contestada

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.

Respuesta :

Answer:

[tex]W = 1.22 \times 10^9 J[/tex]

Explanation:

Initial potential energy of the given spacecraft is given as

[tex]U = -\frac{GM_e m}{r} - \frac{GM_m m}{r}[/tex]

so we have

[tex]U = - \frac{Gm}{r}(M_e + M_m)[/tex]

so we have

[tex]M_e = 5.98 \times 10^{24} kg[/tex]

[tex]M_m = 7.35 \times 10^{22} kg[/tex]

[tex]m = 1160 kg[/tex]

[tex]r = 3.84 \times 10^8 m[/tex]

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

[tex]U = -1.22 \times 10^9 J[/tex]

now total work done to move it to infinite is given

W = 0 - U

[tex]W = 1.22 \times 10^9 J[/tex]

Otras preguntas