(a) 5.65 times the Earth's radius
The escape velocity for a projectile on Earth is
[tex]v_e=\sqrt{\frac{2GM}{R}}[/tex]
where
G is the gravitational constant
M is the Earth's mass
R is the Earth's radius
If the projectile has an initial speed of 0.421 escape speed,
[tex]v=0.421 v_e[/tex]
So its initial kinetic energy will be
[tex]K=\frac{1}{2}m(0.421 v)^2=0.089 m(\sqrt{\frac{2GM}{R}})^2=0.177 \frac{GMm}{R} [/tex]
where m is the mass of the projectile
At the point of maximum altitude, all this energy is converted into gravitational potential energy:
[tex]K=U\\0.177 \frac{GMm}{R}=\frac{GMm}{r}[/tex]
where r is the distance from the Earth's centre reached by the projectile. We can write r as a multiple of R, the Earth's radius:[tex]0.177 \frac{GMm}{R}=\frac{GMm}{nR}[/tex]
And solving the equation we find
[tex]n=\frac{1}{0.177}=5.65[/tex]
So, the projectile reaches a radial distance of 5.65 times the Earth's radius.
b) 2.36 times the Earth's radius
The kinetic energy needed to escape is:
[tex]K=\frac{1}{2}mv_e^2 = \frac{1}{2}m(\sqrt{\frac{2GM}{R}})^2=\frac{GMm}{R}[/tex]
This time, the projectile has 0.421 times this energy:
[tex]K=0.421 \frac{GMm}{R}[/tex]
Again, at the point of maximum altitude, all this energy will be converted into potential energy:
[tex]0.421 \frac{GMm}{R}=\frac{GMm}{nR}[/tex]
and by solving for n we find
[tex]n=\frac{1}{0.421}=2.36[/tex]
So, the projectile reaches a radial distance of 2.36 times the Earth's radius.
c) [tex]E=U=\frac{GMm}{R}[/tex]
The least initial mechanical energy needed for the projectile to escape Earth is equal to the gravitational potential energy of the projectile at the Earth's surface:
[tex]E=U=\frac{GMm}{R}[/tex]
Indeed, the kinetic energy of the projectile must be equal to this value. In fact, if we use the formula of the escape velocity inside the formula of the kinetic energy, we find
[tex]K_e=\frac{1}{2}mv_e^2 = \frac{1}{2}m(\sqrt{\frac{2GM}{R}})^2=\frac{GMm}{R}[/tex]
