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:

Options

A. `t_(1) = 4t_(2)`
B. `t_(1) = 2t_(2)`
C. `t_(1) = t_(2)`
D. `t_(1) gt t_(2)`

Answer

B

Explanation

From Kepler.s second law, areal velocity must be constant
`(dA)/(dt) = K`
`A prop t rArr (A_(1))/(A_(2)) = (t_(1))/(t_(2))` …(i)
Now, from given figure in questions
`A_(1) = SCD " "A_(2) = SAB`
From equation (i)
`t_(1) = 2t_(2)`