Q133 / 160 · AP EAMCET 2022 (05 Jul Shift 2) · Physics

Two planets, A and B orbit around a star such that time period of $A$ is 8 times the time period of $B$. The ratio of orbital velocities of the planets A and B is

  • A
    $4: 1$
  • B
    $1: 4$
  • C
    $2: 1$
  • Correct answer
    D
    $1: 2$

Solution

$\mathrm{T}_{\mathrm{A}}=8 \mathrm{~T}_{\mathrm{B}}$ $\Rightarrow \mathrm{T}_{\mathrm{A}}^2=64 \mathrm{~T}_{\mathrm{B}}^2 \Rightarrow \mathrm{R}_{\mathrm{A}}^3=64 \mathrm{R}_{\mathrm{B}}^3 \quad\left[\because \mathrm{T}^2 \propto \mathrm{R}^3\right]$ $\Rightarrow R_A=4 R_B$ Now, $V=\frac{2 \pi R}{T}$ $\frac{\mathrm{V}_{\mathrm{A}}}{\mathrm{V}_{\mathrm{B}}}=\frac{\mathrm{R}_{\mathrm{A}}}{\mathrm{T}_{\mathrm{A}}} \times \frac{\mathrm{T}_{\mathrm{B}}}{\mathrm{R}_{\mathrm{B}}}=\frac{\mathrm{R}_{\mathrm{A}}}{\mathrm{R}_{\mathrm{B}}} \times \frac{\mathrm{T}_{\mathrm{B}}}{\mathrm{T}_{\mathrm{A}}}=4 \times \frac{1}{8}=\frac{1}{2}$

Hence, the correct answer is option: 'D'.

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