# Mass/escape velocity of a comet

1. Apr 20, 2010

### Quincy

1. The problem statement, all variables and given/known data
On July 4, 2005, the NASA spacecraft Deep Impact fired a projectile onto the surface of Comet Tempel 1. This comet is about 9.0 km across. Observations of surface debris released by the impact showed that dust with a speed as low as 1.0 m/s was able to escape the comet.
How far from the comet's center will this debris be when it has lost 90 % of its initial kinetic energy at the surface?

2. Relevant equations

3. The attempt at a solution

The equation for escape velocity can help determine the mass: 1.0 m/s = sqrt((2*G*M)/4500 m), M = 3.4 x 10^13.

KE at the surface = (1/2)(3.4 x 10^13)(1.0)^2 = 1.7 x 10^13 -- 90% of that is 1.53 x 10^13 J. So 90% of the kinetic energy has been converted into potential energy, right? Potential energy is (GMm)/r... but then I have two unknowns (m & r). How do I solve this? Is there another equation that can make the m cancel?

2. Apr 21, 2010

### AtticusFinch

If you believe in conservation of energy, the potential and kinetic energies of an escaping object will always add to zero. So when an object has lost 90% of it's initial kinetic energy...

$$(1-0.90)\frac{1}{2}mv_e^2 + \frac{-GMm}{x} = 0$$

Hopefully that helps.

Also, there's a couple of things I would like to say about your attempt at a solution.

1) "KE at the surface = (1/2)(3.4 x 10^13)(1.0)^2 = 1.7 x 10^13" Not quite. That's the kinetic energy of the entire comet if it was moving at 1.0 m/s, not the initial kinetic energy of the dust.

2) "So 90% of the kinetic energy has been converted into potential energy, right?" Yes, in a sense. The dust loses kinetic energy in it's effort to do work against the force of gravity. While doing this work, the gravitational potential energy increases (gets less negative) so in a sense the kinetic energy is being converted into potential energy.

3) "Is there another equation that can make the m cancel?" If you use the hint I gave you, this will end up happening.