Question

At a certain instant, the earth, the moon, and a station- ary 1250-kg spacecraft lie at the vertices of an equilateral triangle whose sides are 3.84 * 105 km in length. (a) Find the magnitude and direction of the net gravitational force exerted on the spacecraft by the earth and moon. State the direction as an angle measured from a line connecting the earth and the spacecraft. In a sketch, show the earth, the moon, the spacecraft, and the force vector. (b) 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? Ignore any gravitational effects due to the other planets or the sun.

Answer 1

Answer:

a). F = 3.376 N, θ = 59.18°

b). W = 1.3x $10^{9}$ J      

Explanation:

We know

Gravitational constant, G = 6.673 x $10^{-11}$ N-$m^{2}$/$kg^{-2}$

Mass of the earth, M = 5.97 x $10^{24}$ kg

mass of the moon, m = 7.35 x $10^{22}$ kg  

Mass of the satellite, $m_{s}$ = 1250 kg

Distance between the objects, r = 3.84 x $10^{5}$ km

                                                      = 3.84 x $10^{8}$ m

Now

The force on the satellite due to moon

$F_{m}= \frac{G\times m\times m_{s}}{r^{2}}$

$F_{m}= \frac{6.673\times 10^{-11}\times 7.35\times 10^{22}\times 1250}{(3.84\times 10^{8})^{2}}$

$F_{m}$ = 0.0415 N ( in the positive x direction )

The force on the space craft due to the earth

$F_{m}= \frac{G\times M\times m_{s}}{r^{2}}$

$F_{m}= \frac{6.673\times 10^{-11}\times 5.97\times 10^{24}\times 1250}{(3.84\times 10^{8})^{2}}$

$F_{m}$ = 3.377 N ( at 60° to x axis )

Now component of force of earth along x axis

$F_{e_{x}} = F_{e}\times cos 60$

                     = 3.377 x 0.5

                     = 1.6885 N

Now component of force of earth along y axis

$F_{e_{y}} = F_{e}\times sin 60$

                      = 3.377 x 0.86

                      = 2.90 N

∴ Net force on the space craft due to earth and moon along x axis

$F_{x}$ = $F_{e}$ cos 60+$F_{m}$

                       = 1.3885+0.0415

                        = 1.73 N

Net force on the space craft due to earth and moon along y axis  

$F_{x}$ = $F_{e_{y}}$

                         = 2.90 N

Therefore, total force F = $\sqrt{(F_{x}^{2})+(F_{y}^{2})}$

                                    F = $\sqrt{(1.73^{2})+(2.90^{2})}$

                                    F = 3.376 N

∴ Magnitude of the net gravitational force on the space craft is 3.376 N

Direction of net force on the space craft is given by

$\Theta = \arctan \left (\frac{F_{y}}{F_{x}}\right )$

$\Theta = \arctan \left (\frac{2.90}{1.73}\right )$

$\Theta = 59.18$°

Therefore this direction is 59.18° from the line joining earth and the space craft.

b).

∴ Gravitational potential energy of the space craft is given by

$E = \frac{G.M.m_{s}}{r}+\frac{G.m.m_{s}}{r}$

$E = \frac{G\times m_{s}\left ( M+m \right )}{r}$

$E = \frac{6.673\times 10^{-11}\times 1250\left ( 5.97\times 10^{24}+7.35\times 10^{22} \right )}{3.84\times 10^{8}}$

E = 1312769385 J

E = 1.3 x $10^{9}$ J

Therefore minimum work done is 1.3x $10^{9}$ J