Question

A planet is revolving in a circular orbit with a time period $T$ around the center of a star solely under the gravity of the star. Suppose the distance between the star and the planet is halved. The individual radii of the star and the planet are also halved, keeping their uniform mass densities unchanged. What will be the time period of the new orbit of the planet?

• A. $T$
• B. $2T$
• C. $\frac{T}{2}$
• D. $\frac{T}{4}$

Hint:

For any gravitational system, if all linear dimensions (orbital radius and object sizes) are scaled by a factor $k$ while densities remain constant, the orbital time scale remains invariant ($T \propto \sqrt{\frac{r^3}{M}} \propto \sqrt{\frac{r^3}{r^3}} = \text{const}$).

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
Kepler's Third Law relates the orbital period to the orbital radius and the mass of the central body.
Key Formula or Approach:
\( T^{2} = \frac{4\pi^{2}r^{3}}{GM} \).
Mass \( M = \rho \times \text{Volume} = \rho \times \frac{4}{3}\pi R^{3} \).

Step 2: Detailed Explanation:
1. Initial Case: Period \( T \) at radius \( r \) around star of mass \( M \).
2. Final Case:
$\bullet$ Orbital radius becomes \( r' = r/2 \).
$\bullet$ Star radius becomes \( R' = R/2 \). Since density \( \rho \) is constant, the new mass is \( M' = \rho \frac{4}{3}\pi (R/2)^{3} = M/8 \).
3. New Period \( T' \):
\[ (T')^{2} \propto \frac{(r/2)^{3}}{M/8} = \frac{r^{3}/8}{M/8} = \frac{r^{3}}{M} \]
Since the ratio \( r^{3}/M \) remains the same, \( (T')^{2} = T^{2} \).

Step 3: Final Answer:
The time period remains unchanged as T.