A satellite is orbiting around the earth in a circular orbit of radius `r`. A particle of mass `m` is projected from the satellite in a forward direction with a velocity `v = 2//3` times the orbital velocity (this velocity is given w.r.t. earth). During subsequent motion of the particle, its minimum distance from the centre of earth is

Options

A. `r/2`
B. `r`
C. `(2r)/3`
D. `(4r)/5`

Answer

A

Explanation

As the velocity of the particle is les than the orbital velocity of the satellite, the particle goes in the elliptical orbit of the semi major axis less than r.
Let `r_(1)` be the minimum distance and `v_(1)` be the velocity of the particle at this position, then
`m_(0)xxsqrt(2/3)v_(0)xxr-m_(0)v_(1)r_(1)`
where `m_(0)` is the mass of the particle and `v_(0)` is the orbital speed equal to `sqrt(GM//r)`.
`v_(1)r_(1)=sqrt(2/3)v_(0)r`
From energy conservation
`(m_(0)xx2/3v_(0)^(2))/2-(Gm_(0))/r=(m_(0)v_(1)^(2))/2-(GMm_(0))/(r_(1))` Solving the equations we get `r_(1)=r//2`