Ratio of max height of a projectile to planet radius

[yankans]

Homework Statement

Hint:Disregard any dissipative effects of the atmosphere of the planet.
A projectile is launched from the surface of a planet (mass M, radius R)
with a launch speed equal to 60 percent of the escape speed for that planet.
The projectile will rise to a maximum height and fall back to the surface of
the planet.

What will be the ratio of its maximum height above the surface to the radius of the
planet, h/R?

Homework Equations

escape speed:
KE = E(gravity)
1/2 mv^2 = (GMm)/R
F/m = a
a(gravity) = GM/R^2

The Attempt at a Solution

1/2 mv^2 = (GMm)/R
v(escape) = (2GMm)/R
v=sqrt((2GM)/R)
vi = 0.6 sqrt((2GM)/R)

vf^2 = vi^2 - 2gh
0 = 0.72GM/R - 2(GM/R^2)h
2(GM/R^2)h = 0.72GM/R
h/R = 0.36

 

[yankans]

Sorry I reworked this problem by myself and I solved it!

 

[baba89]

The ratio of the maximum height of the projectile to the radius of the planet is 0.36. This means that the maximum height reached by the projectile is approximately one third of the radius of the planet. This ratio is independent of the mass of the planet and only depends on the escape velocity and the acceleration due to gravity at the surface. Therefore, for any planet with the same escape velocity and surface gravity, the ratio of maximum height to planet radius will be the same. This ratio can also be used to determine the maximum height of a projectile launched from the surface of any planet, as long as the atmospheric effects are disregarded.