Question

In a solar system, the time period of revolution of a planet tracing a circular orbit of radius \(R\) is proportional to:

• A. \(R^3\)
• B. \(R^{1/2}\)
• C. \(R^{3/2}\)
• D. \(R^2\)

Hint:

Remember Kepler's Third Law: \[ T^2\propto R^3 \] For quick questions directly write \[ T\propto R^{3/2} \]

Answer 1

The Correct Option is C

Step 1: Set up the orbiting planet.
For a planet on a circular orbit of radius $R$, gravity supplies the centripetal force.
Step 2: Write the force balance.
$\dfrac{GMm}{R^2} = \dfrac{mv^2}{R}$.
Step 3: Bring in the time period.
The orbital speed is $v = \dfrac{2\pi R}{T}$. Substituting gives $\dfrac{GM}{R^2} = \dfrac{4\pi^2 R}{T^2}$.
Step 4: Solve for T squared.
Rearranging, $T^2 = \dfrac{4\pi^2 R^3}{GM}$.
Step 5: Read off the proportionality.
So $T^2 \propto R^3$, which means $T \propto R^{3/2}$.
Step 6: Choose the option.
This is Kepler's third law and matches option C.
\[ \boxed{ T \propto R^{3/2} } \]