Question

A particle is thrown vertically upwards with velocity 11.2kms${}^{-1}$ from the surface of earth. Calculate its velocity at height 3R, where R is the radius of earth.

• A. $\simeq$ 9.25 kms${}^{-1}$
• B. $\simeq$ 5.6 kms${}^{-1}$
• C. $\simeq$ 11.2 kms${}^{-1}$
• D. $\simeq$ 4.3 kms${}^{-1}$

Answer 1

The correct option is B $\simeq$ 5.6 kms${}^{-1}$

The given velocity equals escape velocity.
So, $\frac{GMm}{R}=$ initial $K.E$
$v_{escape}=\sqrt{\frac{2GM}{R}}$

Initial total energy $=K.E+P.E=-\frac{GMm}{R}+\frac{GMm}{R}=0$

So, at height $3R$ which is equal to distance $4R$ from the centre of the earth, total energy will be zero.

Hence,
$0=-\frac{GMm}{4R}+\frac{1}{2}m{v}^2$

$\Rightarrow v=\sqrt{\frac{GM}{2R}}=\frac{v_{escape}}{2}$

On solving, we get $v=5.6m/s$