Gravitational Force

Two $5.0\ \text{kg}$ bowling balls sitting $0.30\ \text{m}$ apart pull on each other with about $2 \times 10^{-8}\ \text{N}$, roughly the weight of a grain of fine sand. The same law, applied to the $6 \times 10^{24}\ \text{kg}$ of Earth, pulls each of those balls down with $49\ \text{N}$. Gravity is the weakest force you deal with in mechanics, and it dominates anyway because it only attracts, so the pulls from every bit of a planet add up instead of canceling.

Newton's law of universal gravitation

Every pair of objects with mass attracts. For two objects whose centers of mass are a distance $r$ apart, the force each exerts on the other has magnitude

In vector form, with $\hat{r}$ the unit vector pointing from object 1 toward object 2 (topic 1.1), the force on object 2 is $\vec{F}_{12} = -\frac{G m_1 m_2}{r^2}\,\hat{r}$. The minus sign is the attraction: the force on 2 points back toward 1.

The opening numbers come straight from the formula: $F = \frac{(6.67\times10^{-11})(5.0)(5.0)}{(0.30)^2} = 1.9 \times 10^{-8}\ \text{N}$. The force is proportional to each mass and to $\frac{1}{r^2}$, so tripling the separation divides the force by 9, and doubling one mass doubles it. Write the new force as a ratio to the old one and every constant cancels.

The gravitational field and weight

A mass changes the space around it so that any other mass placed there feels a force. The gravitational field is a way to describe that: at each point, it is the force a small test mass $m$ would feel, divided by $m$. For a source of mass $M$ at distance $r$,

The field depends only on the source and on where you are, not on the test mass. Its unit is $\text{N/kg}$, which is the same as $\text{m/s}^2$. That is no accident: if gravity is the only force on an object, Newton's second law gives $m\vec{a} = m\vec{g}$, so the object's acceleration equals the field at its location. A field of $9.8\ \text{N/kg}$ means a falling rock speeds up by $9.8\ \text{m/s}$ every second.

The gravitational force a planet or moon exerts on a small object near it is called the object's weight, $F_g = mg$. Earth, with $M_E = 5.97 \times 10^{24}\ \text{kg}$ and $R_E = 6.37 \times 10^{6}\ \text{m}$, gives $g = \frac{GM_E}{R_E^2} = 9.81\ \text{N/kg}$ at its surface.

When g counts as constant. The field falls off with distance, but slowly compared with Earth's size. A jet at $10\ \text{km}$ altitude is at $r = 6.38 \times 10^6\ \text{m}$, and $g$ there is smaller by a factor $\left(\frac{6.37}{6.38}\right)^2 = 0.997$, a change of $0.3\%$. So whenever an object's distance from the center changes by a tiny fraction of that distance, you treat the gravitational force as constant over the whole motion and use $g \approx 9.8\ \text{m/s}^2$ near Earth's surface. For a satellite whose distance changes by thousands of kilometers, you can't: you use $\frac{GM}{r^2}$ at each $r$.

Apparent weight, weightlessness, and two kinds of mass

A bathroom scale doesn't measure gravity. It measures how hard it pushes up on you, the normal force. That push is your apparent weight: its magnitude is the magnitude of the normal force. Apply Newton's second law to you, with up positive and $a$ your acceleration:

If you are not accelerating, $F_N = mg$ and the scale reads your true weight. If you accelerate upward ($a > 0$), the scale reads more; downward ($a < 0$), less. The sign of the acceleration decides it, not the direction you are moving. An elevator moving up and slowing down has a downward acceleration, so you feel light.

You appear weightless when nothing pushes on you, which happens in two cases: no forces act at all (deep space, far from any mass), or gravity is the only force acting (free fall, $a = -g$, so $F_N = 0$). Astronauts on the International Space Station, about $400\ \text{km}$ up, are in a field of about $8.7\ \text{N/kg}$, close to $90\%$ of the surface value. They float because they and the station fall together around Earth, so the floor never pushes on them.

The equivalence principle goes further. Inside a closed room, you can't tell whether the floor presses on you because the room sits at rest in a gravitational field $g$ or because the room accelerates at $g$ through empty space. Every measurement you could make inside, from a scale reading to a dropped ball, comes out the same. Gravity and acceleration of your reference frame are locally indistinguishable.

That only works because of a striking experimental fact. Mass shows up in two separate laws:

• Inertial mass is the $m$ in $\vec{F}_{\text{net}} = m\vec{a}$: how strongly an object resists having its motion changed by any interaction.
• Gravitational mass is the $m$ in $\frac{GMm}{r^2}$: how strongly an object is pulled by gravity.

Nothing in Newton's laws forces these to be equal. If they weren't, a falling object would have $a = \frac{m_{\text{grav}}}{m_{\text{inertial}}}\,g$, and a lead ball and a wooden ball would fall at different rates. Experiments have compared the two for more than a century, from Eötvös's torsion balances in the early 1900s (agreement to a few parts in a billion) to the MICROSCOPE satellite in 2022 (about one part in $10^{15}$). They are equivalent, which is why every object in a given field has the same free-fall acceleration and why you can write a single $m$ in both places.

Gravity from a sphere: the shell theorem

A planet is not a point. To find its pull on you, you'd split it into tiny pieces of mass $dm$, find each piece's force $\frac{G\,m\,dm}{s^2}$ along the line to that piece, and add the vectors (an integral). Newton did this for a sphere and found results simple enough to use without redoing the integral. You don't need to prove them, only apply them.

Why zero inside a shell? Stand off-center inside it and look along any narrow cone in both directions. The cone cuts a small patch out of the near wall and a larger patch out of the far wall. A patch's area, and so its mass, grows as the square of its distance $s$ from you, while each kilogram's pull falls as $\frac{1}{s^2}$. The two effects cancel, so opposite patches pull equally hard in opposite directions, and the whole shell sums to zero.

Inside a uniform solid sphere. Let the sphere have mass $M$, radius $R$, and density $\rho = \frac{M}{\frac{4}{3}\pi R^3}$. At distance $r < R$, the mass that pulls is the mass inside radius $r$:

The force grows in proportion to $r$ and points toward the center, so as a component along $\hat{r}$ it has the form $F_r = -kr$ with $k = \frac{GMm}{R^3}$. Dividing by $m$ gives the field shown below. At the center it is zero (pulls from every side cancel), it rises linearly to its surface value $g_0 = \frac{GM}{R^2}$, then falls off as $\frac{1}{r^2}$.

Worked examples

Where students lose points

On the exam

Functional-dependence questions ("the separation is tripled; by what factor does the force change?") are answered fastest as ratios. A derivation for a point inside a sphere earns credit for the step that uses only the enclosed mass and for expressing that mass through density or $\frac{r^3}{R^3}$. A claim about weightlessness or apparent weight needs the normal force named and Newton's second law applied, not a description of how the person feels. The value of $G$ and $g$ near Earth are on the equation sheet; the shell theorem is not, so state it in words when you use it.