An artificial satellite of the moon revolves in a circular orbit whose radius exceeds the radius of the moon n times. In the process of motion, the satellite experiences a slight resistance due to cosmic dust. Assuming the resistance force to depend on the velocity of the satellite as F=alpha v^2 , where a is a constant, find how long the satellite will stay in orbit until it falls onto the moon.s surface.

An artificial satellite of the moon revolves in a circular orbit whose

1.	An artificial satellite of the moon revolves in a circular orbit whose radius exceeds the radius of the moon n times. In the process of motion, the satellite experiences a slight resistance due to cosmic dust. Assuming the resistance force to depend on the velocity of the satellite as F=alpha v^2 , where a is a constant, find how long the satellite will stay in orbit until it falls onto the moon.s surface.
Answer» Solution : The energy of the satellite in ORBIT of radius R, E ` = (-GMm)/(2r)` Here m is the mass of satellite and M is the mass of the moon. If dE/dt is the instantaneous rate of decrease of energy of the satellite, then we have `((dE)/(dt)) = Fv` Given `F = alpha v^2 THEREFORE (d[(-GMm)/(2r)])/(dt)= (alpha v^2)v " or " (GMm)/(2r^2) dr= - alpha v^3 dt` Orbital velocity of satellite, v=`sqrt((GM)/(r )) therefore (GMm)/(2r^2) dr =- alpha [ (GM)/(r )]^(3//2) dt " or" (mr^(-1//2))/(2) dr = - alpha (GM)^(1//2) dt` Integrating above expression , we get`m/2 int_(eta R)^( R) r^(-1//2) dt =- alpha (GM)^(1//2) int_0^t dt ` `m \|sqrtr\|_(eta R)^(R ) = - alpha sqrt(GMt) " or " m [sqrt(eta R) - sqrtR] = alpha sqrt(GM) t` ` therefore t = m/alpha ([sqrt(eta R) - sqrtR])/(sqrt(GM)) = m/alpha (sqrtn - 1) sqrt((R)/(GM))`