A certain object is projected vertically from surface of earth of radius R with a velocity equal to

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7.Gravitation

hard

A certain object is projected vertically from the surface of the earth of radius $R$ with a velocity equal to half the escape velocity. The maximum height attained by the object will be

A

$R/3$

B

$2R$

C

$3R$

D

$6R$

Solution

Let $\mathrm{h}$ be the maximum height reached by the particle. Potential energy on the surface of earth

$=-\frac{\mathrm{GMm}}{\mathrm{R}}$

$PE$ at an altitude $\quad \mathrm{h}=-\frac{\mathrm{GMm}}{\mathrm{R}+\mathrm{h}}$

Escape velocity

$\mathrm{v}=\sqrt{\frac{2 \mathrm{GM}}{\mathrm{R}}}$

The $KE$ of the object is,

$\frac{1}{2} \mathrm{m}\left(\frac{1}{2} \mathrm{v}\right)^{2}=\frac{1}{8} \mathrm{mv}^{2}$

$=\frac{1}{8} \mathrm{m} \times \frac{2 \mathrm{GM}}{\mathrm{R}}=\frac{\mathrm{GMm}}{4 \mathrm{R}}$

By Law of conservation of energy,

$\frac{\mathrm{GMm}}{\mathrm{R}}-\frac{\mathrm{GMm}}{\mathrm{R}+\mathrm{h}} \frac{\mathrm{GMm}}{4 \mathrm{R}}$

Solving, we get; $\mathrm{h}=\frac{\mathrm{R}}{3}$

Standard 11

Physics

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