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Question: Answered & Verified by Expert
A planet of mass $m$ moves around the Sun along an elliptical path with a period of revolution $T$. During the motion, the planet's maximum and minimum distance from Sun is $R$ and $\frac{R}{3}$ respectively. If $T^2=\alpha R^3$, then the magnitude of constant $\alpha$ will be
PhysicsGravitationTS EAMCETTS EAMCET 2019 (06 May Shift 1)
Options:
  • A $\frac{10}{9} \cdot \frac{\pi}{\mathrm{Gm}}$
  • B $\frac{20}{27} \cdot \frac{\pi^2}{\mathrm{Gm}}$
  • C $\frac{32}{27} \cdot \frac{\pi^2}{\mathrm{Gm}}$
  • D $\frac{1}{18} \cdot \frac{\pi^2}{\mathrm{Gm}}$
Solution:
1649 Upvotes Verified Answer
The correct answer is: $\frac{32}{27} \cdot \frac{\pi^2}{\mathrm{Gm}}$
Given, mass of planet $=\mathrm{m}$
Maximum and minimum distance of the planet from Sun is $R$ and $\frac{R}{3}$, respectively.
$\therefore$ Semi-major axis of elliptical path of planet around the Sun,
$a=\frac{R+\frac{R}{3}}{2}=\frac{2 R}{3}$
$\therefore$ Time period of planet is given by
$\begin{aligned} & T=2 \pi \sqrt{\frac{a^3}{G m}}=2 \pi \sqrt{\frac{\left(\frac{2 R}{3}\right)^3}{G m}} \\ & T=2 \pi \sqrt{\frac{8 R^3}{27 G m}}\end{aligned}$
Square on the both sides, we get
$\begin{aligned} T^2 & =4 \pi^2 \cdot \frac{8 R^3}{27 G m} \\ T^2 & =\frac{32 \pi^2}{27 G m} \cdot R^3=\alpha R^3 \\ \therefore \quad \alpha & =\frac{32 \pi^2}{27 G m}\end{aligned}$
Hence, the magnitude of constant $\alpha$ will be $\frac{32}{27} \cdot \frac{\pi^2}{G m}$

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