# Homework Help: Another Gravitational potential energy problem

1. Jul 21, 2007

### neo982

1. The problem statement, all variables and given/known data
Two Jupiter-size planets are released from rest 1.30×10^11 m apart.

What are their speeds as they crash together?

2. Relevant equations

3. The attempt at a solution
I have done K_f + U_f = K_i + U_i with K_i being zero because they have crashed together so... K_f + U_f = U_i is the equation I get....when solving for velocity final I get v_f = sqrt(2*G*m*(-1/r_i +1/r_f)) and Im not so sure I am making r_i and r_f the right values, I have made r_i the 1.30×10^11 and the r_f to be 2*radius of Jupiter which is 6.99*10^7 (from book) I keep getting v_f to be 42,566 m/s but thats not right

also from book the mass of jupiter is 1.90*10^27 kg...

Last edited: Jul 21, 2007
2. Jul 21, 2007

### Dick

The r's in your problem have nothing to do with the radius of the planet. They are the distance from the center of mass of the system to the two bodies. And they shouldn't have opposite signs.

3. Jul 21, 2007

### neo982

well if you solve 1/2mv^2 + (-Gm1m2/r_f) = (-Gm1m2/r_i) for velocity you get that (-1/r +1/r) part so they are not opposite signs, but anyway...so the initial r would be half of the distance given which is 1/2 (1.3x10^11) = 6.5x10^10 right...if this is true then im not sure what would be the r final???

4. Jul 22, 2007

### Dick

Hmm. You've got me. You are much more correct than my sloppy thinking. Apologies. Now I'm having some problems figuring out how your answer could be incorrect. So far I'm baffled. But I haven't put numbers in...

5. Jul 22, 2007

### Dick

You have a factor of two missing. If you think in the center of mass system then the central mass is 2*m, then cut your diameters in half to make them radii. The m in the KE remains the same. I get sqrt(2) times what you do. I'm counting on you to figure this out and not make me give an incoherent explanation again.

6. Jul 22, 2007

### neo982

Im not sure if I follow you....what would the v_f eq look like..?

7. Jul 22, 2007

### Staff: Mentor

Your values for r_i & r_f seem OK to me. But you have an extra sqrt(2) in your formula for v_f. I suspect that comes from you setting KE = 1/2mv^2--but realize that there are two planets sharing that KE.

8. Jul 22, 2007

### neo982

so do I need to make the eq 2*K_f +U_f = U_i with the r's 1.3x10^11 and 2*6.99x10^7 or make them ... half of 1.3x10^11 and the other just 6.99x10^7 ? well...I i know i get the right answer from using 1.3x10^11 and (2*6.99x^10^7) but what is the reason to use the diameter instead of cutting them in half ...?

also, is that the right way to set it up because I get the right answer from my 42,556 / sqrt (2)...or is that the same thing by making 2K

Last edited: Jul 22, 2007
9. Jul 22, 2007

### Staff: Mentor

Right--where K_f is the kinetic energy per planet.
Why would you do that?
What's important is the distance between their centers. When they collide, that distance equals twice the radius. Right?

Work it out and see!

10. Jul 22, 2007

### neo982

Thanks for all your help. I have a really good grasp on it now.