Question

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.

Answer 1

Answer:

$W = 1.22 \times 10^9 J$

Explanation:

Initial potential energy of the given spacecraft is given as

$U = -\frac{GM_e m}{r} - \frac{GM_m m}{r}$

so we have

$U = - \frac{Gm}{r}(M_e + M_m)$

so we have

$M_e = 5.98 \times 10^{24} kg$

$M_m = 7.35 \times 10^{22} kg$

$m = 1160 kg$

$r = 3.84 \times 10^8 m$

$U = - \frac{(6.67 \times 10^{-11})(1160)}{3.84 \times 10^8}(5.98 \times 10^{24} + 7.35 \times 10^{22})$

$U = -1.22 \times 10^9 J$

now total work done to move it to infinite is given

W = 0 - U

$W = 1.22 \times 10^9 J$