Answer:
The magnitude of the free-fall acceleration at the orbit of the Moon is [tex]2.728\times 10^{-3}\,\frac{m}{s^{2}}[/tex] ([tex]\frac{2.784}{10000}\cdot g[/tex], where [tex]g = 9.8\,\frac{m}{s^{2}}[/tex]).
Explanation:
According to the Newton's Law of Gravitation, free fall acceleration ([tex]g[/tex]), in meters per square second, is directly proportional to the mass of the Earth ([tex]M[/tex]), in kilograms, and inversely proportional to the distance from the center of the Earth ([tex]r[/tex]), in meters:
[tex]g = \frac{G\cdot M}{r^{2}}[/tex] (1)
Where:
[tex]G[/tex] - Gravitational constant, in cubic meters per kilogram-square second.
[tex]M[/tex] - Mass of the Earth, in kilograms.
[tex]r[/tex] - Distance from the center of the Earth, in meters.
If we know that [tex]G = 6.674\times 10^{-11}\,\frac{m^{3}}{kg\cdot s^{2}}[/tex], [tex]M = 5.972\times 10^{24}\,kg[/tex] and [tex]r = 382.26\times 10^{6}\,m[/tex], then the free-fall acceleration at the orbit of the Moon is:
[tex]g = \frac{\left(6.674\times 10^{-11}\,\frac{m^{3}}{kg\cdot s^{2}} \right)\cdot (5.972\times 10^{24}\,kg)}{(382.26\times 10^{6}\,m)^{2}}[/tex]
[tex]g = 2.728\times 10^{-3}\,\frac{m}{s^{2}}[/tex]
