Gravitation: NEET previous year questions

The work done equals the change in gravitational potential energy: $W = U_2 - U_1 = \left(-\frac{GMm}{2R}\right) - \left(-\frac{GMm}{R}\right) = \frac{GMm}{2R}$. Using $g = \frac{GM}{R^2}$, this becomes $\frac{mgR}{2}$, option (4). It is not the naive $mgR$ because gravity weakens with height, so less work is needed than if g stayed constant.

Escape speed is $v_e = \sqrt{\frac{2GM}{R}}$. With the masses equal, $v_e \propto R^{-1/2}$, so $\frac{v_2}{v_1} = \sqrt{\frac{R_1}{R_2}} = \sqrt{2}$. The denser, smaller planet is harder to escape from, which is the physical sense check. NCERT Class 11, Gravitation (escape speed).

Kepler's third law states $T^2 \propto R^3$ for orbits about a common central mass, so $T \propto R^{3/2}$. It also follows directly from equating gravitational and centripetal force: $\frac{GMm}{R^2} = m\omega^2 R$ with $\omega = 2\pi/T$. NCERT Class 11, Gravitation (Kepler's laws).

The gravitational force at a height h from the Earth's surface is given by $F = \frac{mgR^2}{(R + h)^2}$. For $h = \frac{R}{2}$, $F = \frac{mgR^2}{(R + \frac{R}{2})^2} = \frac{mgR^2}{\left(\frac{3R}{2}\right)^2} = \frac{4}{9} \times mg$. Since $mg = 72\ N$, $F = \frac{4}{9} \times 72 = 32\ N$. Chapter: Gravitation.