Question

Two planets \(P_1\) and \(P_2\) having masses \(M_1\) and \(M_2\) revolve around the Sun in elliptical orbits, with time periods \(T_1\) and \(T_2\), respectively. The minimum and maximum distances of planet \(P_1\) from the Sun are \(R\) and \(3R\), respectively, whereas for planet \(P_2\), these are \(2R\) and \(4R\), respectively, where \(R\) is a constant. Assuming \(M_1\) and \(M_2\) are much smaller than the mass of the Sun, the magnitude of \(\dfrac{T_2}{T_1}\) is:

• A. \(\dfrac{2}{3}\sqrt{\dfrac{2M_1}{3M_2}}\)
• B. \(\dfrac{3}{2}\sqrt{\dfrac{3M_2}{2M_1}}\)
• C. \(\dfrac{3}{2}\sqrt{\dfrac{3}{2}}\)
• D. \(\dfrac{2}{3}\sqrt{\dfrac{2}{3}}\)

Hint:

Kepler's third law states that \(T^2 \propto a^3\), independent of the planet's mass when the Sun's mass is dominant.

Answer 1

The Correct Option is C

Step 1: Apply Kepler's third law.
For any planet revolving around the Sun, \[ T^2 \propto a^3 \] where \(a\) is the semi-major axis of the elliptical orbit.

Step 2: Determine semi-major axes.
For \(P_1\): \(a_1 = \dfrac{R + 3R}{2} = 2R\). For \(P_2\): \(a_2 = \dfrac{2R + 4R}{2} = 3R\).

Step 3: Take ratio using Kepler's law.
\[ \frac{T_2}{T_1} = \left(\frac{a_2}{a_1}\right)^{3/2} = \left(\frac{3R}{2R}\right)^{3/2} = \left(\frac{3}{2}\right)^{3/2} = \frac{3}{2}\sqrt{\frac{3}{2}} \]

Step 4: Conclusion.
Hence, \(\dfrac{T_2}{T_1} = \dfrac{3}{2}\sqrt{\dfrac{3}{2}}\).