Question

A satellite is put in a circular orbit of radius $R$. Another satellite is thrown in a circular orbit of radius $1.01R$. The percentage difference between the periodic time of second satellite with respect to the the first will be

• A. $1$ increased
• B. $1$ decreased
• C. $1.5$ increased
• D. $1.5$ decreased

Answer 1

The correct option is C $1.5$ increased

Kepler's $3$rd law, $T^2\propto R^3$

$⟹T=k{R}^{\frac{3}{2}}$ where $k$ is a constant

$\frac{\Delta T}{T}=\frac{3}{2}\frac{\Delta R}{R}$

Now, $\Delta R=.01R$ $⟹\frac{\Delta T}{T}=0.015$

Thus, percentage change $\frac{\Delta T}{T}\times 100=1.5$ increased