Question

The figure shows elliptical orbit of a planet m about the sun $S$. The shaded area $SCD$ is twice the shaded area $SAB$. If $t_1$ 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 :

• A. $t _{1}=4 t _{2}$
• B. $t _{1}=2 t _{2}$
• C. $t _{1}= t _{2}$
• D. $t _{1}> t _{2}$

Answer 1

Answer: (B)

According to Kepler’s law, the areal velocity of a planet around the sun always remains constant.
SCD : $A_{1}- t_{1}$ (areal velocity constant)
SAB : $A_{2} - t_{2}$
$\frac{A_{1}}{t_{1}}=\frac{A_{2}}{t_{2}},$
$t_{1}=t_{2}. \frac{A_{1}}{A_{2}},\quad$(given $A_{1} = 2A_{2}$)
$=t_{2}. \frac{2A_{2}}{A_{2}}$
$\therefore \,t_{1}=2t_{2}$