Question

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