Question:
A planet \(P_1\) is moving around a star of mass \(2M\) in an orbit of radius \(R\). Another planet \(P_2\) is moving around another star of mass \(4M\) in an orbit of radius \(2R\). The ratio of time periods of revolution of \(P_2\) and \(P_1\) is:
A planet \(P_1\) is moving around a star of mass \(2M\) in an orbit of radius \(R\). Another planet \(P_2\) is moving around another star of mass \(4M\) in an orbit of radius \(2R\). The ratio of time periods of revolution of \(P_2\) and \(P_1\) is:
Updated On: Apr 10, 2026
- \( \frac{1}{2} \)
- \(2\)
- \(4\)
- \( \frac{1}{4} \)
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The Correct Option is B
Solution and Explanation
Concept:
According to Kepler’s third law,
\[
T^2 \propto \frac{r^3}{M}
\]
where \(T\) is the time period, \(r\) is orbital radius and \(M\) is the mass of the central body.
Step 1:Write the proportional relation} \[ T \propto \frac{r^{3/2}}{\sqrt{M}} \]
Step 2:Compute ratio} For \(P_1\): \[ T_1 \propto \frac{R^{3/2}}{\sqrt{2M}} \] For \(P_2\): \[ T_2 \propto \frac{(2R)^{3/2}}{\sqrt{4M}} \]
Step 3:Take the ratio} \[ \frac{T_2}{T_1} = \frac{(2R)^{3/2}}{\sqrt{4M}} \cdot \frac{\sqrt{2M}}{R^{3/2}} \] \[ = \frac{2^{3/2}}{2} \cdot \sqrt{2} \] \[ = \frac{2\sqrt2}{2}\cdot\sqrt2 \] \[ =2 \] Thus \[ \boxed{\frac{T_2}{T_1}=2} \]
Step 1:Write the proportional relation} \[ T \propto \frac{r^{3/2}}{\sqrt{M}} \]
Step 2:Compute ratio} For \(P_1\): \[ T_1 \propto \frac{R^{3/2}}{\sqrt{2M}} \] For \(P_2\): \[ T_2 \propto \frac{(2R)^{3/2}}{\sqrt{4M}} \]
Step 3:Take the ratio} \[ \frac{T_2}{T_1} = \frac{(2R)^{3/2}}{\sqrt{4M}} \cdot \frac{\sqrt{2M}}{R^{3/2}} \] \[ = \frac{2^{3/2}}{2} \cdot \sqrt{2} \] \[ = \frac{2\sqrt2}{2}\cdot\sqrt2 \] \[ =2 \] Thus \[ \boxed{\frac{T_2}{T_1}=2} \]
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