Respuesta :
Answer:
(a) [tex]F_{sm} = 4.327\times 10^{20}\ N[/tex]
(b) [tex]F_{em} = 1.983\times 10^{20}\ N[/tex]
(c) [tex]F_{se} = 3.521\times 10^{20}\ N[/tex]
Solution:
As per the question:
Mass of Earth, [tex]M_{e} = 5.972\times 10^{24}\ kg[/tex]
Mass of Moon, [tex]M_{m} = 7.34\times 10^{22}\ kg[/tex]
Mass of Sun, [tex]M_{s} = 1.989\times 10^{30}\ kg[/tex]
Distance between the earth and the moon, [tex]R_{em} = 3.84\times 10^{8}\ m[/tex]
Distance between the earth and the sun, [tex]R_{es} = 1.5\times 10^{11}\ m[/tex]
Distance between the sun and the moon, [tex]R_{sm} = 1.5\times 10^{11}\ m[/tex]
Now,
We know that the gravitational force between two bodies of mass m and m' separated by a distance 'r' is given y:
[tex]F_{G} = \frac{Gmm'_{2}}{r^{2}}[/tex] (1)
Now,
(a) The force exerted by the Sun on the Moon is given by eqn (1):
[tex]F_{sm} = \frac{GM_{s}M_{m}}{R_{sm}^{2}}[/tex]
[tex]F_{sm} = \frac{6.67\times 10^{- 11}\times 1.989\times 10^{30}\times 7.34\times 10^{22}}{(1.5\times 10^{11})^{2}}[/tex]
[tex]F_{sm} = 4.327\times 10^{20}\ N[/tex]
(b) The force exerted by the Earth on the Moon is given by eqn (1):
[tex]F_{em} = \frac{GM_{s}M_{m}}{R_{em}^{2}}[/tex]
[tex]F_{em} = \frac{6.67\times 10^{- 11}\times 5.972\times 10^{24}\times 7.34\times 10^{22}}{(3.84\times 10^{8})^{2}}[/tex]
[tex]F_{em} = 1.983\times 10^{20}\ N[/tex]
(c) The force exerted by the Sun on the Earth is given by eqn (1):
[tex]F_{se} = \frac{GM_{s}M_{m}}{R_{es}^{2}}[/tex]
[tex]F_{se} = \frac{6.67\times 10^{- 11}\times 1.989\times 10^{30}\times 5.972\times 10^{24}}{((1.5\times 10^{11}))^{2}}[/tex]
[tex]F_{se} = 3.521\times 10^{20}\ N[/tex]
