Question

The time period of a satellite in a circular orbit of radius R is T. The period of another satellite in a circular orbit of radius 9R is :

• A. 3 T
• B. 9 T
• C. 27 T
• D. 12 T

Hint:

For Kepler's law calculations: $(n^2)^{3/2} = n^3$. Here $9 = 3^2$, so the answer is $3^3 = 27$.

Answer 1

The Correct Option is C

Step 1: By Kepler's Third Law, $T^2 \propto R^3$.
Step 2: $\frac{T_2}{T_1} = (\frac{R_2}{R_1})^{3/2} = (\frac{9R}{R})^{3/2}$.
Step 3: $T_2 = T \times (9)^{3/2} = T \times (3^2)^{3/2} = T \times 3^3 = 27T$.