Answers for Part A 3221 ms Part B 4554 kms Part C 133 s I ju

Answers for

Part A) 3221 m/s

Part B) 4.554 km/s

Part C) 133 s

I just want to know how to get these answers and last half of part c. Thanks! :)

Example Example 4 A 10,000-kg space capsule is to be returned to A 10,000-kg space capsule is to be returned to the Earth from the space telescope, which is in a circular orbit at an altitude of 12,800 km. It is desired to have the capsule leave tangentt t , at A, and arive tangent capsule leave tangent to its orbit, at A, and arrive tangernt to the Earth\'s surface, at P. a) Find the required speed of the capsule at A, VA . b) Find the speed of the space telescope, Vs If the thrust of the engine is 105 N, find the \"burn time\" required, to return to the Earth, and the final angular velocity about this CM c) i /V V.

Solution

This part can be solved by applying the angular momentum conservation and total energy conservation(potential + kinetic energy). The first one simply gives us the equation Ra*Va = Rp*Vp. While the second equation,i.e, the energy equation gives us Vp=(Va^2 + G*M*[1/Rp - 1/Ra])^0.5; solving these 2 equations simultaneously we get the required answer for Vp. Here, Ra is radius of orbit at point a, Rp is altitude at point p which is equal to the radius of the earth, Va & Vp are the velocities at points a & p respectively. Speed of satellite in circular orbit after balancing the centripetal force with the gravitational force the satellite experiences comes out to be v = (G*M/r)^0.5 where G is the universal gravitational constant, M is mass of earth and r is radius of orbit.