## Newton’s Universal Law of Gravitation

What do aching feet, a falling apple, and the orbit of the Moon have in common? Each is caused by the gravitational force. Our feet are strained by supporting our weight—the force of Earth’s gravity on us. An apple falls from a tree because of the same force acting a few meters above Earth’s surface. And the Moon orbits Earth because gravity is able to supply the necessary centripetal force at a distance of hundreds of millions of meters.

In fact, the same force causes planets to orbit the Sun, stars to orbit the center of the galaxy, and galaxies to cluster together. Gravity is another example of underlying simplicity in nature. It is the weakest of the four basic forces found in nature, and in some ways the least understood. It is a force that acts at a distance, without physical contact, and is expressed by a formula that is valid everywhere in the universe, for masses and distances that vary from the tiny to the immense.

Sir Isaac Newton was the first scientist to precisely define the gravitational force, and to show that it could explain both falling bodies and astronomical motions. See the figure below. But Newton was not the first to suspect that the same force caused both our weight and the motion of planets. His forerunner Galileo Galilei had contended that falling bodies and planetary motions had the same cause.

Some of Newton’s contemporaries, such as Robert Hooke, Christopher Wren, and Edmund Halley, had also made some progress toward understanding gravitation. But Newton was the first to propose an exact mathematical form and to use that form to show that the motion of heavenly bodies should be conic sections—circles, ellipses, parabolas, and hyperbolas. This theoretical prediction was a major triumph—it had been known for some time that moons, planets, and comets follow such paths, but no one had been able to propose a mechanism that caused them to follow these paths and not others.

The gravitational force is relatively simple. It is always attractive, and it depends only on the masses involved and the distance between them. Stated in modern language, **Newton’s universal law of gravitation** states that every particle in the universe attracts every other particle with a force along a line joining them. The force is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.

### Misconception Alert

The magnitude of the force on each object (one has larger mass than the other) is the same, consistent with Newton’s third law.

The bodies we are dealing with tend to be large. To simplify the situation we assume that the body acts as if its entire mass is concentrated at one specific point called the **center of mass** (CM), which will be further explored in a tutorial on Linear Momentum and Collisions. For two bodies having masses \(m\) and \(M\) with a distance \(r\) between their centers of mass, the equation for Newton’s universal law of gravitation is

\(F=G\cfrac{\text{mM}}{{r}^{2}}\text{,}\)

where \(F\) is the magnitude of the gravitational force and \(G\) is a proportionality factor called the **gravitational constant**. \(G\) is a universal gravitational constant—that is, it is thought to be the same everywhere in the universe. It has been measured experimentally to be

\(G=6\text{.}\text{674}×{\text{10}}^{-\text{11}}\cfrac{\text{N}\cdot {\text{m}}^{2}}{{\text{kg}}^{2}}\)

in SI units. Note that the units of \(G\) are such that a force in newtons is obtained from \(F=G\cfrac{\text{mM}}{{r}^{2}}\), when considering masses in kilograms and distance in meters. For example, two 1.000 kg masses separated by 1.000 m will experience a gravitational attraction of \(6\text{.}\text{674}×{\text{10}}^{-\text{11}}\phantom{\rule{0.25em}{0ex}}\text{N}\). This is an extraordinarily small force. The small magnitude of the gravitational force is consistent with everyday experience. We are unaware that even large objects like mountains exert gravitational forces on us. In fact, our body weight is the force of attraction of the *entire Earth* on us with a mass of \(6×{\text{10}}^{\text{24}}\phantom{\rule{0.25em}{0ex}}\text{kg}\).

Recall that the acceleration due to gravity \(g\) is about \(9.80 m{\text{/s}}^{2}\) on Earth. We can now determine why this is so. The weight of an object *mg* is the gravitational force between it and Earth. Substituting *mg* for \(F\) in Newton’s universal law of gravitation gives

\(\text{mg}=G\cfrac{\text{mM}}{{r}^{2}}\text{,}\)

where \(m\) is the mass of the object, \(M\) is the mass of Earth, and \(r\) is the distance to the center of Earth (the distance between the centers of mass of the object and Earth). See the figure below. The mass \(m\) of the object cancels, leaving an equation for \(g\):

\(g=G\cfrac{M}{{r}^{2}}\text{.}\)

Substituting known values for Earth’s mass and radius (to three significant figures),

\(g=(6\text{.}\text{67}×{\text{10}}^{-\text{11}}\cfrac{\text{N}\cdot {\text{m}}^{2}}{{\text{kg}}^{2}})×\cfrac{5\text{.}\text{98}×{\text{10}}^{\text{24}}\phantom{\rule{0.25em}{0ex}}\text{kg}}{(6\text{.}\text{38}×{\text{10}}^{6}\phantom{\rule{0.25em}{0ex}}\text{m}{)}^{2}}\text{,}\)

and we obtain a value for the acceleration of a falling body:

\(g=9\text{.}\text{80}\phantom{\rule{0.25em}{0ex}}{\text{m/s}}^{2}.\)

This is the expected value *and is independent of the body’s mass*. Newton’s law of gravitation takes Galileo’s observation that all masses fall with the same acceleration a step further, explaining the observation in terms of a force that causes objects to fall—in fact, in terms of a universally existing force of attraction between masses.

### Optional Take-Home Experiment

Take a marble, a ball, and a spoon and drop them from the same height. Do they hit the floor at the same time? If you drop a piece of paper as well, does it behave like the other objects? Explain your observations.

### Making Connections

Attempts are still being made to understand the gravitational force. As we shall see in a tutorial on Particle Physics, modern physics is exploring the connections of gravity to other forces, space, and time. General relativity alters our view of gravitation, leading us to think of gravitation as bending space and time.

In the following example, we make a comparison similar to one made by Newton himself. He noted that if the gravitational force caused the Moon to orbit Earth, then the acceleration due to gravity should equal the centripetal acceleration of the Moon in its orbit. Newton found that the two accelerations agreed “pretty nearly.”

## Example: Earth’s Gravitational Force Is the Centripetal Force Making the Moon Move in a Curved Path

(a) Find the acceleration due to Earth’s gravity at the distance of the Moon.

(b) Calculate the centripetal acceleration needed to keep the Moon in its orbit (assuming a circular orbit about a fixed Earth), and compare it with the value of the acceleration due to Earth’s gravity that you have just found.

**Strategy for (a)**

This calculation is the same as the one finding the acceleration due to gravity at Earth’s surface, except that \(r\)is the distance from the center of Earth to the center of the Moon. The radius of the Moon’s nearly circular orbit is \(3\text{.}\text{84}×{\text{10}}^{8}\phantom{\rule{0.25em}{0ex}}\text{m}\).

**Solution for (a)**

Substituting known values into the expression for \(g\) found above, remembering that \(M\) is the mass of Earth not the Moon, yields

\(\begin{array}{lll}g& =& G\cfrac{M}{{r}^{2}}=(6\text{.}\text{67}×{\text{10}}^{-\text{11}}\cfrac{\text{N}\cdot {\text{m}}^{2}}{{\text{kg}}^{2}})×\cfrac{5\text{.}\text{98}×{\text{10}}^{\text{24}}\phantom{\rule{0.25em}{0ex}}\text{kg}}{(3\text{.}\text{84}×{\text{10}}^{8}\phantom{\rule{0.25em}{0ex}}\text{m}{)}^{2}}\\ & =& 2\text{.}\text{70}×{\text{10}}^{-3}\phantom{\rule{0.25em}{0ex}}{\text{m/s.}}^{2}\end{array}\)

**Strategy for (b)**

Centripetal acceleration can be calculated using either form of

\(\left.\begin{array}{} a_c = \cfrac{v^2}{r} \\ a_c = r\omega^2 \end{array} \right\}.\)

We choose to use the second form:

\({a}_{c}={\mathrm{r\omega }}^{2}\text{,}\)

where \(\omega \) is the angular velocity of the Moon about Earth.

**Solution for (b)**

Given that the period (the time it takes to make one complete rotation) of the Moon’s orbit is 27.3 days, (d) and using

\(1 d×24\cfrac{\text{hr}}{\text{d}}×60\cfrac{min}{\text{hr}}×60\cfrac{s}{\text{min}}=\text{86,400 s}\)

we see that

\(\omega =\cfrac{\text{Δ}\theta }{\text{Δ}t}=\cfrac{2\pi \phantom{\rule{0.25em}{0ex}}\text{rad}}{(\text{27}\text{.}\text{3 d})(\text{86,400 s/d})}=2\text{.}\text{66}×{\text{10}}^{-6}\cfrac{\text{rad}}{\text{s}}.\)

The centripetal acceleration is

\(\begin{array}{} a_c & = r\omega^2 & = (3.84 \times 10^8\text{ m})(2.66 \times 10^{-6}\text{ rad/s})^2 \\ & = 2.72 \times 10^{-3}\text{ m/s}^2. & \end{array}\)

The direction of the acceleration is toward the center of the Earth.

**Discussion**

The centripetal acceleration of the Moon found in (b) differs by less than 1% from the acceleration due to Earth’s gravity found in (a). This agreement is approximate because the Moon’s orbit is slightly elliptical, and Earth is not stationary (rather the Earth-Moon system rotates about its center of mass, which is located some 1700 km below Earth’s surface). The clear implication is that Earth’s gravitational force causes the Moon to orbit Earth.

Why does Earth not remain stationary as the Moon orbits it? This is because, as expected from Newton’s third law, if Earth exerts a force on the Moon, then the Moon should exert an equal and opposite force on Earth (see the figure below). We do not sense the Moon’s effect on Earth’s motion, because the Moon’s gravity moves our bodies right along with Earth but there are other signs on Earth that clearly show the effect of the Moon’s gravitational force as discussed in the next topic on Satellites and Kepler’s Laws: An Argument for Simplicity.