The space shuttle, with mass 86.400 kg. is in a circular-orbit of radius 6.66 × 106 m around the earth. It takes 90.1 min for the shuttle to complete each orbit. On a repair mission, the shuttle is cautiously moving 1.00 m closer to a disabled satellite every 3.00 s. Calculate the shuttle’s kinetic energy (a) relative to the earth and (b) relative to the satellite.

Solution 76P a)Mass of the shuttle is 86400 kg. The radius of the shuttle’s orbit is, R = 6.66 × 10 m. The time to complete 1 complete revolution is, T = 90.1 min = 90.1 × 60 = 5406 s. The velocity of the shuttle we can find by, v = 2R / T = 2××6.66×106= 7740.66 m/s. 5406 The kinetic energy with respect to earth is, 2 K.E earth= 1/2 × m × v = 1/2 × m × × R2 = 1/2 × 86.4 × (7740.66) = 2.59 × 10 jules b)The shuttle moves 1 meters towards the satellite in every 3 s. So, the velocity of the shuttle w.r.t the satellite will be, v = 1/3 = 0.33 m/s. Hence the kinetic energy would be, K.E satellite1/2 mv 2 = 1/2 × 86400 × 0.33 = 4704.48 jules.