Eris, the largest dwarf planet known in the Solar System, has a radius R = 1200 km and an acceleration due to gravity on its surface of magnitude g = 0.77 m/s2.

Question

(a) Use these numbers to calculate the escape speed from the surface of Eris.
(b) If an object is fired directly upward from the surface of Eris with half of this escape speed, to what maximum height above the surface will the object rise? (Assume that Eris has no atmosphere and negligible rotation.)

I know the speed is 1360 m/s but how do you solve question b? The answer is 400 km. Thank you in advance.

Answer 1

A)
Escape vel. = Ue = sqrt(2GM/R) ...(1)
also g = GM/R^2 So GM/R = gR .....(2)

From(1) & (2)
Ue = sqrt(2gR) = sqrt(2*0.77*1200,000)
= 1360 m/s
B)
u = Ue/2 = (1/2)*sqrt(2GM/R)
Now KEi+PEi = PEf+KEf (conservation of mechanical energy)
or (1/2)mu^2 - GMm/R = -GMm/(R+h)+0
or GM/4R - GM/R = -GM/(R+h)
or 3/4R = 1/(R+h)
or h = R/3 = 400 Km

Answer 2

I'm having trouble following this. How do you go from:
GM/4R - GM/R = -GM/(R + h) to 3/4R = 1/(R+h) to
h=R/3?