Key Questions:


Dropped objects fall to Earth because they are pulled by a force.

Newton's idea:

The Moon's "fall" is its deviation from a straight path.
In one second, the Moon falls 0.00136 meters (about 1.4 mm).
An object on the Earth (e.g., an apple), falls 4.9 meters in 1 second.
The distance to the Moon is about 60 Earth radii.
(4.9 meters)/(0.00136 meters) = 3600.


The orbits of planets around the Sun are approximately circular.
For now, ignore the small departures from circularity.

Let mp denote the mass of a planet.
Let vp denote the speed of the planet in its orbit.
Let rp denote the radius of the orbit.
(We use r instead of a because we're going to use a for acceleration.)
As shown in class, Kepler's 3rd law leads to the conclusion

vp2 ~ 1 / rp,
the square of a planet's orbital speed is inversely proportional to its orbital radius.

Recall that the acceleration in a circular orbit at constant speed is

a = v2/r.
Combined with Kepler's third law, this implies
ap = vp2 / rp ~ (1 / rp) / rp = 1 / rp2.

What force is needed to keep planets in circular orbits obeying Kepler's 3rd law?

F = ma = mpap ~ mp / rp2.
The force should be proportional to the planet's mass and proportional to the inverse square of the planet's distance from the Sun.


Newton proposes the following: There is a force of gravitational attraction between any two bodies that is proportional to their masses and inversely proportional to the square of the distance between their centers.

In the form of an equation:

F = G M1 M2 / d2
F = gravitational force
M1 = mass of body 1
M2 = mass of body 2
d = distance between their centers
G = Gravitational Force Constant (a.k.a. Newton's constant).

Double the mass of one body, and the force doubles.
Double the mass of both bodies, and the force goes up by four.
Double the distance between the bodies, and force goes down by four.
Halve the distance between the bodies, and force goes up by four.


The gravitational force is

F = G M1 M2 / d2.
The acceleration of body 2
a2 = F/M2 = G M1/d2
does not depend on the mass of body 2!
The gravitational acceleration of an object does not depend on its mass, because the extra force on a more massive object is canceled out by its greater resistance to acceleration.
This is a deep fact that would lead Einstein to a new theory of gravity two centuries later.

Newton posits the existence of the gravitational force between separated bodies.
He does not try to explain a "cause" for it.
Some scientists of the time considered this "action at a distance" to be a fatal weakness of Newton's theory.


F = G M1 M2 / d2.
M1 = mass of Sun
M2 = mass of planet
d = distance from Sun to planet.

We showed before that Kepler's 3rd law is explained (for circular orbits) if the force from the Sun is proportional to the planet's mass and inversely proportional to the square of the orbital radius.

Newton's law of gravity is just what's needed.

Newton proves that planets accelerated by an inverse square force from the Sun

Equal area rule would hold for any force directed towards the Sun.
Principle of "conservation of angular momentum."
Amount of "circular" motion (velocity perpendicular to Sun-planet line) cannot be changed by a force towards the Sun.


F = G M1 M2 / d2.
M1 = mass of Earth
M2 = mass of falling object.

What is d?
Newton shows: what matters is distance to center of the Earth.

d = radius of Earth.

a2 = F/M2 = G MEarth / REarth2.

Same for all masses M2.
No dependence on height above ground, as long as height is much smaller than REarth.

Acceleration at Earth's surface is usually denoted g.
Value is g = 9.8 m/sec2, can be measured by dropping objects.


F = G M1 M2 / d2.
M1 = mass of Earth
M2 = mass of Moon
d = Earth-Moon distance.

Can compare acceleration at Earth's surface (9.8 m/sec2) to acceleration of Moon in its orbit (v2/r = 0.0027 m/sec2).
Earth-Moon distance = 60 REarth.
9.8/0.0027 = 3600 = 602.
It works!

A complication:



Consider an object in circular orbit held by gravity of central mass M.

Scaling to Earth-Sun system:
(r / 1 AU) x (v / vearth)2 = (M / Msun).
In terms of period P = 2\pi r / v,
(r / 1 AU)3 = (P / 1 year)2 x (M / Msun).
Textbook gives a still more general form of Kepler's 3rd Law (box 4-4), which applies in case where orbiting object is not much smaller in mass than central object.
Gives same result as above when central object is dominant (e.g., Sun for planets, Jupiter for its moons).

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Updated: 2005 April 17[dhw]