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Question: Figure shows elliptical orbit of a planet P about the sun S. The shaded area SCD is twice the shaded...

Figure shows elliptical orbit of a planet P about the sun S. The shaded area SCD is twice the shaded area SAB. If is the timer for the planet to move from C to D and is the time to move from A to B, then

A

t1=t2\mathrm { t } _ { 1 } = \mathrm { t } _ { 2 }

B

t1=2t2\mathrm { t } _ { 1 } = 2 \mathrm { t } _ { 2 }

C

t1=4t2\mathrm { t } _ { 1 } = 4 \mathrm { t } _ { 2 }

D

t1>t2\mathrm { t } _ { 1 } > \mathrm { t } _ { 2 }

Answer

t1=2t2\mathrm { t } _ { 1 } = 2 \mathrm { t } _ { 2 }

Explanation

Solution

According to Kepler’s seconds law, equal areas are swept in equal intervals of time.

As area SCD = 2 area SAB, hence t1=2t2\mathrm { t } _ { 1 } = 2 \mathrm { t } _ { 2 }