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 t1 is the time for the planet to move from C to D and t2 is the time to move from A to B then

Question

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 t1 is the time for the planet to move from C to D and t2 is the time to move from A to B then (A) t1 = t2 (B) t1 = 2t2 (C) t1 = 4t2 (D) t1 > t2

Tardigrade Admin · Accepted Answer

Equal areas are swept in equal time. t1, the time taken to go from C to D=2 t2 where t2 is the time taken to go from A to B. As it is given that area S C D=2 S A B.

Q. The figure shows elliptical orbit of a planet $m$ about the sun $S$ . The shaded area $SC D$ is twice the shaded area $S A B$ . 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

$t_{1} = t_{2}$

$t_{1} = 2 t_{2}$

$t_{1} = 4 t_{2}$

$t_{1} > t_{2}$

Equal areas are swept in equal time. $t_{1}$ , the time taken to go from $C$ to $D = 2 t_{2}$
where $t_{2}$ is the time taken to go from $A$ to $B$ .
As it is given that area $SC D = 2 S A B$ .

Q. The figure shows elliptical orbit of a planet m about the sun S. The shaded area SCD is twice the shaded area SAB. If t1​ is the time for the planet to move from C to D and t2​ is the time to move from A to B then

t1​=t2​

t1​=2t2​

t1​=4t2​

t1​>t2​