Two satellites A and B are revolving around the earth circular orbits of radius `r_(1)` and `r_(2)` respectively with `r_(1) lt r_(2)`. Plane of motion of the two are same. At position 1, A is given an impulse in the direction of velocity by firing a rocket so that it follows an elliptical path to meet B at position 2 as shown. A?t position 2, A is given another impluse so that velocities of A and B at 2 become equal and the move together.

For any elliptical path of the satellite of the time period of revolution is given by Kepler's planetary law as `T^(2)alpha r^(3)` where a is semi major axis of the ellipse which is `(r_(1)+r_(2))/(2)` in this case. Also angular mopmentum of any satellite revolving around the Earth will remain a constant about EArth's centre as force of gravity on the satellite which keeps it in elliptical path is along its position vector relative to the earth centre.
If `r_(2)=3r_(1)` and time period of revolution for B be T than time taken by A in moving from position 1 to position 2

Options

A. `T(sqrt(3))/(sqrt(2))`
B. `T(sqrt(3))/(2)`
C. `(Tsqrt(2))/(3sqrt(3))`
D. `(Tsqrt(2))/(3)`

Answer

C

Explanation

In the continuation with same comprehesion
`r=((r_(2)+r_(2)//3))/(2)=2r_(2)//3`
Let time period of A in elliptical path is `T_(0)`
`therefore ((T_(0))/(T))^(2) = ((r )/(r_(2)))^(3)`
`therefore T_(0) = (T(2sqrt(2)))/(sqrt(3))`
Time taken by A to move from 1 to 2,
`t = T_(0)//2=(Tsqrt(2))/(3sqrt(3))`