Question

A 30.8 kg satellite circles planet Cruton every 8.66 h. The magnitude of the gravitational force exerted on the satellite by Cruton is 72.5 N. (a) What is the radius of the orbit? (b) What is the kinetic energy of the satellite? (c) What is the mass of planet Cruton?

Answer 1

Time period of rotation (T) = 2π/ω, where ω is the angular velocity

=> ω = 2π/T =  2π/(8.66*60*60) = 2.016*10^-4 rad/s

(a) gravitation force = centripetal force

=> 72.5 = mrω², where r is the radius of the orbit

=> 72.5 = (30.8)(r)(2.016*10^-4)²

=> r = 5.8*10⁷ m

(b) kinetic enrgy = (1/2)mr²ω²

=> KE = (1/2)(30.8)(5.8*10⁷)²(2.016*10^-4)²

=> KE = 2.1*10⁹ J

(c) Gravitational force = centripetal force

=>GMm/r² = mrω², where G=constant of gravitation, M=mass of planet cruton

=>M = r³ω²/G = (5.8*10⁷)³(2.016*10^-4)²/(6.67*10^-11)

=>M = 1.18*10²⁶ kg