Escape Velocity Physics?

(a) What is the escape speed on a spherical asteroid whose radius is 540 km and whose gravitational acceleration at the surface is 2.6 m/s2?





Ve = √(2 R g)


i found that to be 1680 m/s





(b) How far from the surface will a particle go if it leaves the asteroid's surface with a radial speed of 1000 m/s?





used const. acc. equation





Vf^2-Vo^2 = 2 * a * x





-(1000m/s)^2 = 2 * (-2.6m/s^2) * x


x= 192000 m


^^^getting a wrong answer





(c) With what speed will an object hit the asteroid if it is dropped from 1000 km above the surface?

Escape Velocity Physics?
(a) correct, although you obviously need a refresher on WHY this equation works.





(b)


You can't use constant acceleration, because the gravitational force isn't constant. We only use that near the earth's surface where the radius isn't changing much.





You have to use conservation of energy.





Potential energy = -GmM/r


Kinetic energy = 1/2mv^2





So the particle starts with potential and kinetic energy (which is zero if the particle has exactly enough speed to escape to infinity and negative if it doesn't have enough to escape--notice how if you set the total energy equal to zero, you get the equation you used in part (a)).





The particle ends with only potential energy (negative since it didn't escape). Set the energies equal and solve for final position.





You may have noticed that the equations call for M, but your problem only gives you g at the surface. No problem. F = GmM/r^2 = mg. Solve that to get g = GM/r^2.





(c) Just do the same thing in reverse. You have initial position and speed (0). You have final position. Set energies equal. Solve for final speed.