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 0.520 of the escape speed from Earth and (b) its initial kinetic energy is 0.520 of the kinetic energy required to escape Earth

Answer 1

Answer:

a)$r_1=1.36R$

b)$r_2=2.083R$

Explanation:

Given:

a) when the initial velocity of the projectile is 0.520 times the escape velocity from the earth.

Let r be the radial distance from the earth's surface Let M be the mass of the Earth and R be the radius of the Earth

Now using conservation of Energy at earths surface and at distance r we have

$\dfrac{-GMm}{R}+\dfrac{m(0.52V_e)^2}{2}=\dfrac{-GMm}{r_1}\\\dfrac{-GMm}{R}+\dfrac{m\times 0.52^2\times \dfrac{2GM}{R}}{2}=\dfrac{-GMm}{r_1}\\r_2=1.36\ R$

b) when the Initial kinetic Energy of the projectile is 0.52 times the Kinetic Energy required to escape the Earth

Conservation of Energy we have

$\dfrac{-GMm}{R}+0.52\times KE_{escape}=\dfrac{-GMm}{r_2}\\\dfrac{-GMm}{R}+0.52\times\dfrac{m\times \dfrac{2GM}{R}}{2}=\dfrac{-GMm}{r_2}\\r_2=2.083\ R$