A spacecraft of mass 8.00 104 kg is in a circular orbit at an altitude of 600 km

Question

A spacecraft of mass 8.00 104 kg is in a circular orbit at an altitude of 600 km above the Earth's surface. Mission Control wants to fire the engines in a direction tangent to the orbit so as to put the spacecraft in an elliptical orbit around the Earth with an apogee of 2.00 104 km, measured from the Earth's center. How much energy must be used from the fuel to achieve this orbit? (Assume that all the fuel energy goes into increasing the orbital energy. This model will give a lower limit to the required energy because some of the energy from the fuel will appear as internal energy in the hot exhaust gases and engine parts.)

Explanation / Answer

Energy required to place spacecraft in an orbit is,

E = KE + PE

= -GMm/(r+h) + GMm/r + 1/2mv^2

Since, v^2 = GM/(r+h)

So, PE = GMm/2r(r+h)

And E = GMm.2r(r+h)/2r(r+h) - GMm.2r/2r(r+h) + GMmr/2r(r+h)

Or E = GMm(r+2h)/2r(r+h)

Here, G = 6.67*10^-11

M = 5.98*10^24 kg

m = 8*10^4 kg

Apogee, r = 2*10^4 km

h = 0.06*10^4 km

So, E = 6.67*10^-11*5.98*10^24*8*10^4(2+2*0.06)*10^4 / 2*2*10^4(2+0.06)*10^4

= 8.2097*10^14 J

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