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# 1.1: Discovering Gravity

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## Terrestrial Gravity: Galileo Analyzes a Cannonball Trajectory

From the earliest times, gravity meant the tendency of most bodies to fall to earth. In contrast, things that leaped upwards, like flames of fire, were said to have “levity”. Aristotle was the first writer to attempt a quantitative description of falling motion: he wrote that an object fell at a constant speed, attained shortly after being released, and heavier things fell faster in proportion to their mass. Of course this is nonsense, but in his defense, falling motion is pretty fastit’s hard to see the speed variation when you drop something to the ground. Aristotle most likely observed the slower motion of things falling through water, where buoyancy and fluid resistance dominate, and assumed that to be a slowed-down version of falling through airwhich it isn’t.

Galileo was the first to get it right. (True, others had improved on Aristotle, but Galileo was the first to get the big picture.) He realized that a falling body picked up speed at a constant ratein other words, it had constant acceleration (as he termed it, the word means “addition of speed” in Italian). He also made the crucial observation that, if air resistance and buoyancy can be neglected, all bodies fall with the same acceleration, bodies of different weights dropped together reach the ground at the same time. This was a revolutionary ideaas was his assertion that it should be checked by experiment rather than by the traditional method of trying to decipher what ancient authorities might have meant.

Galileo also noted that if a ball rolls without interference on a smooth horizontal surface, and friction and air resistance can be neglected, it will move with constant speed in a fixed directionin modern language, its velocity remains constant.

He considered the motion of an object when not subject to interference as its “natural” motion.

Using his terminology, then, natural horizontal motion is motion at constant velocity, and natural vertical motion is falling at constant acceleration.

But he didn’t stop therehe took an important further step, which made him the first in history to derive useful quantitative results about motion, useful that is to his boss, a duke with military interests. The crucial step was the realization that for a cannonball in flight, the horizontal and vertical motions can be analyzed independently. Here’s his picture of the path of a horizontally fired cannonball:

The vertical drop of the cannonball at the end of successive seconds, the lengths of the vertical lines ci, df, eh are the same vertical distances fallen by something dropped from rest. If you drop a cannonball over a cliff it will fall 5 meters in the first second, if you fire it exactly horizontally at 100 meters per second, it will still fall 5 meters below a horizontal line in the first second. Meanwhile, its horizontal motion will be at a steady speed (again neglecting air resistance), it will go 100 meters in the first second, another 100 meters in the next second, and so on. Vertically, it falls 5 meters in the first second, 20 meters total in two seconds, then 45 and so on.

Galileo drew the graph above of the cannonball’s position as a function of time, and proved the curve was parabolic. He went on to work out the range for given muzzle velocity and any angle of firing, much to the gratification of his employer.

## Moving Up: Newton Puts the Cannon on a very High Mountain

Newton asked the question: what if we put the cannon on a really high (imaginary, of course!) mountain above the atmosphere and fired the cannon really fast? The cannonball would still fall 5 meters in the first second (ignoring the minor point that g goes down a bit on a really high mountain), but if it’s going fast enough, don’t forget the curvature of the earth! The surface of the earth curves away below a horizontal line, so if we choose the right speed, after one second the cannonball will have reached a point where the earth’s surface itself has dropped away by 5 meters below the originally horizontal straight line. In that case, the cannonball won’t have lost any height at all defining “height” as distance above the earth’s surface.

Furthermore, “vertically down” has turned around a bit (it means perpendicular to the earth’s surface) so the cannonball is still moving “horizontally”, meaning moving parallel to the earth’s surface directly beneath it. And, since it’s above the earth’s atmosphere, it won’t have lost any speed, so exactly the same thing happens in the next second, and the nextit therefore goes in a circular path. Newton had foreseen how a satellite would movehere’s his own drawing, with VD, VE and VF representing the paths of successively faster shots:

Newton’s brilliant insightthe above pictureis fully animated in my applet here.

We can find how fast the cannonball must move to maintain the circular orbit by using Pythagoras’ theorem in the diagram below (which grossly exaggerates the speed so that you can see how to do the proof).

The cannonball fired from point P goes v meters horizontally in one second and drops 5 meters vertically, and, if v has the right value, the cannonball will still be the same distance R from the earth’s center it was at the beginning of the second. (Bear in mind that v is actually about a thousandth of R, so the change in the direction of “down” will be imperceptible, not like the exaggerated figure here.)

Knowing that the radius of the earth R is 6400 km, there is enough information in the above diagram to fix the value of v. Notice that there is a right angled triangle with sides R and v and hypoteneuse $$R+5$$. Applying Pythagoras’ theorem

(R + 5)2 = R2 + v2,

R2 + 10R + 25 = R2 + v2.

Newton knew (in different units) that R = 6400 km, so the 25 in the above equation can be neglected to give:

v2 = 10R = 10×6400×1000 so v = 8000.

The units for v are of course meters per second, on our diagram we show v as a distance, that traveled in the first second.

So the cannonball must move at 8 km per second, or 5 miles per second if its falling is to match the earth’s curvaturethis is 18,000 mph, once round the earth in a little less than an hour and a half. This is in fact about right for a satellite in low earth orbit.

## Onward into Space: The Cannonball and the Moon

It occurred to Newton one day (possibly because of a falling apple) that this familiar gravitational force we experience all the time here near the surface of the earth might extend outwards as far as the moon, and in fact might be the reason the moon is in a circular orbit. The radius of the moon’s orbit (384,000 km) and its speed in orbit (about 1 km per second) had long been known (see my notes here if you’re interested in how it was measured), so it was easy to find, using the same Pythagorean arguments as used for the cannonball above, that the moon “falls” 1.37 millimeters below a straight line trajectory in one second.

That is to say, the ratio of the moon’s acceleration towards the center of the earth to the cannonball’s is 1.37/5000, or about 1/3600.

But the radius of the moon’s orbit is about 60 times greater than the cannonball’s (which is just the radius of the earth, approximately). Since 60×60 = 3600, Newton concluded that the gravitational force decreased with distance as 1/r2.

## Newton’s Universal Law of Gravitation

Newton then boldly extrapolated from the earth, the apple and the moon to everything, asserting his Universal Law of Gravitation:

Every body in the universe attracts every other body with a gravitational force that decreases with distance as 1/r2.

But actually he knew more about the gravitational force: from the fact that bodies of different masses near the earth’s surface accelerate downwards at the same rate, using F = ma (his Second Law) if two bodies of different masses have the same acceleration they must be feeling forces in the same ratio as their masses (so a body twice as massive feels twice the gravitational force), that is, the gravitational force of attraction a body feels must be proportional to its mass.

Now suppose we are considering the gravitational attraction between two bodies (as we always are), one of mass m1, one of mass m2. By Newton’s Third Law, the force body 1 feels from 2 is equal in magnitude (but of course opposite in direction) to that 2 feels from 1. If we think of m1 as the earth, the force m2 feels is proportional to m2, as argued aboveso this must be true whatever m1 is. And, since the situation is perfectly symmetrical, the force must also be proportional to m1.

Putting all this together, the magnitude of the gravitational force between two bodies of masses m1 and m2 a distance r apart

$F = G\dfrac{m_1m_2}{r2} \tag{1}$

The constant G = 6.67×10-11 N.m2/kg2.

It is important to realize that G cannot be measured by any astronomical observations. For example, g at the surface of the earth is given by

$g = \dfrac{Gm_E}{rE^2} \tag{1}$

where mE is the mass and rE the radius of the earth. Notice that by measuring g, and knowing rE, we can find GmE. But this does not tell us what G is, since we don’t know mE! It turns out that this same problem arises with every astronomical observation. Timing the planets around the sun will give us GmSun. So we can figure out the ratio of the sun’s mass to the earth’s, but we can’t find an absolute value for either one.

The first measurement of G was made in 1798 by Cavendish, a century after Newton’s work. Cavendish measured the tiny attractive force between lead spheres of known mass. For details on how an experiment at the University of Virginia in 1969 improved on Cavendish’s work, click on the UVa Physics site here. Cavendish said he was “weighing the earth” because once G is measured, he could immediately find the mass of the earth mE from g = GmE/rE2, and then go on the find the mass of the sun, etc.

## Describing the Solar System: Kepler’s Laws

Newton’s first clue that gravitation between bodies fell as the inverse-square of the distance may have come from comparing a falling apple to the falling moon, but important support for his idea was provided by a detailed description of planetary orbits constructed half a century earlier by Johannes Kepler.

Kepler had inherited from Tycho Brahe a huge set of precise observations of planetary motions across the sky, spanning decades. Kepler himself spent eight years mathematically analyzing the observations of the motion of Mars, before realizing that Mars was moving in an elliptical path.

To appreciate fully how Kepler’s discovery confirmed Newton’s theory, it is worthwhile to review some basic properties of ellipses.

## Mathematical Interlude: Ellipses 101

A circle can be defined as the set of all points which are the same distance R from a given point, so a circle of radius 1 centered at the origin O is the set of all points distance 1 from O.

An ellipse can be defined as the set of all points such that the sum of the distances from two fixed points is a constant length (which must obviously be greater than the distance between the two points!). This is sometimes called the gardener’s definition: to set the outline of an elliptic flower bed in a lawn, a gardener would drive in two stakes, tie a loose rope between them, then pull the rope tight in all different directions to form the outline.

In the diagram, the stakes are at F1, F2, the red lines are the rope, P is an arbitrary point on the ellipse.

• CA is called the semimajor axis length a, CB the semiminor axis, length b.
• F1, F2 are called the foci (plural of focus).

Notice first that the string has to be of length 2a, because it must stretch along the major axis from F1 to A then back to F2, and for that configuration there’s a double length of string along F2A and a single length from F1 to F2. But the length A' F1 is the same as F2A, so the total length of string is the same as the total length A'A = 2a.

Suppose now we put P at B. Since F1B = BF2, and the string has length 2a, the length F1B = a

We get a useful result by applying Pythagoras’ theorem to the triangle F1BC,

$(F_1C)^2 = a^2 - b^2$

(We shall use this shortly.)

Evidently, for a circle, F1C = 0. The eccentricity - e of the ellipse is defined as the ratio of F1C to a, $e = \dfrac {F_1C}{ \alpha } = \sqrt {1- \left(\dfrac {b}{a} \right)^2}$ .

The eccentricity of a circle is zero. The eccentricity of a long thin ellipse is just below one.

F1 and F2 on the diagram are called the foci of the ellipse (plural of focus) because if a point source of light is placed at F1, and the ellipse is a mirror, it will reflectand therefore focusall the light to F2. (This can be proved using the string construction.)

An ellipse is essentially a circle scaled shorter in one direction: in (x, y) coordinates it is described by the equation $\dfrac {x^2}{a^2} + \dfrac {y^2}{b^2} = 1$ a circle being given by a = b.

In fact, in analyzing planetary motion, it is more natural to take the origin of coordinates at the center of the Sun rather than the center of the elliptical orbit. It is also more convenient to take $$( r, \theta )$$ coordinates instead of (x, y) coordinates, because the strength of the gravitational force depends only on r. Therefore, the relevant equation describing a planetary orbit is the $$( r, \theta )$$ equation with the origin at one focus. For an ellipse of semi major axis a and eccentricity e the equation is: $\dfrac {a(1-e^2)}{r} = 1 + e \cdot cos \theta$ .

It is not difficult to prove that this is equivalent to the traditional equation in terms of x, y presented above.

Kepler summarized his findings about the solar system in his three laws:

1. The planets all move in elliptical orbits with the Sun at one focus.

2. As a planet moves in its orbit, the line from the center of the Sun to the center of the planet sweeps out equal areas in equal times, so if the area SAB (with curved side AB) equals the area SCD, the planet takes the same time to move from A to B as it does from C to D.

For my Flashlet illustrating this law, click here.

3. The time it takes a planet to make one complete orbit around the sun T (one planet year) is related to its average distance from the sun R: $T^2 \propto R^3$ .

In other words, if a table is made of the length of year T for each planet in the solar system, and its average distance from the sun R, and $$\dfrac {T^2}{R^3}$$ is computed for each planet, the numbers are all the same.

These laws of Kepler’s are precise, but they are only descriptiveKepler did not understand why the planets should behave in this way. Newton’s great achievement was to prove that all this complicated behavior was the consequence of one simple law of attraction.

## How Newton’s Law of Universal Gravitation Explains Kepler’s Laws

Surprisingly, the first of Kepler’s lawsthat the planetary paths are ellipticalis the toughest to prove beginning with Newton’s assumption of inverse-square gravitation. Newton himself did it with an ingenious geometrical argument, famously difficult to follow. It can be more easily proved using calculus, but even this is nontrivial, and we shall not go through it in class. (The proof is given later in these notes, if you’re curious to see how it’s done.)

The best strategy turns out to be to attack the laws in reverse order.

## Kepler’s Third Law (for circular orbits)

It is easy to show how Kepler’s Third Law follows from the inverse square law if we assume the planets move in perfect circles, which they almost do. The acceleration of a planet moving at speed $$v$$ in a circular orbit of radius $$R$$ is $$v^2/R$$ towards the center. (Of course you already know this, but it is amusing to see how easy it is to prove using the Pythagoras diagram above: just replace the 5 meters by $$\dfrac{1}{2}$$ at2, the “horizontal” distance v by vt, write down Pythagoras’ theorem and take the limit of a very small time.)

Newton’s Second Law F = ma for a planet in orbit becomes:

$\dfrac {mv^2}{R} = G \dfrac {Mm}{R^2}. \tag{1.1.4}$

Now the time for one orbit is $$T = \dfrac {2 \pi R}{v}$$, so dividing both sides of the equation above by $$R$$, we find:

$\left( \dfrac {T}{2 \pi} \right)^2 = \dfrac {R^3}{GM} \tag{1.1.4}$

so

$\dfrac {T^2}{R^3} = \dfrac {4 \pi^2}{GM}. \tag{1.1.4}$

This is Kepler’s Third Law: $$\dfrac {T^2}{R^3}$$ has the same numerical value for all the sun’s planets.

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## Contributors

This page titled 1.1: Discovering Gravity is shared under a not declared license and was authored, remixed, and/or curated by Michael Fowler.

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