Question

A satellite is launched into a circular orbit of radius $R$ around the earth. A second satellite is launched into an orbit of radius $1.01R$. By how much approximately the time period of the second satellite is larger than that of the first one?

• A. $0.5\%$
• B. $1.5\%$
• C. $1\%$
• D. $3\%$

Answer 1

The correct option is B $1.5\%$

The time period of revolution of a satellite and radius of circular orbit is given by Kepler’s law of time periods, which is

$T^2\propto R^3$

$\Rightarrow \frac{T_2}{T_1}={\left(\frac{R_2}{R_1}\right)}^{3/2}$

$\Rightarrow \frac{T_2}{T_1}={\left(\frac{1.01R}{R}\right)}^{3/2}$

$\Rightarrow \frac{T_2}{T_1}=(1+0.01{)}^{3/2}$

Since, $0.01<<1$, we can use binomial approximation

$(1+x{)}^n=1+nx;$ if $x<<1$

$\therefore \frac{T_2}{T_1}=1+(\frac{3}{2}\times 0.01)$

$\Rightarrow \frac{T_2}{T_1}=\left(1.015\right)$

Hence, $T_2$ is greater than $T_1$ by a factor of $0.015$ or $1.5\%$.

Hence, option $\left(b\right)$ is correct.