From Kepler’s Law to Newton’s Gravity, Yourself — Part 2

Sometimes, when you’re doing physics, you have to make a wild guess, do a little calculating, and see how things turn out.

In a recent post, you were able to see how Kepler’s law for the planets’ motions (R3=T2 , where R the distance from a planet to the Sun in Earth-Sun distances, and T is the planet’s orbital time in Earth-years), leads to the conclusion that each planet is subject to an acceleration a toward the Sun, by an amount that follows an inverse square law

  • a = (2π)2 / R2

where acceleration is measured in Earth-Sun distances and in Earth-Years.

That is, a planet at the Earth’s distance from the Sun accelerates (2π)2 Earth-distances per Earth-year per Earth-year, which in more familiar units works out (as we saw earlier) to about 6 millimeters per second per second. That’s slow in human terms; a car with that acceleration would take more than an hour to go from stationary to highway speeds.

What about the Moon’s acceleration as it orbits the Earth?  Could it be given by exactly the same formula?  No, because Kepler’s law doesn’t work for the Moon and Earth.  We can see this with just a rough estimate. The time it takes the Moon to orbit the Earth is about a month, so T is roughly 1/12 Earth-years. If Kepler’s law were right, then R=T2/3 would be 1/5 of the Earth-Sun distance. But we convinced ourselves, using the relation between a first-quarter Moon and a half Moon, that the Moon-Earth distance is less than 1/10 othe Earth-Sun distance.  So Kepler’s formula doesn’t work for the Moon around the Earth.

A Guess

But perhaps objects that are orbiting the Earth satisfy a similar law,

  • R3=T2 for Earth-orbiting objects

except that now T should be measured not in years but in Moon-orbits (27.3 days, the period of the Moon’s orbit around the Earth) and R should be measured not in Earth-Sun distances but in Moon-Earth distances?  That was Newton’s guess, in fact.

Newton had a problem though: the only object he knew that orbits the Earth was the Moon.  How could he check if this law was true? We have an advantage, living in an age of artificial satellites, which we can use to check this Kepler-like law for Earth-orbiting objects, just the way Kepler checked it for the Sun-orbiting planets.  But, still there was something else Newton knew that Kepler didn’t. Galileo had determined that all objects for which air resistance is unimportant will accelerate downward at 32 feet (9.8 meters) per second per second (which is to say that, as each second ticks by, an object’s speed will increase by 32 feet [9.8 meters] per second.) So Newton suspected that if he converted the Kepler-like law for the Moon to an acceleration, as we did for the planets last time, he could relate the acceleration of the Moon as it orbits the Earth to the acceleration of ordinary falling objects in daily life.

Geosynchronous Satellites

Let’s check first if satellites satisfy Kepler’s law, appropriately modified for Earth and Moon. Among artificial satellites, some of the most important for communications and the GPS system (nope, bad example — the GOES weather satellites are a better one) are called “geosynchronous.”  That means they orbit Earth once a day — so T, for them, is about 1/27.3 = 0.037 of a Moon orbit period.  Meanwhile they are launched to a height of 22,000 miles (36000 kilometers) above the Earth’s surface.  That’s about 1/11 = 0.091 of the Moon-Earth distance of 240,000 miles (384000 km) or so, as we confirmed in this post.  So let’s see:  does it work? If R=0.091, T = R3/2 =0.027 of a Moon orbit period. Hmm. That’s only 17 hours. Something is off..

Oh, wait a minute.  Maybe instead of measuring R as the distance from the Earth’s surface to the satellite, we should measure R as the distance from the Earth’s center to the satellite?  For the Moon, this difference isn’t very important — it’s only 2% — but for these satellites its a 20% difference! Then instead of 22000 miles we’d have 26000 miles (42000 km), 0.11 of the Moon-Earth distance. And if R=0.11, T = R3/2 =0.036 of a Moon orbit period — about 24 hours. Now it all works!

The International Space Station

Let’s check also the international space station, which is a mere 250 miles (410 km) above the Earth’s surface, and completes an orbit every 90 minutes   If you live at mid-latitudes, you can actually check this directly: sometimes you can see it pass across the sky twice within two hours!  (And you can check its height, too, by seeing how long it’s visible given how fast it orbits.)  That means it orbits our planet 16 times per day, or about 16*27.3=437 times per Moon orbit. So T = 1/437 = 0.0023 or so.

Now if we thought R shouldbe the distance from Earth’s surface, things would not work at all.  250 miles is about 1/1000 the Moon-Earth distance, and that would imply T = R3/2 = (1/1000)3/2 = 1/32000 of a Moon-orbit roughly, which is about once a minute! WAY too fast.

But if instead we use the distance from the Earth’s center to the altitude of the space station for R, that’s about 4250 miles (6800 km), or about 1/57 = 0.018 of the Moon-Earth distance.  Then

  • T = (0.018)3/2 = 0.0024 Moon orbits,

which as we saw is about 90 minutes. Bingo.

Falling Keys and Apples

Poor Newton didn’t have these satellites to help him check his ideas. But there was one thing he could do: he could consider the acceleration of an apple as it falls to the ground.  Or he could have used his keys, or a coin, or a steel ball, because they all accelerate at the same rate.  Try it; drop a heavy book and a penny at the same time.  You’ll see they land on the ground at the same time.  Check it yourself.

Now that we’ve established a Kepler-like law for objects that orbit the Earth, the same logic used for the planets, which taught us that they accelerate toward the Sun by an amount given by an inverse square law (measured in units of the Earth-Sun distance per Earth-year per Earth-year), will tell us that the Moon and artificial satellites are accelerated toward to Earth by ther own inverse square law:

  • a = (2π)2 / R2 (in units of Moon-Earth distance per Moon-orbit per Moon-orbit)

These units aren’t very intuitive, so let’s convert the Moon’s acceleration to familiar units along similar lines for what we did for the planets. Since R=1 for the Moon, its acceleration is

  • aMoon = (2π)2 Moon-Earth distances per Moon-orbit per Moon-orbit = (2π)2 (384,000 km)/(27.3 days)2 =(2π)2 (384,000,000 meters)/(2,350,000 seconds)2 = 0.0027 m/s2,

where “m/s2” is short for “meters per second per second”.

If this acceleration is due to the same gravity that holds us to the ground and makes dropped objects fall, then when we take R=4000 miles (6400 km)=0.0166 of the Moon-Earth distance, which is our distance from Earth’s center, we should get the acceleration Galileo measured: 32 feet (9.8 meters) per second per second.

a = (2π)2 / R2 = aMoon /R2 = ( 0.0027 m/s2 ) / (0.0166)2 = 9.9 m/s2

It works!

Gravity is Indeed Responsible

So what have we learned (and not learned)?

  • Not only does gravity cause ordinary objects around us on Earth to fall, it is also responsible for the acceleration toward Earth that keeps the Moon and artificial satellites in near-circular orbits.
  • That acceleration decreases with distance from the center of the Earth as 1/(distance)2.
  • Since a similar acceleration toward the Sun, also decreasing as 1/(distance)2, holds the planets in near-circular orbits around the Sun, it appears that this acceleration is also due to gravity — gravity that pulls objects toward the Sun.
  • We do not yet know why we need a slightly different Kepler formula for objects orbiting the Earth and for objects orbiting the Sun, or what determines how they are different.
  • We do not yet know whether certain properties of objects (masses? magnetic fields?) are involved in these formulas.
  • We do not know why some objects orbit the Earth and others orbit the Sun.
  • We still don’t know how far the Earth is from the Sun

We are well on our way to retracing Newton’s steps, though there are still a few more to go.

Might the Sun orbit the Earth?

By the way, if we thought gravity might cause the Sun to orbit the Earth (remember we still haven’t proven which one orbits the other yet), we might think the Sun would satisfy our Earth-orbiting Kepler law. But we can see this doesn’t work. We showed the Sun is at least 10 times further away from the Earth as the Moon is, so R is at least 10 in Moon-Earth distances. But then T = R3/2 would have to be at least 30 Moon orbits, which would be at least two and a half years. That’s too long, so we can be sure: if the Sun orbits the Earth, it is not held in place by gravity in the same way that the Moon is.

This is to be contrasted with what we found a couple of posts ago — where we saw that Kepler’s law for the planets would apply to the Earth, if the Earth orbited the Sun… and so, were that true, it would seem that the Earth would be held in orbit by the Sun’s gravity, just like the other planets.

We just need a simple, intuitive proof that the Earth goes round the Sun and not the other way round…

6 responses to “From Kepler’s Law to Newton’s Gravity, Yourself — Part 2

  1. This is a great series. Thanks. By the way, Newton imagined earth skimming satellites, as per the famous image found in every textbooks.

  2. The fundamental cause for Inverse-square law can be understood as geometric dilution corresponding to point-source radiation into three-dimensional space.?

    Here the physical reality is the Time dilation not the Mass (rest) ?
    The feeble “g” is canceld from both sides of the equation, thus we percept (feeling the weight of moving the Ship with hands) only the Energy as Mass not the constant momentum?

  3. Sir I have not read all of the article yet, but what I have read contains a serious error. You stated that the GPS satellites are in synchronous orbit (24 hours). They are not. The orbit time is 12 hours.

    • Oops; thanks for pointing out the error. I should instead have mentioned the GOES weather satellites. The error doesn’t affect the physics arguments, fortunately.

  4. I mean, 6mm/s^2 Isn’t that slow an acceleration on human terms, it’s one you can see measurably increase. Taken as a round figure of 1cm/s per 2 seconds that gets you a speed of 30cm/s after just a minute which is hardly snail speeds. It’s quite astounding honestly how it adds up even on timescales of days.

    How interesting to see how Newton and others (may have) worked his way through these steps. Reminds me of the old tales of th ancient Greeks, deciphering the world with geometry and reason.

  5. At *high-energy, the quantized gauge bosons levitate (local maximum) from metastable to stable (dV/dF = 0), defying the Phenomenological gravity, ‘being is correlated quantum systems’ (like tethered balloon). The variation of V with respect to T depends on H, and the variation of V with respect to H depends on T (not, only depends on the Rest-mass).?

    “Equations of motion”. Equation means Proportions or Symmetries. Increase in W mass break this, example in “F = m a“ if “m (momentum or the potential-energy density V)” changes in the particular quantum field theory at a ColdSpot (“vev” = 0) or ‘metastable’?

    Any variation with space or time typically requires extra energy… changing things requires more *energy than leaving things unchanged… thus the StandardModel duplicating the Supersymmetry?
    V = 1/2 mT2 T2 = Momentum = rest-mass (non-zero as long as r isn’t zero).
    Added mass will increase the momentum.

    Cartesian like Pythagorean, Materialize or localized the Mass-factor.

    “Unbroken phase“ : = = ‘mirror image is digitally you’ or holographic Supersymmetry duplicating the StandardModel?
    “Broken phase“ : and are not equal = ‘mirror image is not you’ or Radio-activity?

    The lesson here is that in the broken phase for H, T has a vev as well; in the unbroken phase, neither field does. For r small enough, it’s H that’s driving everything, with T just following along: when becomes non-zero, it triggers to be slightly non-zero.