A particle is fired vertically upwards from the surface of earth and reaches a height 6400 km. The initial velocity of the particle is ($R = 6400\ km,\ g = 10ms^{- 2}$)

Question

Dr. Abubacker · Accepted Answer

(b) If a body is projected from the surface of earth with a velocity $v$ and reaches a height $h$, applying conservation of energy (relative to surface of earth)
$\frac{1}{2}mv^{2} = \frac{mgh}{\lbrack 1 + (h/R)brack}$
$h = R = 6400$ km, $g = 10ms^{- 2}$
So, $v^{2} = gh\ \ ie,\ v = \sqrt{10 	imes 6400 	imes 10^{3}} = 8\ kms^{- 1}$

Q. A particle is fired vertically upwards from the surface of earth and reaches a height 6400 km. The initial velocity of the particle is ($R = 6400\ km,\ g = 10ms^{- 2}$)

Explanation

Reviewed by Dr. Abubacker

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