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Figure shows elliptical orbit of a plane...

Figure shows elliptical orbit of a planet P 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 from A to B , then

A

`t_(1)=t_(2)`

B

`t_(1)=8t_(2)`

C

`t_(1)=4t_(2)`

D

`t_(1)=2t_(2)`

Text Solution

Verified by Experts

The correct Answer is:
D
From kepler's law areal velocity = constant so `("area "SCD)/(t_(1))=("area "SAB)/(t_(2))impliest_(1)=2t_(2)`
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