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At a certain instant, the earth, the moon, and a stationary 1160 kg spacecraft lie at the vertices of an equilateral triangle whose sides are 3.84 x 10^5 km in length.-What is the minimum amount of work that you would have to do to move the spacecraft to a point far from the earth and moon? You can ignore any gravitational effects due to the other planets or the sun.

User Ibread
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1 Answer

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Answer:


W = 1.22 * 10^9 J

Step-by-step explanation:

Initial potential energy of the given spacecraft is given as


U = -(GM_e m)/(r) - (GM_m m)/(r)

so we have


U = - (Gm)/(r)(M_e + M_m)

so we have


M_e = 5.98 * 10^(24) kg


M_m = 7.35 * 10^(22) kg


m = 1160 kg


r = 3.84 * 10^8 m


U = - ((6.67 * 10^(-11))(1160))/(3.84 * 10^8)(5.98 * 10^(24) + 7.35 * 10^(22))


U = -1.22 * 10^9 J

now total work done to move it to infinite is given

W = 0 - U


W = 1.22 * 10^9 J

User Gabac
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