Question

A particle is fired vertically upward from earth's surface and it goes up to a maximum height of 6400 km. Find the initial speed of particle.

Answer 1

The particle attain maximum height = 6400 km
On earth's surface, its P.E. and K.E.
$E_e=\frac{1}{2}M{V}^2+\left(\frac{-gmM}{r}\right)$
In space, its P.E. and K.E.
$E_3=\left(\frac{-GMm}{R+h}\right)+0E_3=\left(\frac{-GMm}{2R}\right)$
Equation (1) and (2)
$-\frac{GMm}{R}+\frac{1}{2}m{v}^2=-\frac{GMm}{2R}or\left(\frac{1}{2}\right)m{v}^2=GMm(-\frac{1}{2R}+\frac{1}{R})or{v}^2=\frac{GM}{R}=\frac{6.67\times {10}^{-11}\times 6\times {10}^{24}}{6400\times {10}^3}=\frac{40.02+{10}^{13}}{6.4\times {10}^6}=6.2\times {10}^7=0.62\times {10}^{8}OrV=\sqrt{0.62\times {10}^8}=0.79\times {10}^{4}m/s=79km/s$