An artificial satellite of a planet revolves in a circular orbit whose radius exceeds the radius of the planet 7 times. In the process of motion, the satellite experiences slight resistance due to cosmic dust. Assuming the resistive force depends on the speed of the satellite as F=kv2 (where k is a constant), The time taken by the satellite will stay in orbit until it falls on to the planet’s surface. is mkgR(x−1) (Given m= mass of the satellite, g=acceleration due to gravity at the surface of planet and R= radius of the planet). Find the value of x.

Question

An artificial satellite of a planet revolves in a circular orbit whose radius exceeds the radius of the planet 7 times. In the process of motion, the satellite experiences slight resistance due to cosmic dust. Assuming the resistive force depends on the speed of the satellite as F=kv2 (where k is a constant), The time taken by the satellite will stay in orbit until it falls on to the planet’s surface. is  mkgR(x−1) (Given m= mass of the satellite, g=acceleration due to gravity at the surface of planet and R= radius of the planet). Find the value of x.

Infinity Learn · Accepted Answer

Let r be the orbital radiusGMmr2=mV2r⇒V=GMrTotal energy of the satelliteE=−GMmr+12mV2⇒E=−GMm2rdEdr=GMm2r2......(1)dEdt=p=F→⋅V→⇒dEdt=−F⋅V=−kV2V=−kV3dEdt=−k(GMr)3/2........(2)From (1) and (2) GMm2r2dr=−k(GMr)3/2. dt∴t=∫dt=m2kGM∫7RRdrr; t=mkgR[7−1]

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