Accurate Mass of the Earth
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, by maybe a thousand times.
Why it is not currently known more accurately
When Newton developed his gravitation theory, he arrived at a relatively simple equation,
F = Gm1m2/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2 * a3 / ,
where T is the (sidereal) orbital period, a is the semi-major axis (essentially the average distance
between the two in elliptic orbits), and  is the product of the Gravitational Constant and the
total mass in the system. (This mathematically confirmed one of Kepler’s earlier Laws).
It is easy to get T extremely accurately by careful observation and a quite accurately, and so
this equation can give a very accurate value for . 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 , no really precise value for either the actual mass in the system or
the Gravitational Constant has been possible.
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 / 
or
5.5723922 * 10+12 = 2.2485124 * 10+27 / 
We can solve for  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  / 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 digits! 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!
This example, and ALL other available experiments, can result in an excellently accurate value
for  (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 could 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 an extremely accurate value for the acceleration
due to gravity is 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  again! Even though our
experiments can determine  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). A mid-course correction rocket may burn to adjust 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 (precisely 1-kilogram, for example). We will use it as a
‘satellite’ to orbit around the spacecraft! 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 10meter distance to the main spacecraft, and it is given a slight velocity, it will orbit the spacecraft
and the tether would be released and discarded.
If the spacecraft mass was 1,000 kilograms, then the equation above gives:
T2 = 4 * 9.8696044 * 103 / 
where  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 semimajor axis distance and the orbital period. This again gives an extremely accurate value for , 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 is always having a very accurate fuel gauge,
the mass of the spacecraft should always be known to possibly the nearest gram. Out of a 1,000kilogram spacecraft, that is one part in a million, which would then allow G to be determined 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.