Solution:
Ms = 1.99 × 1030 kg− mass of Sun;
Me = 5.98 × 1024kg− mass of Earth;
Mm = 7.35 × 1022kg − mass of Moon;
rSM = 1.50 × 1011m − distance to the Moon from the Sun;
rEM = 3.85 × 108m − distance to the Moon from the Earth;
The gravitational force that acts on the Moon by the Earth (Law of Gravity):
[tex]F_{e} = G \frac {M_{e} * M_{m} } {r^{2}_{EM}} = 6.67 x 10^{-11} N * (\frac {m} {kg})^{2}*\frac {5.98 * 10^{24} kg * 7.35 * 10^{22} kg} {(3.85 x 10^{8}m)^{2}} = 1.98 * 10^{20} N[/tex]
The gravitational force that acts on the Moon by the Sun (Law of Gravity):
[tex]F_{S} = G \frac {M_{s} * M_{m} } {r^{2}_{EM}} = 6.67 x 10^{-11} N * (\frac {m} {kg})^{2}*\frac {1.99 * 10^{30} kg * 7.35 * 10^{22} kg} {(1.50 x 10^{11}m)^{2}} = 4.34 * 10^{20} N[/tex]
Net gravitational force on the moon:
[tex]F = F_{e} + F_{s} [/tex]
Pythagorean theorem for a right triangle ABC:
[tex]F = \sqrt {F^{2}_{S} + F^{2}_{e}} = \sqrt {(1.98 *
10^{20}N)^{2} + (4.34 * 10^{20} N)^{2}} = 4.77 * 10^{20}N[/tex]
Answer: Answer: magnitude of the net gravitational force on the moon is
4.77 × [tex]10^{20} [/tex]N.
