The figure shows eliptical orbit of a planet m about the Sun S. The shaded area SCD is twice the shaded area SAB. If `t_(1)` is the time for the planet of 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

SCD : `A_(1) prop t_(1)" "` (a real velocity constant)
SAB : `A_(2) prop t_(2)`
`(A_(1))/(t_(1))=(A_(2))/(t_(2)), t_(1)=t_(2). (A_(1))/(A_(2)), A_(1)=2A_(2)`
`therefore t_(1)=2t_(2)`
Thus correct choice is (b).