# Energy in Meteor Impact

1. Dec 4, 2007

### odie5533

1. The problem statement, all variables and given/known data
A meteor of mass about 3.4 x 10^12 kg is heading straight for Jupiter. When it hits there will be a huge release of energy, visible her on earth. Assuming it has fallen from far away, how much energy will be released when it hits Jupiter? The radius of Jupiter is 7 x 10^7 m and its mass is 1.9 x 10^27 kg.
A) 3.4 x 10&19 J
B) 5.8 x 10^18 J
C) 5.4 x 10^22 J
D) 6.8 x 10^19 J
E) 6.1 x 10^21 J

3. The attempt at a solution
I'm somewhat lost as to how to approach this problem. When I've worked with energies, it has always involved velocity. In the chapter on gravitation in my text I only found mention of gravitational potential energy, $$U = -\frac{Gm_{e}m}{r}$$, but I don't have the distance between the masses. I'm guessing the collision would be inelastic, but I'm not sure how that helps. I can only think of ways of solving this if I'm given the velocity of the asteroid. Any help on this would be greatly appreciated.

2. Dec 4, 2007

### odie5533

After looking it over again, I think maybe I could use an integral.
$$W_{grav} = -Gm_{e}m\int_{r_{1}}^{r_{2}}\frac{dr}{r^2}$$
That is listed in our text.
So maybe...
$$W_{grav} = -Gm_{e}m\int_{\infty}^{0}\frac{dr}{r^2}$$
I don't think that's at all right. Just thinking aloud

If anyone has any ideas, any at all, please share. I'm turning this in completed or not in a few minutes. =/

Last edited: Dec 4, 2007
3. Dec 4, 2007

### Dick

Compare the gravitational potential energy of the meteor on Jupiter's surface with it's energy at r=infinity (where U=0). You don't need to integrate anything.

Last edited: Dec 4, 2007
4. Dec 4, 2007

### odie5533

I asked my physics prof before class and he said to use the escape velocity in the kinetic energy equation:

$$V = \sqrt{\frac{2GM_{J}}{R_{J}}}$$
$$K = \frac{1}{2}MV^2 = \frac{1}{2}M\frac{2GM_{J}}{R_{J}} = \frac{GMM_{J}}{R_{J}}$$

Your way works better, and is easier to understand. Thanks :)

5. Dec 4, 2007

### Dick

Sure. But it's the same thing. The calculation of escape velocity comes from potential energy. Which in turn is also the same thing as the integral you proposed earlier. Integrating the force over distance gives you the potential.