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7.Gravitation
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If an object is projected vertically upwards with speed, half the escape speed of earth, then the maximum height attained by it is [ $R$ is radius of earth]
A
$R$
B
$\frac{R}{2}$
C
$2 R$
D
$\frac{R}{3}$
Solution
(d)
$V_\theta=\sqrt{\frac{2 G M}{R}}$
$M \rightarrow \text { mass of earth }$
$R \rightarrow \text { Radius of earth }$
Now, conserving potential energy at the surface of earth and highest point,
$-\frac{G M m}{R}+\frac{1}{2} m\left(\frac{1}{2} \sqrt{\frac{2 G M}{R}}\right)^2=-\frac{G M m}{r}$
$-\frac{G M m}{R}+\frac{G M m}{4 R}=-\frac{G M m}{r}$
$-\frac{3 G M m}{4 R}=-\frac{G M m}{r}$
$\Rightarrow r=\frac{4 R}{3}$
$\Rightarrow R+h=\frac{4 R}{3}$
$\Rightarrow h=\left(\frac{R}{3}\right)$
Standard 11
Physics
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