Question

Part A

Find the kinetic energy of a 1.84×10³ kg satellite in a circular orbit about the Earth, given that the radius of the orbit is 1.46×10⁴ miles.

Part B

How much energy is required to move this satellite to a circular orbit with a radius of 2.72×10⁴ miles?

Answer 1

Part A -

First of all convert 'mile' into meter.

Radius, r = 1.46 x 10^4 mile = 1.46 x 10^4 x 1609 m = 2.35 x 10^7 m

Therefore, the orbital velocity of the satellite (v) -  

v^2 = G Me / r

=> v^2 = (6.67x 10^-11 x 5.98 x 10^24) / (2.35 x 10^7)
=> v^2 = 16.97 x 10^6

The expression for the kinetic energy of the satellite is -

KE = (1/2)*m*v^2 = (1/2) * (1.84 x 10^3) * (16.97 x 10^6)

= 15.61 x 10^9 J = 1.561 x 10^10 J

Part B -

Now calculate the orbital velocity of the satellite when r' = 2.72 x 10^4 mile = 2.72 x 10^4 x 1609 m

= 4.38 x 10^7 m

v'^2 = G Me / r'

=> v'^2 = (6.67x 10^-11 x 5.98 x 10^24) / (4.38 x 10^7)
=> v'^2 = 9.11 x 10^6

So,the new kinetic energy of the satellite is -

KE' = (1/2)*m*v'^2 = (1/2) * (1.84 x 10^3) * (9.11 x 10^6)

= 8.38 x 10^9 J = 0.838 x 10^10 J

Therefore, the energy required to move the satellite in new circular orbit -

W = KE - KE' = 1.561 x 10^10 J - 0.838 x 10^10 J = 0.723 x 10^10 J = 7.23 x 10^9 Joule (Answer).