Question

Suppose radius of the moons orbit around the earth is doubled. Its period around the earth will become :

• A. $1/2$ times
• B. $\sqrt{2}$ times
• C. $2^{2/3}$ times
• D. $2^{3/2}$ times

Answer 1

Answer: (D)

According to Keplers law $T^2\propto R^3$
(Here : R is orbital radius and T is time period)
Now according to question when orbital radius is doubled, then period will be
$\frac{T_1}{T_2}={\left(\frac{R}{2R}\right)}^{3/2}$
$T_2=2^{3/2}{T}_1$
(Here $:R_1=R,R_2=2R)$
Hence, the time period will become $2^{3/2}$ times.