Gravitational Constant

An Important Gravitation Experiment

Modern science has extremely accurate values for most 'constants' such as the speed of light, the value of pi, etc, but the Gravitational Constant is only known to around three significant figures! There is a relatively simple and inexpensive experiment that could be done to greatly increase the accuracy of this important constant.

Why it is not currently known better

When Newton developed his gravitation theory, he arrived at a relatively simple equation,

F = G * m1 * m2 / r2,

where F is the gravitational force acting between two objects, the m's are the masses of the two objects, and r is the distance between the centers of the two objects.

With the exception of relativistic factors, as far as science knows, that equation is exact.

Newton used that equation to derive the equations of motion for two objects orbiting (due to gravitational force alone) each other, and he got:

T2 = 4 * PI2 * a3 / mu,

where T is the (sidereal) orbital period, a is the semi-major axis (essentially the average distance between the two in elliptic orbits), and mu is the product of the Gravitational Constant and the total mass in the system.

It is easy to get T and a extremely accurately by careful observation, and so this equation can give an extremely accurate value for mu. The problem is that NEITHER the actual exact mass involved NOR the Gravitational Constant is known very accurately, and so even with a very precise value for mu, no really precise value for either the actual mass in the system or the Gravitational Constant has been possible.

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We can use an example. The Moon and Earth revolve around a common point in space, which happens to be within the Earth. This period takes one sidereal month, or 27.3216610 days, or 2,360,591.5 seconds, T. We know that the Moon's orbit has a semi-major axis of 384,749,900 meters. The equation above then gives:

(2,360,591.5 sec)2 = 4 * 9.8696044 * 384,749,9003 / mu

or

5.5723922 * 10+12 = 2.2485124 * 10+27 / mu

We can solve for mu and get:

4.0350935 * 10+14

This number is probably accurate to its eight significant digits.

In the best laboratory experiments, done in a vacuum with the most perfect equipment available, the value of G, the Gravitational Constant has been determined as being 6.67 * 10-11 and no more accurately, due to equipment limitations on such ultra-sensitive experiments. With this value for G, we can get the total mass of the Earth-Moon system to be mu / G, or:

4.0350935 * 10+14 / 6.67 * 10-11

or

6.05 * 10+24 kg.

The Moon is accurately known to be 1 / 81.270 of the mass of the Earth, so the Moon accounts for 0.07 * 10+24 kg which leaves

5.98 * 10+24 kg

as the best available estimate for the mass of the Earth! Only to three significant figures! As a Physicist, I am ashamed that after three hundred years of having the gravitational equations, and all the equipment that modern science has, that's the best we can do!

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This example, and ALL other available experiments, can result in excellently accurate values for mu (which is G times the total mass involved), but there has been no way of accurately determining either G or the total mass. Even though we know the product of those numbers really accurately, we only have a very poor idea of what either precisely is!

If you are a thoughtful person, you might think, Aha, I can simply drop a precisely known mass from a tall building and really accurately measure the speed (and therefore acceleration) it experiences while it falls. This is harder than it sounds, since air friction slows it down, but such an experiment can be done in a near vacuum, and extremely accurate values for the acceleration due to gravity are known. Since we know the mass of that object extremely well and also the acceleration, Newton's F = m * a means we can also know the exact force acting on it due to the Earth's gravitational attraction.

F = G * m1 * m2 / r2,

This is again Newton's universal gravitation equation, and we now know the left side extremely accurately, being m * a. So now this equation can be written:

m2 * a = G * m1 * m2 / r2,

where m2 is the mass of our object, m1 is the mass of the Earth, and r is the distance between the two, the radius of the Earth. Continuing, we have:

a = G * mE / r2,

Notice that again we have extremely accurate (measured) values for a and for r, and so we can solve for an extremely accurate value for the product G * mE. That's mu again! Even though our experiments can determine mu extremely accurately, we still do not accurately know either G or the mass of the Earth!

A New Experiment

Whenever we send spacecraft to Mars or the other outer planets, once it has entirely escaped Earth's gravity (a few million miles out) it generally just coasts for maybe nine months (for Mars trips, longer for Jupiter and beyond) until a mid-course correction rocket burn adjusts the trajectory to arrive exactly where we want it to go. During those nine months, and also in the nine months after that rocket burn, virtually nothing happens!

Why don't we include a small object (1 kilogram, for example). After the spacecraft is a few million miles out from Earth, that object would be released, possibly on a temporary tether. Once it was at around a 10 meter distance to the main spacecraft, and it is given a slight velocity, it will orbit the spacecraft and the tether would be discarded.

If the spacecraft mass was 1,000 kilograms, then the equation above gives:

T2 = 4 * 9.8696044 * 103 / mu

where mu is now 1001 * 6.67 * 10-11 or 6.67 * 10-8

T then is 769,000 seconds, or 8.90 days.

If the spacecraft had this satellite, it would be easy to determine the distance by radar ranging to many significant figures and a very accurate orbit could be determined, specifically the semi-major axis distance and the orbital period. This again gives an extremely accurate value for mu, as before, which should be at least eight significant digits.

The mass of the spacecraft is rather accurately known, because we built it! As long as fuel load remaining is accurately known, and good practice always having a very accurate fuel gauge, the mass of the spacecraft should be known to possibly the nearest gram. Out of a 1,000 kilogram spacecraft, that is one part in a million, which would then allow G to be known to an accuracy of one part in a million, six significant figures. That's a whole lot better than the three significant figures that three hundred years of science has gotten us so far, a thousand times more accurate.

With G being known one thousand times more accurately, then the actual mass of the Earth, Moon, Sun and everything else would also be known one thousand times more accurately. I would think that would be tremendous incentive to include this very simple experiment on one or more long distance spacecraft in the future.!


This presentation was first placed on the Internet in February 2004.

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C Johnson, Theoretical Physicist, Physics Degree from Univ of Chicago