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planet. (Although the sun does accelerate in response to the attractions of
the planets, its mass is so great that the effect had never been detected by
the prenewtonian astronomers.) Since the outer planets were moving slowly
along more gently curving paths than the inner planets, their accelerations
were apparently less. This could be explained if the sun’s force was deter-
mined by distance, becoming weaker for the farther planets. Physicists were
also familiar with the noncontact forces of electricity and magnetism, and
knew that they fell off rapidly with distance, so this made sense.
In the approximation of a circular orbit, the magnitude of the sun’s
force on the planet would have to be
F=ma=mv
2
/r .(1)
Now although this equation has the magnitude, v, of the velocity vector in
it, what Newton expected was that there would be a more fundamental
underlying equation for the force of the sun on a planet, and that that
equation would involve the distance, r, from the sun to the object, but not
the object’s speed, v — motion doesn’t make objects lighter or heavier.
Self-Check
If eq. (1) really was generally applicable, what would happen to an object
released at rest in some empty region of the solar system.
Equation (1) was thus a useful piece of information which could be
related to the data on the planets simply because the planets happened to be
going in nearly circular orbits, but Newton wanted to combine it with other
equations and eliminate v algebraically in order to find a deeper truth.
To eliminate v, Newton used the equation
v=
circumference
T
=2
p
r/T .(2)
Of course this equation would also only be valid for planets in nearly
circular orbits. Plugging this into eq. (1) to eliminate v gives
F=
4
p
2
mr
T
2
.(3)
This unfortunately has the side-effect of bringing in the period, T, which we
expect on similar physical grounds will not occur in the final answer. That’s
where the circular-orbit case, T
.
r
3/2
, of Kepler’s law of periods comes in.
Using it to eliminate T gives a result that depends only on the mass of the
planet and its distance from the sun:
F
.
m/r
2
.[ force of the sun on a planet of
mass m at a distance r from the sun;
same proportionality constant for all
the planets ]
(Since Kepler’s law of periods is only a proportionality, the final result is a
proportionality rather than an equation, and there is this no point in
hanging on to the factor of 4
p
2
.)
It would just stay where it was. Plugging v=0 into eq. (1) would give F=0, so it would not accelerate from rest, and
would never fall into the sun. No astronomer had ever observed an object that did that!
Chapter 10Gravity