# Energy transfer question

1. Jan 7, 2008

### BosonJaw

Hello guys!!!

Would an object about 6 ft in length and about 180 lbs hitting the Earth at 99% of light speed destroy the Earth? If not, what would it do? Obviously assuming it did not disintegrate due to frictional forces first. Can this be solved by a simple energy transfer equation?

Thanks!!!

2. Jan 7, 2008

### Shooting Star

The earth has a certain amount of energy due to self-gravity, which is holding it together. If at least that amount of energy can be transferred by a collision, there's a chance that that the earth would come apart (partly if not wholly, in reality). So, you just have to compare the two energies.

There are simple formulas to find both the energies.

3. Jan 7, 2008

### AFG34

right when it hits the earth, it only has kinetic energy<--- in an ideal world
Ek = .5mv^2
m=mass
v=velocity

4. Jan 7, 2008

### Shooting Star

> Ek = .5mv^2

For v=0.99 c, perhaps a different formula would be more apt.

5. Jan 7, 2008

### Shooting Star

The PE of a self-gravitating sphere of uniform density is -(3/5)Gm^2/r.

The KE would be m*c^2*(gamma(0.99) -1), provided there is no conversion of mass to energy.

You can plug in the values now and see.

Last edited: Jan 7, 2008
6. Jan 7, 2008

### MaWM

Yes, this can be solved pretty simply. At impact, the object would have kinetic energy equal to 9mc^2. Where m is the mass of the object. That's about as much energy as the use of 810 *pounds* of antimatter. You'd have to work out the exact numbers (plug in for c and convert 180lbs to grams or kilograms), but my sense is that that is a ridiculuously large amount of energy. You also have to decide what you mean be "destroy the Earth" (crack it in half? blow it to bits? or just melt the crust so that nothing survives?). You can then intergrate the gravity potential to find the total amount of energy you'd need. Again, I could be wrong.. but that is a whole lot of energy.

7. Jan 7, 2008

### Danger

It would pretty much be "I have to find a new home."

8. Jan 7, 2008

### Shooting Star

> At impact, the object would have kinetic energy equal to 9mc^2.
Why?

> That's about as much energy as the use of 810 *pounds* of antimatter.
Why?

9. Jan 8, 2008

### MaWM

Well.. as you said, the kinetic energy is (gamma-1)mc^2. For v=.99c, this works out to 9mc^2. That is also equal to the energy that would be liberated if 9 such asteriods were converted directly to energy. This could be accomplished if half that amount was antimatter. Since the asteriod weighs 180lbs, we'd need 810 lbs of antimatter.

10. Jan 8, 2008

### Shooting Star

I'd given the formula for KE, that is, total energy minus rest energy. I considered it to be made of matter, and had mentioned it. It's the impact which is "breaking" up the earth, not matter-antimatter reaction.