Question

Exoplanet 'Y' has a mass of 8.0x10^24 kg and a radius of 7.5x10^6 m. A satellite is in circular orbit about exoplanet Y at a distance of 3.0x10^7 m above the surface. What is the period of the satellite's orbit in hours. Also mass of satelite is 4000kg.

Answer 1

For uniform circular motion,centripetal force= mv2/r, where m is mass, v is velocity and r is radius of the circular path.

Also, gravitational force between 2 mass m1 and m2 is given by: Gm1*m2 /r2 , where G is universal gravitational constant, m1 and m2 are the masses and r is distance between them.

Let mass of planet be M and let mass of satellite be m.

Gravitational force provides the necessary centripetal force to the satellite. So,

So, GMm/r2 = mv2/r , where r is radius of the circular path of the satellite.

=>v2 = GM/r => v = (GM/r)1/2

Here,M=8*10^24 kg and r=7.5x10^6+ 3.0x10^7 m = 3.75*10^7 m

So, v=(GM/r)1/2 =[6.67*10-11*8*1024/(3.75*107)]1/2 = 3772.18 m/s

Now,period of orbit = distance/time = 2r/v = 2*3.75*107/( 3772.18) = 62462.41 seconds

= 626462.41/(60*60) hours = 17.35 hours.