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 0.351 of the escape speed from Earth and
(b) its initial kinetic energy is 0.351 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?

Answer 1

Answer:

Explanation:

We shall apply law of  conservation of mechanical energy for projectile being thrown .

Total energy on the surface = total energy at height h required

a ) At height h , velocity = .351 x ( 2 GM/R x h )

$\frac{-GM}{R} + \frac{m\times(.351\times\sqrt{2GM})^2 }{2R } = \frac{-GMm}{R+h} + 0$

$\frac{-GMm}{R} +\frac{1}{2}\times  \frac{-2GMm}{R} \times0.123=\frac{-GMm}{R+h}$

$\frac{0.877GMm}{R} =\frac{-GMm}{R+h}$

h = .14 R

b )

$\frac{-GM}{R} + \frac{m\times(.351\times2GM) }{2R } = \frac{-GMm}{R+h} + 0$

$\frac{-0.649GMm}{R} = \frac{-GMm}{R+h}$

h = .54 R

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

= GMm / R + 1/2 m (2GM/R)

= 0