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 $1.01R$. The time period of the second satellite is larger than that of the first one by approximately

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

Answer 1

The correct option is B $1.5\%$

According to Kepler's 3rd law of planetary motion, the time period of revolution $T$ and radius of circular orbit $R$ are related as:$$T^2\propto R^3$$So, we can write

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

Substituting the value of $R_1$ and $R_2$,

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

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

Since, $0.01<<1$, we can use binomial approximation $(1+x{)}^n=1+nx,$ if $x<<1$

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

$\Rightarrow \frac{T_2}{T_1}=1.015$

$\Rightarrow \frac{T_2-T_1}{T_1}\times 100=(1.015-1)\times 100=1.5\%$

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

Hence, option (b) is the correct answer.