The figure shows elliptical orbit of a planet $m$ about the sun $\mathrm{S}$. The shaded area SCD is twice…

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
  1. $\mathrm{t}_1=\mathrm{t}_2$
  2. $t_1 > t_2$
  3. $t_1=4 t_2$
  4. $t_1=2 t_2$

Solution

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

Asked in: NEET 2009 (Mains)

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