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

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 t1 is the time for the planet to move from C to D and t2 is the time to move A to B then <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/p-tr3sp9masfnvbqop.jpg" /> (A) t1=4t2​ (B) t1=2t2​ (C) t1=t2​ (D) t1>t2​

Tardigrade Admin · Accepted Answer

According to Kepler's second law, 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 of

SC D = 2 S A B

.

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

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

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

$t_{1} = t_{2} $

$t_{1} > t_{2} $

According to Kepler's second law, 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 of $SC D = 2 S A B$ .

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

t1​=4t2​​

t1​=2t2​​

t1​=t2​​

t1​>t2​​