Question

A uniform thin rod of mass $m$ and length $R$ is placed normally on surface of earth as shown. The mass of earth is $M$ and its radius is $R$. Then the magnitude of gravitational force exerted by earth on the rod is :

• A. $\frac{G M m}{4 R^{2}}$
• B. $\frac{G M m}{2 R^{2}}$
• C. $\frac{4 G M m}{9 R^{2}}$
• D. $\frac{G M m}{8 R^{2}}$

Answer 1

Answer: (B)

Force on element $d x$ on rod is
$d F=\left(\frac{m}{R} d x\right) \frac{G M}{x^{2}}$

Total force on rod is
$F=\int d F=\frac{G M m}{R} \int\limits_{R}^{2 R} \frac{1}{x^{2}} d x$
$=\frac{G M m}{R}\left[-\frac{1}{x}\right]_{R}^{2 R}$
$=\frac{G M m}{2 R^{2}}$