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 (R^{3}=T^{2} , 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 = (2***π*)^{2}/ R^{2}

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 **(2 π)^{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=T^{2/3} 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,

- R
^{3}=T^{2}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.**