravitational potential at radius R.

The condition for distant space (escape) is:

$$PE \ge 0$$

You should see from (1) that $2KE_{orbit} + PE_{orbit} = 0$, which means that the kinetic energy required for escape is ______ the KE for orbit. I think that should help you answer the question.

AM

8. Jan 31, 2005

### krypt0nite

So ratio would be 1:2?

9. Jan 31, 2005

### Andrew Mason

If you mean: Kinetic energy of escape = 2 KE of orbit, then yes.

AM

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