Question

Consider a 475 kg satellite in a circular orbit at a distance of 3.06 x 104 km above the Earth's surface. What is the minimum amount of work W the satellite's thrusters must do to raise the satellite to a geosynchronous orbit? Geosynchronous orbits occur at approximately 3.60 x 104 km above the Earth's surface. The radius of the Earth and the mass of the Earth are RE = 6.37 x 109 km and Me = 5.97 x 1024 kg, respectively. The gravitational constant G= 6.67 x 10-11 m3/kg.s?. Assume the change in mass of the satellite is negligible.

Answer 1

Solution:- •.a Satellite Earth The total mechanical Earth-Satellite system energy of the is given by E = - GMAM 7 Here, the E= total mechanical energy of Earth-Satellite system ME = mass of the Earth ms = mass of the satellite g = distance of the satellite brom the centre of the Earth G = gravitational constant

The minimum amount of work w the satellites thrusters do to raise the satellite from roro to r=%2 is given by Concept W = - change in energy of the total mechanical the Eartha Satellite system - Germany. - (- Colome) . Go + Gems :WE GMem: (+,-) Here, 8,=6.37x10²km +3.06x10 km = 36970000 m r₂ = 6.378103 km + 3.608 10 km § 42370000m ME = 5.97X1024 kg Ms = 475kg G= 6.67 810lm /kg-s?

:: W = 6.67X10-" X 5.97810**x 475x (76920000 - 42340000) .' TW6.52x10 J Answer