Law of Gravity

• Newton derived the law of gravity by combining Kepler's third law with the mathematical description of the centripetal force.
• The gravitational force acting on two masses $M$ and $m$ is proportional to both and inversely proportional to the squared distance $r$ between them: $$F_G = G\frac{Mm}{r^2}$$
• The proportional factor $G=6.67\cdot 10^{-11}\,\mathrm{N}$ is called the gravitational constant and can be measured for example with the Cavendish experiment.

One of the great achievements of Isaac Newton was to find out that the force which lets objects fall to the ground and the force which keeps celestial bodies on an elliptical track around a central body is basically identical. With the help of Kepler's laws and the formula for the centripetal force, he was then able to determine a relation between the gravitational force, the masses of the two attracting objects, and their distance to each other. The gravitational acceleration is independent of the mass $m$ of the accelerated object. Taking the second law of Newton $F=ma$ into account, we can therefore assume that the gravitational force has to be proportional to $m$: $$F_G \propto m$$ Due to Newton's third law, it makes sense to state that the gravitational force should also be proportional to the mass of the central body because of reaction: $$F_G \propto M$$ In the next step, we assume that gravitation acts like a centripetal force that keeps the planets in the solar system on their trajectory around the sun or the moon around the earth. This can be written as $$F_C = m\omega^2 r$$ Inserting $\omega=2\pi/T$ leads to $$F_C = m \frac{4\pi^2}{T^2} r$$ Kepler's third law states that $T^2$ over $r^3$ is a constant $k$ and can therefore be written as $$\frac{T^2}{r^3} = k$$ We want to eliminate now the time from the formula of $F_G$. Expanding the right side of $F_G$ with $r^2$ results in $$F_C = m 4\pi^2 \frac{T^2}{r^3} \frac{1}{r^2}$$ We can now insert Kepler's law into this equation which finally leads to $$F_C = m4\pi^2k\frac{1}{r^2}$$ From this equation can immediately deduce that the gravitational force must be inversely proportional to the squared distance between the interacting objects: $$F_C \propto \frac{1}{r^2}$$ Putting all together leads to $$F_G\propto\frac{Mm}{r^2}$$ Introducing a proportional factor $G$ results in the well-known Newton's law of gravity: $$\boxed{F_G = G\frac{Mm}{r^2}}$$ The value of the gravitational constant can be determined with the Cavendish Experiment.