# Energy required to go from one planet to another

1. Apr 26, 2013

1. Description of Problem
A binary planet system comprises two identical planets of mass M and radius R with their centers a distance 10 R apart. The minimum energy that the engines of a spacecraft need to supply to get a rocket of mass m from the surface of one planet to the surface of the other is of the form XGMm/R. What is X?

2. Relevant equations
Escape Velocity v= √2GM/R
Gravitational Kinetic Energy KE=GMM/2R
Gravitational Potential Energy U = GmM/R
Kinetic Energy = 1/2mv2

3. Attempt at a solution

use escape velocity to find the KE of escape velocity is KE = mMG/R

I'm not sure where to go from here...

subtract the potential energy that the second planet supplies once the rocket reaches halfway?

this gives x=.8

2. Apr 26, 2013

### Staff: Mentor

I think you are supposed to make some approximations, otherwise the problem becomes really complicated.
- neglect orbital mechanics and how rockets work
- neglect the motion of the planets around each other?

If you just consider the potential energy at the surface of a planet and at the center between the planets, you get some nice answer (it is not 0.8, however - don't forget the potential of the other planet).

3. Apr 26, 2013

What we need to find is the amount of energy to get the rocket half way to the other planet, right? this can be found y finding the difference between the potential energies at the surface and half way through?

Potential at the surface is MmG/R - MmG/9R = 8MmG/9R
The second term is the Potential E from the second planet - I didn't include this in my previous attempt

potential halfway is MmG/5R

this gives X=.688

This is close to the answer which is listed as X=0.7

Is this correct?

Last edited: Apr 26, 2013
4. Apr 26, 2013

### barryj

I understand this "Potential at the surface is MmG/R - MmG/9R = 8MmG/9R"

How did you get this "potential halfway is MmG/5R"

5. Apr 26, 2013

Halfway between the center of the two planets is 5R, so the potential E from the second planet at half way is MmG/5R

Conceptually, the potential at that point is 0 because the planets pull on it equally?

6. Apr 26, 2013

### Staff: Mentor

And the potential from the other planet?

The force is 0, the potential is not (with your definition of the potential).

7. Apr 26, 2013

the potential from the other planet is the same...

so the potential at the midpoint is 2MmG/5R?

8. Apr 27, 2013

### Staff: Mentor

I think there is a minus sign missing everywhere, but apart from that: right.

9. Apr 27, 2013

### barryj

How can there be a potential if there is no force?
Also, consider one of your previous equations,.. Potential at the surface is MmG/R - MmG/9R = 8MmG/9R ,
using this, the potential halfway between would be zero because the first term would change to MmG/5R as well as the second term and the difference would be zero it seems.

10. Apr 27, 2013

### Staff: Mentor

There is nothing wrong with that. Actually, every potential minimum, maximum* and critical point has no (net) force, while every point in space has a potential.

*does not exist in gravity
There is a sign error, too.

11. Apr 27, 2013

### barryj

If you do the calculus and integrate the f dl from R to 5R you get 8MnG/9R, so I think this would be the correct answer. In a similar manner if you integrate from R to infinity you will get 0,8, not 8/9. I am a bit confused but will figure this out eventually, unless someone shows me the way.