Newton's Law of Universal Gravitation

The law of universal gravitation

Newton's big idea was that the force that makes an apple fall is the same kind of force that holds the Moon in orbit. Every object with mass attracts every other object with mass. For you and your desk the force is far too small to notice. For planets it is enormous.

Why this form? The force grows with each mass: more mass means more pull. It weakens as the objects move apart, and it weakens fast, with the square of the distance. The tiny value of $G$ is why you only feel gravity from something as massive as a planet.

The force is attractive, and it comes as a pair. Sphere 1 pulls on sphere 2 with the same magnitude that sphere 2 pulls on sphere 1, in opposite directions. This is Newton's third law, and it holds even if one object is much bigger than the other.

Both arrows have the same length: the two forces are equal in magnitude.

Inverse-square reasoning

Often you don't need $G$ at all. You can compare two situations with ratios. Since $F_g \propto \dfrac{m_1 m_2}{r^2}$:

• Double one mass: the force is $2\times$ as large.

• Double the distance: the force is $\dfrac{1}{2^2} = \dfrac14$ as large.

• Cut the distance to one third: the force is $3^2 = 9$ times as large.

An inverse-square law also shows up in data. If you graph $F$ versus $r$, you get a curve. If you graph $F$ versus $\dfrac{1}{r^2}$, you get a straight line through the origin whose slope is $Gm_1m_2$. A straight line is the easiest shape to check by eye, which is why scientists linearize their data this way.

Gravitational field strength

Take a small object of mass $m$ at the surface of a planet of mass $M$ and radius $R$. The distance from the planet's center is $R$, so the gravitational force on the object is $F_g = \dfrac{GMm}{R^2}$. Newton's second law says $F_g = mg$. Set them equal:

The same steps work at any distance $r$ from the center, which gives $g = \dfrac{GM}{r^2}$.

The mass $m$ cancels. This works because the $m$ in Newton's second law (inertial mass, how much an object resists acceleration) and the $m$ in the law of gravitation (gravitational mass, how strongly gravity pulls on it) are equal. That is why a bowling ball and a marble fall with the same acceleration (if you ignore air resistance). The quantity $g$ is the gravitational field strength: the force per kilogram at that location, in N/kg, which is the same as m/s². It depends only on the planet's mass $M$ and on the distance $r$ from its center.

Because $g \propto M/r^2$, you can also compare planets with ratios. A planet with twice Earth's radius and the same mass has $\dfrac14$ of Earth's $g$. A planet with 3 times Earth's mass and the same radius has $3g$.

Weight, mass, and "weightlessness"

Mass (in kg) measures how much matter an object has and how strongly it resists changes in motion. It does not change from place to place. Weight is the gravitational force on the object: $F_g = mg$, in newtons. It changes with the local $g$.

Now back to the hook. Take an astronaut with $m = 70\ \text{kg}$. On the ground, $F_g = mg \approx 686\ \text{N}$. At the station's altitude, $r$ is only about 6% larger than $R_E$, so $g$ is about 11% smaller (so about 89% of its surface value). The force is still roughly $600\ \text{N}$. Gravity has hardly gone away.

Then why the floating? The station and the astronaut are both falling around Earth with the same acceleration (the local $g$). The floor never pushes up on the astronaut, because the floor is accelerating at the same rate. A bathroom scale would read zero. This is called apparent weightlessness: the astronaut still has weight in the sense of the gravitational force, but nothing is pushing back on them, so they feel no weight.

Net gravitational force from several objects

When more than two masses are involved, each one pulls on your object separately. The net force is the vector sum of those individual pulls. Do it in three steps:

1. Find the magnitude of each pull with $F = Gm_1m_2/r^2$, using that pair's distance.

2. Draw each force as an arrow pointing toward the object that causes it (gravity always attracts).

3. Add the forces by components, using a sign for each direction.

If the objects are not in a line, split each force into $x$ and $y$ components with $F\cos\theta$ and $F\sin\theta$ (see your force-components notes), then add components separately. Symmetry can save work: pulls that are mirror images in $x$ cancel in $x$.

When the net force on an object is exactly zero, the object is in equilibrium. For two bodies, such a point lies where the pulls match in magnitude. Setting $\dfrac{GM_1m}{r_1^2} = \dfrac{GM_2m}{r_2^2}$ shows that $m$ and $G$ cancel, so the location depends only on the two big masses and the distances.

Back to the start: the astronaut floats not because gravity has gone away, but because the astronaut and the station are falling together.

Summary