A projectile is shot directly away from Earth\'s surface. Neglect the rotation o

Question

A projectile is shot directly away from Earth's surface. Neglect the rotation of Earth.

(a) As a multiple of Earth's radius RE, what is the radial distance a projectile reaches if its initial speed is two-fifths of the escape speed from Earth?

(b) As a multiple of Earth's radius RE, what is the radial distance a projectile reaches if its initial kinetic energy is three-fifths of the kinetic energy required to escape Earth?

(c) What is the least initial mechanical energy required at launch if the projectile is to escape Earth?

Explanation / Answer

Using energy conservation, we have
KEi + Ui = KEf + Uf
1/2 mv2 - GMm/RE = 0 - GMm/r

v =2/5 2 GME / RE
1/2 m(2/5 2 GME / RE)2 - GMm/RE = 0 - GMm/r
GMm/RE ( ( -1 + 0.32) = - GMm/r

-0.68 GMm/R= - GMm/r

r = 1.47 R

(b)

energy conservation, we have
KEi + Ui = KEf + Uf
1/2 mv2 - GMm/RE = 0 - GMm/r

(3/5) 1/2 mv2 - GMm/RE = 0 - GMm/r

3/10 mv^2 -GMm/RE = 0 - GMm/r

3 m/10 ( 2 GME / RE)^2 -GMm/RE = 0 - GMm/r

6 GMm/10RE - GMm/RE = 0 - GMm/r

6/10 RE - 1/RE = 1/r

5/10 RE = 1/r

r = 2RE

(c)

When the projectile reaches infinity, it stops and thus has no kinetic energy.
It also has no gravitational potential energy at infinity. Its total mechanical energy at infinity
is therefore ZERO.

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