The international space center is 1500km above the earths surface. Determine the escape velocity of a rocket launched into space at this altitude if the mass of the earth is 6*10^24kg and the radius is 6.4*10^6m.

Answer 1

(1/2) m v^2 = difference in potential energy between rocket at infinity and at r =( 6.4*10^6 + 1.5*10^6) meters = G m M/r
where G = 6.67*10^-11
v^2 = 2 (6.67*10^-11)(6*10^24)/7.9*10^6