How does newton's law of gravitation agree with kepler's observations?
Answers
To show how Kepler’s law comes from Newton’s laws of motion and his law of gravitation, we will first of all make two simplifying assumptions, to make the mathematics easier. First we will assume that the orbits are circular, rather than elliptical. Secondly, we will assume that the Sun is at the centre of a planet’s circular orbit. Neither of these assumptions is strictly true, but they will make the derivation much simpler.
Newton’s law of gravity states that the gravitational force between two bodies of masses M \text{ and } m is given by
F = \frac { G M m }{ r^{2} } \text{ (equation 2)}
where r is the distance between the two bodies and G is a constant, known as Newton’s universal gravitational constant, usually called “big G”. In the case we are considering here, r is of course the radius of a planet’s circular orbit about the Sun.
When an object moves in a circle, even at a constant speed, it experiences an acceleration. This is because the velocity is always changing, as the direction of the velocity vector is always changing, even if its size is constant. From Newton’s 2nd law, F=ma, which means if there is an acceleration there must be a force causing it, and for circular motion this force is known as the centripetal force. It is given by
F = \frac{ m v^{2} }{ r } \text{ (equation 3)}
where m is the mass of the moving body, v is its speed, and r is the radius of the circular orbit. This centripetal force in this case is provided by gravity, so we can say that
\frac{ G M m }{ r^{2} } = \frac{ m v^{2} }{ r }
With a little bit of cancelling out we get
\frac{ G M }{ r } = v^{2} \text{ (equation 4)}
But the speed v is given by the distance the body moves divided by the time it takes. For one full circle this is just
v = \frac{ 2 \pi r }{ T }
where 2 \pi r is the circumference of a circle and T is the time it takes to complete one full orbit, its period. Substituting this into equation (4) gives
\frac{ G M }{ r } = \frac{ 4 \pi^{2} r^{2} }{ T^{2} }
Doing some re-arranging this gives
\boxed{ T^{2} = \frac{4 \pi^{2} }{ GM } a^{3} } \text{ (equation 5)}
where we have substituted a for r. This, as you can see, is just Kepler’s 3rd law, with the constant of proportionality k found to be (4 \pi^{2})/(GM). So, Kepler’s 3rd law can be derived from Newton’s laws of motion and his law of gravity. The value of k above is true if we express a in metres and T in seconds. But, if we express a in Astronomical Units and T in Earth years, then k actually comes out to be 1!
Newton’s form of Kepler’s 3rd law
A web search for Newton’s form of Kepler’s 3rd law will turn up the following equation
(M + m) T^{2} = a^{3} \text{ (equation 6)}
How can we derive this? I will show how it is done in part 2 of this blog, as we will need to learn about something called “reduced mass”, and also the “centre of mass”.