Question

A satellite revolves around a planet in a circular orbit of radius R with the time period t. If the orbital radius is 3R, then its time period of revolution is

• A. t
• B. 2t
• C. $\sqrt{27}t$
• D. $\sqrt{3}t$
• E. $2\sqrt{2}t$

Hint:

$\sqrt{27}$ is the same as $3\sqrt{3}$, which is approximately $5.2$. This tells you that tripling the distance doesn't just triple the time—it increases the time period by more than 5 times!

Answer 1

The Correct Option is C

Concept:
Physics - Kepler's Third Law of Planetary Motion.
The square of the time period ($T$) of a satellite is directly proportional to the cube of the radius ($R$) of its orbit: $T^2 \propto R^3$.
Step 1: Establish the ratio from Kepler's Law.
$$ \left(\frac{T_2}{T_1}\right)^2 = \left(\frac{R_2}{R_1}\right)^3 $$
Step 2: Substitute the given values.

• $R_1 = R, \quad T_1 = t$
• $R_2 = 3R, \quad T_2 = ?$

$$ \left(\frac{T_2}{t}\right)^2 = \left(\frac{3R}{R}\right)^3 $$
Step 3: Simplify the right side.
$$ \left(\frac{T_2}{t}\right)^2 = (3)^3 = 27 $$
Step 4: Solve for $T_2$.
Take the square root of both sides: $$ \frac{T_2}{t} = \sqrt{27} $$ $$ T_2 = \sqrt{27}t $$ The new time period is $\sqrt{27}t$.