Question

The escape velocity of projection from the earth is approximately (R = 6400 km) ​

Answer 1

$\sf{V}_{escape} = \sqrt{ \dfrac{2GM}{R} }$

$\sf{=\sqrt{ \dfrac{2(6.64 \times {10}^{ - 11} \times 5.98 \times {10}^{24)} }{6.38 \times {10}^{6} } {ms}^{ - 1} }}$

$\sf{=\sqrt{1.251 \times {10}^{8} \: {ms}^{ - 1} }}$

$\sf{= 11184 \: {ms}^{ - 1}}$

$\sf{≈11.2K{ms}^{-1}}$