Question

When an object is in a circular orbit of radius $r$, its time period of revolution about the earth is $T$ and orbital velocity is $v$ when its orbit is $r+$$\Delta$r. Find the change in time period $\Delta$T and orbital velocity $\Delta$v.

• A. $\frac{3\pi \Delta r}{v},\frac{\pi \Delta r}{T}$
• B. $\frac{2\pi \Delta r}{v},\frac{\pi \Delta r}{2T}$
• C. $\frac{2\pi \Delta r}{3v},\frac{2\pi \Delta r}{T}$
• D. none

Answer 1

The correct option is A $\frac{3\pi \Delta r}{v},\frac{\pi \Delta r}{T}$

$v_0=\sqrt{\frac{GM}{r}},v_0-\Delta v=\sqrt{\frac{GM}{r+\Delta r}}$

$=\sqrt{\frac{GM}{r}}{(1+\frac{\Delta r}{r})}^{-1/2}or\Delta v=\sqrt{\frac{GM}{r}}\frac{\Delta r}{2r}$

$=v_0\frac{\Delta r}{2r}$

$=\frac{2\pi r}{T}\frac{\Delta r}{2r}=\frac{\pi \Delta r}{T}$

$T=\frac{2\pi r^{3/2}}{\sqrt{GM}}$, and $T^{\prime }=\frac{2\pi (r+\Delta r{)}^{3/2}}{\sqrt{GM}}$

$=\frac{2\pi r^{3/2}}{\sqrt{GM}}{(1+\frac{\Delta r}{r})}^{3/2}$

or $T^{\prime }=T(1+\frac{3}{2}\frac{\Delta r}{r})$

$\Delta T=\frac{3T\Delta r}{2r}=\frac{3}{2}\frac{\left(2\pi r\right)}{v}\frac{\Delta r}{r}=\frac{3\pi \Delta r}{v}$