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The figure shows elliptical orbit of a p...


The figure shows elliptical orbit of a planet m about the sun S. the shaded area SCD is twice the shaded area SAB. If `t_(1)` be 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)=4t_(2)`

B

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

C

`t_(1)=t_(2)`

D

`t_(1)gtt_(2)`

Text Solution

Verified by Experts

The correct Answer is:
B
According to kepler.s second law equal areas are swept in equal time `t_(1),` the time taken togo from C to `D=2t_(2)` where `t_(2)` is the time taken to go from A to B. As it is given that area `SCD=2SAB`.
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