Question

What is the escape speed on a spherical asteroid whose radius is 387 km and whose gravitational acceleration at the surface is 0.595 m/s²? (b) How far from the surface will a particle go if it leaves the asteroid's surface with a radial speed of 568 m/s? (c) With what speed will an object hit the asteroid if it is dropped from 2376 km above the surface?

Answer 1

A) escape speed is Ve = sqrt(2*g*R) = sqrt(2*0.595*387000) = 678.62 m/s

B) From v^2-u^2 = 2*g*h

here v is final speed = 0 m/s

u is initial speed = 568 m/s

g is acclaration due to gravity on surface of earth = 0.595 m/s^2

h is the required distance

then h = u^2/(2*g) = (568*568)/(2*0.595) = 271112.6 m = 271.11 km

C) From v^2-u^2 = 2*g*h

since it was dropped initial speed is u = 0 m/s

v = sqrt(2*g*h)
v = sqrt(2*0.595*2376000) = 1681.5 m/s