An artificial satellite (mass m) of a planet (mass M) revolves in a circular orbit whose radius is n times the radius R of the planet in the process of motion the satellite experiences a slight resistance due to cosmic dust. Assuming the force of resistance on satellite to depend on velocity as `F=av^(2)` where 'a' is a constant caculate how long the satellite will stay in the space before it falls on to the planet's surface.

Answer

`(msqrt(R)(sqrt(n)-1))/(asqrt((GM)))`

Explanation

`F=m(dv)/(dt)=av^(2)`……….i
At `t=0`, when the satellite is at a distance `nR` from the centre of the earth. Velocity of the satellite is
`v_(i)=sqrt((GM)/(nR))`
At time `t` when the satellite is
`v_(i)=sqrt((GM)/(nR))`
`v_(f)=sqrt((GM)/R)`
from eqn i `m(dv)/(v^(2))=a dt`

Integrating both the sides we get
`mint_(v_(i))^(v_(f))(dv)/(v^(2))=a int_(0)^(t)dtimpliesm[v^(-2+1)/(-2+1)]_(v_(i))^(v_(f))=at`
on solving we get `t=(msqrtR(sqrtn-1))/(asqrt((GM))`