Q107 / 197 · NEET 2009 (Mains) · Physics

The figure shows elliptical orbit of a planet $m$ about the sun $\mathrm{S}$. The shaded area SCD is twice the shaded are $\mathrm{SAB}$. It $t_1$ is the time for the planet to move from $C$ to $D$ and $t_2$ is the time to move from $A$ to $B$ then Figure

  • A
    $\mathrm{t}_1=\mathrm{t}_2$
  • B
    $t_1>t_2$
  • C
    $t_1=4 t_2$
  • Correct answer
    D
    $t_1=2 t_2$

Solution

Kepler's $2^{\text {nd }}$ law $$ \begin{aligned} & \Rightarrow\left(\frac{\Delta \mathrm{A}}{\Delta \mathrm{t}}\right)_{\text {planet }}=\text { constant } \\ & \frac{\mathrm{A}_1}{\mathrm{t}_1}=\frac{A_2}{t_2} \\ & \Rightarrow \frac{2 \mathrm{~A}}{\mathrm{t}_1}=\frac{A}{t_2} \\ & \Rightarrow \mathrm{t}_1=2 \mathrm{t}_2 \end{aligned} $$

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

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