Question

A spaceship is sent to investigate a planet of mass $M$ and radius $R$. While hanging motionless in space at a distance $5R$ from the centre of the planet, the spaceship fires an instrument package of mass $m$, which is much smaller than the mass of the spaceship. The angle $\theta$ for which the package just grazes the surface of the planet is

• A. ${\sin }^{-1}\left(\frac{1}{2}\sqrt{1+\frac{2GM}{5v_0^{2}R}}\right)$
• B. ${\sin }^{-1}\left(\frac{1}{5}\sqrt{1+\frac{8GM}{5v_0^{2}R}}\right)$
• C. ${\sin }^{-1}\left(\frac{1}{8}\sqrt{1+\frac{3GM}{5v_0^{2}R}}\right)$
• D. ${\sin }^{-1}\left(\frac{1}{3}\sqrt{1+\frac{6GM}{5v_0^{2}R}}\right)$

Answer 1

The correct option is B ${\sin }^{-1}\left(\frac{1}{5}\sqrt{1+\frac{8GM}{5v_0^{2}R}}\right)$

Since no external torque is present around the centre of the planet. angular momentum about the centre will be conserved.
$\Rightarrow m{V}_o\sin \theta \left(5R\right)=mVR$
$\Rightarrow V=5V_{o}sin\theta .............\left(i\right)$
Energy conservation (potential+kinetic) of the system
$\frac{1}{2}m{V}_o^2-\frac{GMm}{5R}=\frac{1}{2}m{V}^2-\frac{GMm}{R}$
Substituting $V$ from eq. $\left(i\right)$
$\Rightarrow \frac{V_o^2}{2}-\frac{GM}{5R}=\frac{25V_o^2{\sin }^2\theta }{2}-\frac{GM}{R}$

$\Rightarrow \frac{25{\sin }^2\theta -1}{2}{V}_o^2=\frac{4GM}{5R}$

$\Rightarrow \sin \theta =\frac{1}{5}\sqrt{1+\frac{8GM}{5V_o^{2}R}}$

or $\Rightarrow \theta ={\sin }^{-1}\left(\frac{1}{5}\sqrt{1+\frac{8GM}{5V_o^{2}R}}\right)$
Thus (B) is the correct option.