Question

If due to air drag, the orbital radius of the earth decreases from $R$ to $R$ $-\Delta R$, $\Delta R<<R$, the change in $KE$ , $\Delta K$ is

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

Answer 1

The correct option is C $\frac{GM}{{2R}^2}\times \Delta R$

Orbital velocity, $v=\sqrt{\frac{GM}{R}}$ and $K.E=\frac{1}{2}m{v}^2$

Now for $R^{\prime }=R-\Delta R$, new orbital velocity, $v^{\prime }=\sqrt{\frac{GM}{R-\Delta R}}$

Therefore $K.E^{\prime }=\frac{1}{2}m\frac{GM}{R-\Delta R}$

$\Delta K=K.E^{\prime }-K.E=\frac{1}{2}m[-\frac{GM}{R}+\frac{GM}{R-\Delta R}]=\frac{1}{2}mGM[(R-\Delta R{)}^{-1}-(R{)}^{-1}]$

$\Delta K=\frac{1}{2}mGM\left(R{)}^{-1}[\right(1-\frac{\Delta R}{R}{)}^{-1}-\left(1\right)]$

Using bionomial expansion, $(1+x{)}^n=1+nx$ for $x<<1$

$\Delta K=\frac{1}{2}GMm\left(R{)}^{-1}[\right(1+\frac{\Delta R}{R})-(1)]$

$\Delta K=\frac{1}{2}m\frac{GM}{R^2}\Delta R$