A planet moves around the sun on an elliptical path as shown in the figure. If t1 and t2 are the time taken by the planet to cover equal area S A B and S C D respectively, then

Question

A planet moves around the sun on an elliptical path as shown in the figure. If t1 and t2 are the time taken by the planet to cover equal area S A B and S C D respectively, then (A) t1>t2 (B) t1<t2 (C) t1=t2 (D) t1=√t2

Tardigrade Admin · Accepted Answer

According to Kepler's second law of motion, the line that joins any planet to the sun sweeps equal area in equal intervals of time. Given, areas S A B and S C D are equal, so t1=t2

Q. A planet moves around the sun on an elliptical path as shown in the figure. If $t_{1}$ and $t_{2}$ are the time taken by the planet to cover equal area $S A B$ and $SC D$ respectively, then

$t_{1} > t_{2}$

$t_{1} < t_{2}$

$t_{1} = t_{2}$

$t_{1} = t_{2}$

According to Kepler's second law of motion, the line that joins any planet to the sun sweeps equal area in equal intervals of time.
Given, areas $S A B$ and $SC D$ are equal,
so $t_{1} = t_{2}$

Q. A planet moves around the sun on an elliptical path as shown in the figure. If t1​ and t2​ are the time taken by the planet to cover equal area SAB and SCD respectively, then

t1​>t2​

t1​<t2​

t1​=t2​

t1​=t2​​

According to Kepler's second law of motion, the line that joins any planet to the sun sweeps equal area in equal intervals of time. Given, areas SAB and SCD are equal, so t1​=t2​

Q. A planet moves around the sun on an elliptical path as shown in the figure. If $t_{1}$ and $t_{2}$ are the time taken by the planet to cover equal area $S A B$ and $SC D$ respectively, then

$t_{1} > t_{2}$

$t_{1} < t_{2}$

$t_{1} = t_{2}$

$t_{1} = t_{2}$

According to Kepler's second law of motion, the line that joins any planet to the sun sweeps equal area in equal intervals of time.
Given, areas $S A B$ and $SC D$ are equal,
so $t_{1} = t_{2}$