Question

The drawing (not to scale) shows one alignment of the sun, earth, and moon. the gravitational force vector f sm that the sun exerts on the moon is perpendicular to the force vector f em that the earth exerts on the moon. the masses are: mass of sun = 1.99 1030 kg, mass of earth = 5.98 1024 kg, mass of moon = 7.35 1022 kg. the distances shown in the drawing are rsm = 1.5 1011 m and rem = 3.85 108 m. determine the magnitude of the net gravitational force on the moon.

Answer 1

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):

 

$F_{e} = G \frac {M_{e} * M_{m} } {r^{2}_{EM}} = 6.67 x 10^{-11} N * (\frac {m} {kg})^{2}*\frac {5.98 * 10^{24} kg * 7.35 * 10^{22} kg} {(3.85 x 10^{8}m)^{2}} = 1.98 * 10^{20} N$

The gravitational force that acts on the Moon by the Sun (Law of Gravity):

$F_{S} = G \frac {M_{s} * M_{m} } {r^{2}_{EM}} = 6.67 x 10^{-11} N * (\frac {m} {kg})^{2}*\frac {1.99 * 10^{30} kg * 7.35 * 10^{22} kg} {(1.50 x 10^{11}m)^{2}} = 4.34 * 10^{20} N$

Net gravitational force on the moon:

$F = F_{e} + F_{s}$

Pythagorean theorem for a right triangle ABC:

 

$F = \sqrt {F^{2}_{S} + F^{2}_{e}} = \sqrt {(1.98 * 10^{20}N)^{2} + (4.34 * 10^{20} N)^{2}} = 4.77 * 10^{20}N$

Answer: Answer: magnitude of the net gravitational force on the moon is 4.77 × $10^{20}$N.