Q73 / 200 · NEET 2009 (Screening) · Chemistry

The figure shows elliptical orbit of a planet $m$ about the sun $S$. The shaded area $S C D$ is twice the shaded area $S A B$. If $\mathrm{t}_1$ is the time for the planet to move from $C$ to $D$ and $\mathrm{t}_2$ is the time to move from $A$ to $B$, then Figure

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

Solution

Key Idea Apply Kepler's law of area of planetary motion. The line joining the sun to the planet sweeps out equal areas in equal time interval $i e$, areal velocity is constant. $$ \frac{\mathrm{dA}}{\mathrm{dt}}=\text { constant } $$ or $$ \frac{\mathrm{A}_1}{\mathrm{t}_1}=\frac{\mathrm{A}_2}{\mathrm{t}_2} $$ $$ \begin{array}{llrl} \Rightarrow & \mathrm{t}_1 & =\frac{\mathrm{A}_1}{\mathrm{~A}_2} \mathrm{t}_2 \\ \text { Given } & \mathrm{A}_1 & =2 \mathrm{~A}_2 \\ \therefore & \mathrm{t}_1 & =2 \mathrm{t}_2 \end{array} $$

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

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