Earth-Moon gravitational force

Problem

Calculate the Earth-Moon gravitational force given $M_T = 5{,}97\cdot 10^{24}$ kg, $M_L = 7{,}35\cdot 10^{22}$ kg, $r = 3{,}84\cdot 10^8$ m, $G = 6{,}67\cdot 10^{-11}$ N m$^2$/kg$^2$.

The mutual gravitational forces between the Earth and the Moon, separated by the distance $r$.

Solution

Applying the law of universal gravitation between the two bodies: $F = G\frac{M_T\, M_L}{r^2}$ Substituting the data: $F = 6{,}67\cdot 10^{-11}\cdot \frac{5{,}97\cdot 10^{24}\cdot 7{,}35\cdot 10^{22}}{(3{,}84\cdot 10^8)^2}$ Numerator of the masses: $5{,}97\cdot 10^{24}\cdot 7{,}35\cdot 10^{22} = 4{,}39\cdot 10^{47}$ kg$^2$. Denominator: $(3{,}84\cdot 10^8)^2 = 1{,}47\cdot 10^{17}$ m$^2$. Hence: $F = 6{,}67\cdot 10^{-11}\cdot \frac{4{,}39\cdot 10^{47}}{1{,}47\cdot 10^{17}} = 6{,}67\cdot 10^{-11}\cdot 2{,}98\cdot 10^{30}$ $F \approx 1{,}99\cdot 10^{20}~\text{N}$ This is the force that keeps the Moon in orbit: huge in absolute value, but distributed across gigantic masses. $\ev{F \approx 1{,}99\cdot 10^{20}~\text{N}}$

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Topics: Gravitation Concepts: Law of universal gravitation Skills: Symbolic set-up