1.

Two satellites A and B are revolving around the earth in circular orbits of radius r_(1) and r_(2) respectively with r_(1)ltr_(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. focal lengths of the elliptical path are r_(1) and r_(2) respectively. at position 2, A is given another impulse so that velocities of A and B at 2 become equal and the two move together. for any elliptical path of the satellite time period of revolution is given Kepler's planetry law as T^(2)alphar^(3) where a is semi-major axis of the ellipse which is (r_(1)+r_(2))/2 in this case. also angular momentum 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 given its position vector relative to the earth centre. If r_(2)=3r_(1) and time period of revolution for B be T then time taken by A in moving from position 1 to position 2 is

Answer»

`T(sqrt(3))/(sqrt(2))`
`T(sqrt(3))/2`
`(Tsqrt(2))/(3sqrt(3))`
`(Tsqrt(2))/3`

SOLUTION :`U_(A)=sqrt(2gh)=sqrt((2xxGm)/(2R^(2))H)=sqrt((Gmh)/R)`
From `A` to `B`, field due to shell is zero, but fiedl due to SPHERE is non-zero.
Hence, `t_(AB)ltR/(U_(A))lt(R^(2))/(sqrt(GMh))`


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