Question

If due to air drag, the orbital radius of the earth decreases from $R$ to $R$ $-\Delta R$, $\Delta R<<R$, the work done by air friction force $W$ is

• A. $-\frac{GMm}{R^2}\times \Delta R$
• B. $\frac{GMm}{R^2}\times \Delta R$
• C. $\frac{GMm}{{2R}^2}\times \Delta R$
• D. $-\frac{GMm}{{2R}^2}\times \Delta R$

Answer 1

The correct option is D $-\frac{GMm}{{2R}^2}\times \Delta R$

As $\Delta U=-\frac{GMm}{R^2}\Delta R$ and $\Delta K.E=\frac{GMm}{2R^2}\Delta R$ (solved earlier)
Work-energy theorem, $W_{gravitational}+W_{friction}=\Delta K.E$

$-\Delta U+W_{friction}=\Delta K.E$

$\frac{GMm}{R^2}\Delta R+W_{friction}=\frac{GMm}{2R^2}\Delta R$

$⟹W_{friction}=-\frac{GMm}{2R^2}\Delta R$