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 $PE$, $\Delta U$ 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 A $-\frac{GMm}{R^2}\times \Delta R$

$P.E=-\frac{GMm}{R}$

Now for $R^{\prime }=R-\Delta R$

Therefore $P.E^{\prime }=-\frac{GMm}{R-\Delta R}$

$\Delta U=P.E^{\prime }-P.E=[\frac{GMm}{R}-\frac{GMm}{R-\Delta R}]=-GMm[(R-\Delta R{)}^{-1}-(R{)}^{-1}]$

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

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

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

$\Delta U=-\frac{GMm}{R^2}\Delta R$