Projectile Escape Trajectory: Solving for Earth's Radius at Maximum Height

[popo902]

Homework Statement

A projectile is shot directly away from Earth's surface. Neglect the rotation of the Earth. What multiple of Earth's radius RE gives the radial distance (from the Earth's center) the projectile reaches if (a) its initial speed is 0.225 of the escape speed from Earth and (b) its initial kinetic energy is 0.225 of the kinetic energy required to escape Earth? (Give your answers as unitless numbers.)

Homework Equations

The Attempt at a Solution

ok so i have no idea what to do now
i found the escape speed of earth
but what now?
If i had to guess i'd multiply.225 to my KE and the Escape V

so i'd end up with this energy equation
-GMm/R + 1/2m(.225)v^2 = 0
then cancel small m
so

-GM/R + 1/2(.225)V^2 = 0
then solve for...R?
is that right? or am i not on track?

 

[zachzach]

Homework Statement

A projectile is shot directly away from Earth's surface. Neglect the rotation of the Earth. What multiple of Earth's radius RE gives the radial distance (from the Earth's center) the projectile reaches if (a) its initial speed is 0.225 of the escape speed from Earth and (b) its initial kinetic energy is 0.225 of the kinetic energy required to escape Earth? (Give your answers as unitless numbers.)

Homework Equations

The Attempt at a Solution

ok so i have no idea what to do now
i found the escape speed of earth
but what now?
If i had to guess i'd multiply.225 to my KE and the Escape V

so i'd end up with this energy equation
-GMm/R + 1/2m(.225)v^2 = 0
then cancel small m
so

-GM/R + 1/2(.225)V^2 = 0
then solve for...R?
is that right? or am i not on track?

So E_i = \frac{-GMm}{R} + \frac{1}{2}mv_{0}^2

v_{esc} = [\frac{2GM}{R}]^{1/2} \rightarrow v_0 = (0.225)[\frac{2GM}{R}]^{1/2}

E_i = \frac{-GMm}{R} + \frac{1}{2}m[\frac{2GM}{R}(0.051)]

When at its max height, V = 0 so

E_f = \frac{-GMm}{r}

Set equal and use algebra to solve for r.