Question

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 0.421 of the escape speed from Earth and (b) its initial kinetic energy is 0.421 of the kinetic energy required to escape Earth? (Give your answers as unitless numbers.) (c) What is the least initial mechanical energy required at launch if the projectile is to escape Earth? (a) Number Enter your answer for part (a) in accordance to the question statement 1.215 Units Choose the answer for part (a) from the menu in accordance to the question statement

Answer 1

(a) 5.65 times the Earth's radius

The escape velocity for a projectile on Earth is

$v_e=\sqrt{\frac{2GM}{R}}$

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,

$v=0.421 v_e$

So its initial kinetic energy will be

$K=\frac{1}{2}m(0.421 v)^2=0.089 m(\sqrt{\frac{2GM}{R}})^2=0.177 \frac{GMm}{R}$

where m is the mass of the projectile

At the point of maximum altitude, all this energy is converted into gravitational potential energy:

$K=U\\0.177 \frac{GMm}{R}=\frac{GMm}{r}$

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:$0.177 \frac{GMm}{R}=\frac{GMm}{nR}$

And solving the equation we find

$n=\frac{1}{0.177}=5.65$

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:

$K=\frac{1}{2}mv_e^2 = \frac{1}{2}m(\sqrt{\frac{2GM}{R}})^2=\frac{GMm}{R}$

This time, the projectile has 0.421 times this energy:

$K=0.421 \frac{GMm}{R}$

Again, at the point of maximum altitude, all this energy will be converted into potential energy:

$0.421 \frac{GMm}{R}=\frac{GMm}{nR}$

and by solving for n we find

$n=\frac{1}{0.421}=2.36$

So, the projectile reaches a radial distance of 2.36 times the Earth's radius.

c) $E=U=\frac{GMm}{R}$

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:

$E=U=\frac{GMm}{R}$

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

$K_e=\frac{1}{2}mv_e^2 = \frac{1}{2}m(\sqrt{\frac{2GM}{R}})^2=\frac{GMm}{R}$