Question

If due to air drag, the orbital radius of the earth decreases from $R$ to $R$ $-\Delta R$, $\Delta R<<R$, then the expression for increase in the orbital velocity $\Delta v$ is

• A. $\frac{\Delta r}{2}\sqrt{\frac{GM}{R^3}}$
• B. $-\frac{\Delta R}{2}\sqrt{\frac{GM}{R^3}}$
• C. $\Delta R\sqrt{\frac{GM}{R^3}}$
• D. $-\Delta R\sqrt{\frac{GM}{R^3}}$

Answer 1

Answer: (A)

$\frac{\Delta r}{2}\sqrt{\frac{GM}{R^3}}$

Orbital velocity, $v=\sqrt{\frac{GM}{R}}$

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

$v^{\prime }-v=\sqrt{\frac{GM}{R-\Delta R}}-\sqrt{\frac{GM}{R}}=\sqrt{GM}[(R-\Delta R{)}^{-1/2}-(R{)}^{-1/2}]$

$v^{\prime }-v=\sqrt{GM}\left(R{)}^{-1/2}[\right(1-\frac{\Delta R}{R}{)}^{-1/2}-\left(1\right)]$

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

$v^{\prime }-v=\sqrt{GM}\left(R{)}^{-1/2}[\right(1+\frac{\Delta R}{2R})-(1)]$

$v^{\prime }-v=\frac{\Delta R}{2}\sqrt{\frac{GM}{R^3}}$