a) The gravitational force between the Earth and the Moon is [tex]1.98\cdot 10^{20} N[/tex]
b) The acceleration of the Moon is [tex]2.70\cdot 10^{-3} m/s^2[/tex]
c) The acceleration of gravity on the Moon is [tex]1.62 m/s^2[/tex], and the weight of a person on the moon is [tex]W=1.62 m[/tex] [N]
Explanation:
a)
The gravitational force between the Earth and the Moon is given by:
[tex]F=G\frac{m_e m_m}{r^2}[/tex]
where
[tex]G=6.67\cdot 10^{-11} m^3 kg^{-1}s^{-2}[/tex] is the gravitational constant
[tex]m_e = 5.97\cdot 10^{24} kg[/tex] is the mass of the Earth
[tex]m_m = 7.34 \cdot 10^{22} kg[/tex] is the mass of the Moon
[tex]r=3.84\cdot 10^8 m[/tex] is the distance between the Earth and the Moon
Substituting the values, we find:
[tex]F=(6.67\cdot 10^{-11})\frac{(5.97\cdot 10^{24})(7.34\cdot 10^{22})}{(3.84\cdot 10^8)^2}=1.98\cdot 10^{20} N[/tex]
b)
We can find the Moon's acceleration by using Newton's second law:
[tex]F=m_m a[/tex]
where:
[tex]F=1.98\cdot 10^{20} N[/tex] is the net force exerted on the Moon by the Earth
[tex]m_m = 7.34 \cdot 10^{22} kg[/tex] is the mass of the Moon
a is the acceleration of the Moon
Re-arranging the equation,
[tex]a=\frac{F}{m_m}[/tex]
And solving,
[tex]a=\frac{1.98\cdot 10^{20}}{7.34\cdot 10^{22}}=2.70\cdot 10^{-3} m/s^2[/tex]
c)
We start by calculating the acceleration of gravity (g) on the Moon, which is given by
[tex]g_m = \frac{Gm_m}{R^2}[/tex]
where:
[tex]G=6.67\cdot 10^{-11} m^3 kg^{-1}s^{-2}[/tex] is the gravitational constant
[tex]m_m = 7.34 \cdot 10^{22} kg[/tex] is the mass of the Moon
[tex]R=1740 km = 1.74\cdot 10^6 m[/tex] is the radius of the Moon
Substutiting,
[tex]g_m = \frac{(6.67\cdot 10^{-11})(7.34\cdot 10^{22})}{(1.74\cdot 10^6)^2}=1.62 m/s^2[/tex]
And therefore, the gravitational force of a person of mass m standing on the surface of the Moon is:
[tex]W=mg_m = 1.62 m[/tex] [N]
Learn more about gravitational force:
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