Newton's Law of Gravitation

One force for the sky and the ground

Isaac Newton answered yes in 1687. The pull that makes the mango fall also reaches the Moon. It acts between every pair of bodies in the universe. This attraction between masses is called gravitation. The force you know as weight is one example: the gravitational pull of the whole Earth on a body near it.

Newton stated the rule for two small bodies. Take two particles of masses \(m_1\) and \(m_2\), a distance \(r\) apart. Each is small enough to be treated as a point mass, a body with mass but no size. Each particle pulls the other towards itself. The size of the pull grows with each mass and falls with the square of the distance.

The square of the distance

The distance enters as \(r^2\), so the law is an inverse-square law. Double the distance and the force becomes four times smaller, not two. Triple it and the force is nine times smaller. Halve the distance and the force grows four times.

In Fig 1.1 the distance grows in equal steps, but the force does not fall in equal steps. Moving from \(1\,\text{m}\) to \(2\,\text{m}\) costs three quarters of the force. Moving from \(2\,\text{m}\) to \(3\,\text{m}\) costs only a further \(5/36\) of it. The pull weakens fast at first and then slowly. It never reaches zero, however far apart the bodies are.

What kind of force it is

Gravitation always attracts. No mass is known that pushes another mass away. It is a central force: it acts along the line joining the two bodies, and its size depends only on their distance. Nothing placed between the bodies changes it. A wall, water or a vacuum between two masses leaves their pull the same.

The two forces of the law form a third-law pair. The Earth pulls the mango down with \(F\), and the mango pulls the Earth up with the same \(F\). The mango moves visibly because its mass is small. The Earth's acceleration, \(F\) divided by \(6 \times 10^{24}\,\text{kg}\), is far too small to notice.

The law as a vector

The size \(F\) is only half the story; the direction is the rest. Let \(\vec r_{12}\) be the vector from \(m_1\) to \(m_2\), with length \(r\). Its unit vector \(\hat r_{12} = \vec r_{12}/r\) points from \(m_1\) towards \(m_2\). The force on \(m_1\) points along \(\hat r_{12}\), and the force on \(m_2\) points the opposite way.

A very weak force

The number \(G\) is the gravitational constant. It is the same everywhere in the universe, for every pair of masses. Its value is \(6.67 \times 10^{-11}\,\text{N m}^2/\text{kg}^2\): the force between two \(1\,\text{kg}\) masses \(1\,\text{m}\) apart. Its dimensions follow from \(G = Fr^2/m_1m_2\). They are \([\text{M}^{-1}\text{L}^3\text{T}^{-2}]\).

Gravitation dominates only when at least one body is enormous. The huge mass of a planet outweighs the tiny \(G\). Between two books on a desk the pull exists, but friction and the smallest breeze are far stronger.

Newton's test with the Moon

Now return to the mango and the Moon. Near the ground the Earth gives the mango an acceleration \(g = 9.8\,\text{m/s}^2\). The mango is one Earth radius, \(R_E = 6.4 \times 10^6\,\text{m}\), from the Earth's centre. The Moon is \(3.84 \times 10^8\,\text{m}\) from the centre, which is \(60\) Earth radii. If the pull falls as \(1/r^2\), the Moon's acceleration must be \(9.8/60^2 = 2.72 \times 10^{-3}\,\text{m/s}^2\).

The Moon moves on a nearly circular path, so its acceleration is centripetal, \(a = 4\pi^2 r/T^2\). Here \(r = 3.84 \times 10^8\,\text{m}\) and the period is \(T = 27.3\) days, or \(2.36 \times 10^6\,\text{s}\). Then \(a = 39.5 \times 3.84 \times 10^8/(2.36 \times 10^6)^2 = 2.72 \times 10^{-3}\,\text{m/s}^2\). The two numbers agree. The Moon is falling towards the Earth, like the mango, but it moves sideways fast enough to keep missing it.