Question

The figure shows the 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 $A$ to $B$ then

• A. $t_{1}=4t_{2}​$
• B. $t_{1}=2t_{2}​$
• C. $t_{1}=t_{2}​$
• D. $t_{1}>t_{2}​$

Answer 1

Answer: (B)

According to Kepler's second law, equal areas are swept in equal time.
$t_{1}$ , the time taken to go from C to D = $2t_{2}$ where $t_{2}$ is the time taken to go from A to B. As it is given that area of $SCD=2SAB$ ._​