- Home
- Standard 11
- Physics
An object is propelled vertically to a maximum height of $4 R$ from the surface of a planet of radius $R$ and mass $M$. The speed of object when it returns to the surface of the planet is
$2 \sqrt{\frac{2 G M}{5 R}}$
$\sqrt{\frac{G M}{2 R}}$
$\sqrt{\frac{3 G M}{2 R}}$
$\sqrt{\frac{G M}{5 R}}$
Solution
(a)
Let $v=$ velocity of object on reaching surface of planet.
Applying energy conservation between points $A$ and $B$, we have
Total energy at $A=$ Total energy at $B$
$\Rightarrow$ Potential encrgy at separation
$(4 R+R)=$ Potential energy at separation
$R+$ Kinctic energy at surface
$\Rightarrow \quad \frac{-G M m}{5 R}=\frac{-G M m}{R}+\frac{1}{2} m v^2$
where, $m=$ mass of object.
$\Rightarrow \frac{G M m}{R}(1-1)-\frac{1}{5}{ }^5 v^2 \Rightarrow \frac{8 M}{5 R}=v^2$
$\therefore$ Velocity at surface, $v=2 \sqrt{\frac{2}{5} \frac{G M}{R}}$
