Question

With what velocity must a body be thrown from the earth's surface so that it may reach a height 4Re above
th's surface ? (Radius of the Earth Re = 6400 km, g=9.8 m/s2)​

Answer 1

Answer:

Explanation:

correct answer

Answer 2

Answer:

10 km/s

Explanation:

By using Conservation of mechanical energy-:

$= > \frac{1}{2} m_{0} {v}^{2} - \frac{GMm_{0}}{R_{e}} = 0 - \frac{GMm_{0}}{(R_{e} + 4R_{e})} \\ = > > \frac{1}{2} m_{0} {v}^{2} = - \frac{GMm_{0}}{ 5R_{e}} + \frac{GMm_{0}}{R_{e}} \\ = > \frac{1}{2} m_{0} {v}^{2} = \frac{4}{5} \frac{GMm_{0}}{R_{e}} \\ = > {v}^{2} = \frac{8}{5} \frac{GM}{R_{e}} \\ = > {v}^{2} = \frac{4}{5} \frac{gR_{e}}{R_{e} } \\ = > {v}^{2} = \frac{8}{5} \times 6400 \times {10}^{3} \\ = > {v}^{2} = \frac{8}{ \cancel5} \times 9.8 \times \cancel6400 \times {10}^{3} \\ = > {v}^{2} = 8 \times 9.8 \times 1280 \times {10}^{3} \\ = > {v}^{2} = {10}^{8} approx \\ = > \boxed{v = 10 \: km/s}$