Question

A 42.9 kg satellite circles planet Cruton every 8.80 h. The magnitude of the gravitational force exerted on the satellite by Cruton is 76.6 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

let

m = 429 kg

Time periode, T = 8.8 h

= 8.8*60*60

= 31680 s

M is mass of planet

let v is the speed of satellite and r is the radius of orbit.

T = 2*pi*r/v

==> v = 2*pi*r/T

centripetal force on satellite, F = m*v^2/r

F = m*(2*pi*r/T)^2/r

= m*4*pi^2*r/T^2

==> r = F*T^2/(m*4*pi^2)

= 76.6*31680^2/(42.9*4*pi^2)

= 4.5392*10^7 m or 45392 km

b) v = 2*pi*r/T

= 2*pi*4.5392*10^7/31680

= 9002.7 m/s

so, KE = 0.5*m*v^2

= 0.5*42.9*9002.7^2

= 1.7385*10^9 J

c) use, T = 2*pi*r^(3/2)/sqrt(G*M)

T^2 = 4*pi^2*r^3/(G*M)

M = 4*pi^2*r^3/(G*T^2)

= 4*pi^2*(4.5392*10^7)^3/(6.67*10^-11*31680^2)

= 5.516*10^25 kg