A satellite is orbiting around earth in a circular orbit of radius `r`. A particle of mass `m` is projected from satellite in forward direction with velocity `v=sqrt(2/3)` times orbital velocity (this velocity is with respect to the 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)/3`

Answer

A

Explanation

As the velocity of particle is less than orbital velocity of satellite, the particle goes in elliptical orbit semimajor axis less then `r`.

Let `r_(1)` be the maximum distance and `v_(1)` be velocity of particle at this position, then `m_(0)xxsqrt(2/3)xxv_(0)r=m_(0)v_(1)r_(1)`, where `m_(0)` is the mass of particle and `v_(0)` is orbital speed equal to
`sqrt((GM)/r)rArr v_(1)r_(1)=sqrt(2/3)v_(0)r`
From energy conservation,
`(m_(0)xx2//3v_(0)^(2))/2-(GMm_(0))/r=(m_(0)v_(1)^(2))/2-(GMm_(0))/(r_(1))`
Solving above equations, we get `r_(1)=r/2`.