A satellite (pictured in brown) of mass 2000kg is in a circular orbit around the

Question

A satellite (pictured in brown) of mass 2000kg is in a circular orbit around the earth at a distance 4 R_e from the earth's center (R_e is the earth's radius which is about 6.37 times 10^6 m. Another satellite (pictured in grey) of the same mass as the first and traveling faster but going in the same direction, crashes into it as shown. After the crash, the two objects stick together. What velocity must the second grey satellite have, right before the collision, in order that the two objects end up escaping from the earth's gravitational field? (Don't forget that the acceleration due to gravity at r = R_e is g = 9.81 m/s^2. If you do the problem using symbols instead of numerical values as far as possible, you can use this information to simplify your equations.)

Explanation / Answer

The velocity of the satellite rotating in the orbit v=(GME/r) --eqn 1

Let the unknown velocity of the grey satellite is V and after collision, their combined velocity is V'. Appling the momentum conservation

mv + mV = (m+m) V' => V' = (v + V)/2. -- eqn 2

The velocity V' is the escape velocity. This shows that the kinetic energy of the combined system will be equal to the potential energy at that height

0.5 (m+m) V'2 -GME m/r = 0 => V' = (2GME/r) -- eqn 3

Putting the value of V' and vin eqn 2 , then we get

V = 2(2GME/r) - v = 2(2GME/r) - (GME/r) = 7231.59 m/s = 7.232 km/s

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