From Kepler’s Law to Newton’s Gravity, Yourself — Part 1

Now that you’ve discovered Kepler’s third law — that T, the orbital time of a planet in Earth years, and R, the radius of the planet’s orbit relative to the Earth-Sun distance, are related by

  • R3=T2

the question naturally arises: where does this wondrous regularity comes from?

We have been assuming that planets travel on near-circular orbits, and we’ll continue with that assumption to see what we can learn from it. So let’s look in more detail at what happens when any object, not just a planet, travels in a circle at a constant speed.

Circular Motion at Constant Speed

If an object is on a circle of radius R (and therefore diameter 2R and circumference 2πR), and takes time T to go round the circle at a constant speed v, then its speed is its distance covered (the circle’s circumference) divided by the time it takes to cover it, namely

  • v = (2π R) / T

In Figure 1 I’ve depicted an object at four moments (marked 1, 2, 3, 4) during its circular motion. At each location, I’ve drawn an arrow indicating its position, relative to the center of the circle, in blue, and an arrow in red that indicates its velocity — its speed and the direction of its motion. Notice the speed of the object does not change as it moves from point 1 to point 2 and so on — all the red arrows have the same length. But as the object’s position goes round in a circle, the direction of its motion is going round in a circle too.

Figure 1: An object going round in a circle (from point 1 to point 2 to point 3 to point 4 and back to 1) at constant speed. Its position is given by the blue arrows; its distance from the center does not change even as its direction from the center changes. Its motion is indicated by the red arrows; its speed does not change even though the direction of its motion changes.

Velocity measures the change in an object’s position. Similarly, acceleration measures the change in an object’s velocity. We can carry the analogy further. Just as the position arrow (the blue one) has length R and goes round in a circle at a rate v in time T, the velocity arrow (the red one) has length v and goes round in a circle at a rate a in time T. So if v = (2π R)/ T , it must be that

  • a = (2π v)/ T

This is indicated in Figure 2, which shows the red velocity arrows at the moments 1, 2, 3, 4 (notice how they are shifted from Figure 1), and also the corresponding acceleration arrows in green, which indicate how the velocity arrow is changing as time goes by.

Figure 2: The velocity (red arrows) and acceleration, the change in velocity (green arrows) of the same object; the acceleration, too, is of a constant amount but varying direction. Compare the red arrows and the point locations, as well as the figure as a whole, to Figure 1.

Compare Figures 1 and 2 carefully. Acceleration is to velocity as velocity is to position, so a = (2π v)/ T just as v =(2π R)/ T . The only differences are in the directions of the arrows at the moments 1, 2, 3 and 4.

Now let’s combine Figures 1 and 2. First, let’s use our formula for v to rewrite our formula for a

  • a = (2π v)/ T = (2π [2π R/T] / T = (2π)2 R / T2

That tells us how much acceleration (how much change in the velocity’s direction) is going on for an object on a circle with radius R and period T. As far as I know, this formula was first pointed out by Christiaan Huygens around 1659, after Galileo’s death and before Newton’s appearance on the proto-scientific scene.

Now let’s put all three arrows — distance, velocity and acceleration — on the same figure. You can see in Figure 3 that the acceleration arrow points opposite to the position arrow; at all times, the acceleration points inward toward the center of the circle.

Figure 3: Combining Figures 1 and 2 reveals that as the object moves around the circle at constant speed, its acceleration (green arrows) always points toward the center of the circle, opposite to the position arrows (blue) and perpendicular to the velocity arrows (red.).

This may initially seem counter-intuitive. But imagine swinging a ball on a rope around above your head in a circle. To keep the ball from flying away from you, and maintain the motion on a circle, the rope has to pull inward, toward your hand. That is a hint that the rope is creating an inward-tending force, which in turn creates an inward acceleration, just enough to keep the ball moving in a circle at constant speed.

Figure 4: If you take a ball on a string and swing it in a circle over your head, at each moment the string stretches from your hand to the ball, pulling (and accelerating) the ball toward your hand.

Circular Motion for the Planets

For planets, we need not stop here, because Kepler’s third law relates R and T, so we can replace our formula for the acceleration with a simpler one. Because for a planet T2 = R3, where T is measured in Earth-years and R is measured relative to the Earth-Sun distance, we may substitute for the T2 in our acceleration formula above

  • a = (2π)2 R / T2 = (2π)2 R / ( R3 )= (2π)2 / R2

where R is expressed in units of the Earth-Sun distance. (The units look funny, and they are, but I’ll explain that in a moment for those who are interested.) Remarkably, the acceleration of each planet is inward toward the Sun, as though the Sun were pulling on each one with an invisible rope, by an amount given by what we call an inverse square law: acceleration is proportional to 1/R2 .

This is our first hint of Newton’s law of gravity, as known also by Robert Hooke and perhaps others. But so far it’s just a hint.

Aside: Interpreting this Formula for Acceleration

[This section can be skipped if you’re not interested in the details.]

How exactly should we interpret the units in our formula for a and R? Acceleration tells us the change in velocity that occurs over time. When we used Kepler’s law, we were measuring time in Earth-years (not in seconds or minutes or days, as we often do) and length in Earth-Sun distances (not in feet or meters or miles or kilometers, as we often do.) So whereas we often speak of speed in miles-per-hour or meters-per-second, here we are speaking of speed in Earth-Sun distances per Earth-year. And a change in velocity — the acceleration a in the formula above — is the change in the velocity per Earth-year. Putting that together, the acceleration (2π)2 / R2 is in units of Earth-Sun distances per Earth-year per Earth-year!

Let’s put this in more familiar units that make sense in daily life, focusing on a planet at the Earth-Sun distance, with R=1. Its acceleration is then (2π)2 Earth-Sun distances per Earth-year per Earth-year. We can write that in more familiar terms, such as meters per second per second, by using the fact that the Earth-Sun distance is about 150,000,000,000 meters and a year lasts about 31,500,000 seconds. That gives, for R=1,

  • a = (2π)2 (1.50 1011 meters)/(3.15 107 seconds)2 = 0.006 meters per second per second = 6 millimeters per second per second

This is a very small acceleration in terms of daily life. When a car, initially stationary, accelerates to highway speeds over a period of 10 seconds, it accelerates at about 3 meters per second per second. If a car only accelerated at 6 millimeters per second per second, it would take it more than an hour to reach highway speed!

Why Didn’t Kepler Know the Inverse Square Law?

Why didn’t Kepler immediately know, when he realized T2=R3, that this was due to an inward acceleration satisfying an inverse square law? Kepler suspected the Sun was moving the planets, but the problem was that he didn’t imagine that something might be pulling each planet toward the Sun itself. He had a conceptual bias that kept him from realizing what was going on.

In daily life, objects that aren’t pushed soon come to a stop. It seems Kepler was under the misapprehension that this applied to the planets as well, and that the Sun would have to continually push them forward in the direction of their velocity, the way a parent might push a baby stroller along a sidewalk. Otherwise, he imagined, they would soon cease to move at all.

But Kepler had it wrong. Ordinary objects in daily life stop moving because of frictional forces, which arise when objects are rubbing up against each other. If space were full of a fluid like honey, which exerts a strong frictional force on objects that try to move through it, then Kepler’s thinking would have made sense: without a force pushing the planets forward around their orbits, they would quickly come to a stop in the honey. But as was gradually guessed by Galileo and Newton and others in the 17th century, objects in truly empty space do not slow down of their own accord. They will continue to move in straight lines at a constant speed, unless some external effect causes them to deviate from that steady motion. And the planets do indeed move in almost perfectly empty space. In empty space, something pulling inward with just the right strength is enough to change a planet’s preferred steady, straight-line motion to steady circular motion — or to more complex motions.

That an inward force can create circular motion out of straight-line motion may seem deeply counter-intuitive. But it is also correct. You can catch a glimpse of this when you swing a ball on a circle over your head, and let go of the rope. When you do so, the force of the string on the ball suddenly ceases, and instantly the ball will start moving in a straight line away from you. (The ball will also start to fall, to be sure, but this is a distraction. Without the Earth’s gravity, the released ball would have moved in a straight line forever.)

Figure 5: If you are spinning a ball over your head as in Figure 4, and you let go of the ball at the point where the blue arc meets the red line, then the ball will move initially along the red dashed line. It would remain on that line indefinitely were it not for gravity, which causes the ball to fall as it moves horizontally. (Try this yourself.)

Onward to Newton and Gravity?

Once you realize that something is making the planets accelerate toward the Sun, it’s easy to imagine that a similar effect is causing the Moon to accelerate toward the Earth, and thus might explain the Moon’s near-circular orbit of our planet.

But we already know of something that causes objects to accelerate toward the Earth. We call it “gravity,” and we imagine it as pulling our bodies and our dropped keys toward the ground. Since these two effects point in the same direction, perhaps they could actually be the same? Could it be gravity that holds the Moon to the Earth, and, by extension, the planets to the Sun?

If so, then the acceleration of a dropped ball at Earth’s surface should be related to the acceleration of the Moon as it circles the Earth. But how exactly are they related? That was the question that faced Isaac Newton.

10 responses to “From Kepler’s Law to Newton’s Gravity, Yourself — Part 1

  1. this series is fantastic!! i teach economics and i might try to use these examples and the interplay of empirical work, measurement tools (the lens!), and theoretical insight to introduce the importance of each to modeling on the first day of class next year. thank you thank you!!

  2. Slobodan Nedic

    Even if Kepler might have been aware of this implication of his third law/rule (by possibly having been familiar with the Huygens’s centrifugal acceleratin formula), that would not be sufficient for him, since he was ‘biased’ by the physicallity of presence of both attracting and repulsung tendencies. It is true that the last minute modified Brahe’s measurements brought him to arrive at the Kepler’s Equatinn, he subsequently attributed it – along the Second law – to merely Solar action. And Hook might have been aware of these issues …On the other side, Newton had not account for the proper angle-time dependence implied by the Kepler’s Equation, and was just lucky enough ;(

    • This is an interesting comment, but I’m not sure I understand all of it — maybe there is some history here that I do not know. Can you point me to anything to read about this?

  3. Inverse square law applies in-between weights NOT Masses?

  4. The acceleration is always (constant) 6 millimeters per second per second , if it is go beyond, will go out of the Orbit?

    • If the acceleration (inward) is greater, given the same speed and location, it will fall inward from that point, and end up on an elliptical orbit which is inside the circle. If the acceleration is less, then it will end up on an elliptical orbit which is outside the circle. If the acceleration is sufficiently small, then the object will escape — if it is zero, it will escape on a straight line.

      • In a curved spacetime if the acceleration is conserved?

        • Phenomenology as physical law:
          Tethering of balloon (the Rest Mas).
          PUshing a harboured Ship with hands.
          The equivalence of gravitational and inertial mass (constant 6 millimeters per second per second).
          Time dilation (in one way act like the Photon in Space).
          Allmost non-repeating pattern (Diffeomorphism).
          “Time dilation and rest-Mass cannot co-exist”.

  5. Slobodan Nedic

    Could you, please, let me know if you have received my response with links to materials related to my comment – please write to nedic.slbdn@aol.com ?!