Q: The escape speed from a very small asteroid is only 28 m/s. If you throw a rock away from the asteroid at a speed of 38 m/s, what will be its final speed?

 

I know it's not 10 m/s.

 

I was thinking of saying 28 = sqrt(2GM/Rinitial) to find the mass of the asteroid and then using that to find the final speed when it's thrown at 38 m/s, but then I also realized that I don't have the mass of the rock or radius of the rock, so that's out the window. Really, I'm not sure how to proceed, but this is what I know:

 

K_{ball final} + K_{asteroid final} + U_final = K_{ball initial} + K_{asteroid initial} + U_initial + W

Consider assuming Ufinal=0 (what does that mean?), then do the same calculation assuming an initial velocity of 28 m/s, then compare.

Think about the energy of the rock.

 

at some point it has an amount, x, of potential energy and "28 m/s"worth of kinetic energy.

 

as the rock moves away into space, what happens to its Kinetic Energy?

 

If it is at escape velocity , where will it end up, and how much kinetic energy will it have?

 

See the way yet?

Total Energy decreasing with Potential energy is very small.

G is rapidly decrease with the distance of the asteriod.

 

So

Initial kinetic energy =~ Final kinetic energy

 

Initial velocity =~ Final velocity

 

In the Earth case it is not negligible.

If you can just barely escape, what is your final (i.e. leftover) kinetic energy?

Escape kinetic energy = Total energy - Potential increase energy.

1/2 m V_escape² = 1/2 m V₀² - Int (form r₀ to r_escape ) G m m_e /r² dr

V₀ ; initial velocity