The Formulas of Gravity
When Galileo performed his experiments on gravity over 300 years ago, he found that the
acceleration of gravity above the surface of the earth varies as to the earth's radius
distance squared. Expressed in a formula this is:
K = a * r 2
Acceleration of Gravity formula
Where
K = a planetary constant of gravity
r = the planetary radius
a = the acceleration of gravity
With this formula, the acceleration of gravity can be obtained at any distance from a
planet's surface.
The gravitation formula was developed from the Acceleration of Gravity formula. Solving
for a:
a = K / r 2
This was substituted into the momentum force formula,
F = M * a
Momentum Force Formula
making it:
F = M * K / r 2
Since K is a planetary constant, a universal gravitation constant was necessary to
calculate the gravitation force of any body in space. K was then replaced in the formula
by G, the universal gravitation constant to make it:
F = G M / r 2
For any planet of radius r, F = a, the surface acceleration of gravity, therefore,
a = G M / r 2
Since GM has replaced K in the Acceleration of Gravity formula, then:
K = GM
The gravitation constant is the gravity force exerted by a unit amount of matter. The
gravity force exerted by a whole planet would be:
F = GM
Since the acceleration of gravity varies as to the distance squared that bodies are from
each other, then it is assumed that the gravitation force between two bodies in space
varies as to their inverse distance squared.
Let's return to the Momentum formula. For a small body falling towards the surface of a
planet the momentum force is,
F = m * a
If we substitute a for the Acceleration of Gravity formula we get,
F = m * K / r 2
Since we have determined that the gravity force for a planet, K = GM, we can substitute
GM for K to arrive at the gravity force acting between the small mass and the gravity of
the whole planet,
F = m * GM / r 2
Thus, the final gravitation force formula is:
F = GmM / r 2
The Gravitation Force Formula
How does the gravity force relate to orbital velocity? To understand this, we must
analyze how the acceleration of gravity relates to the momentum of a mass in orbit. You
can learn how orbital velocity relates to orbital radius by observing a bucket whirled about
yourself on the end of a rope. It is found that as the radius distance increases, the
velocity of the bucket decreases.
In fact, it has been found that squaring the orbital velocity of a satellite in orbit about a
planet divided by the orbital radius equals the acceleration of gravity towards that planet
at the orbital radius distance. This keeps the satellite in orbit. Like the bucket, if the
orbital radius is increased, the orbital velocity of the satellite decreases. The first
satellites put up into orbit orbited the earth in about 60 minutes. Later telecommunication
satellites were placed further out, at the distance called the geosynchronous orbit at
22,243.528 miles. At that orbital distance from the earth, the satellite stays directly over
one spot on the ground below giving it an orbital period of about 24 hours.
So the relationship of the acceleration of gravity to orbital velocity is,
a = v 2 / R
When we substitute this into the momentum formula,
F = m * a
we get the centrifugal force formula:
F = m * v 2 / R The Centrifugal Force Formula
When we substitute a = v 2 / R into the Acceleration of Gravity formula, K = a * r 2 , for a
theoretical surface orbiting satellite where the planetary radius r = R, the orbital radius, we
get:
K = R 2 * v 2 / R, which reduces to:
K = v 2 * R
The Orbital Velocity Formula
To summarize, in our study of gravity, we have to work with,
three planetary constant formulas:
K = a * r 2
K = GM
K = v 2 * R
The Acceleration of Gravity Formula
The Planetay Mass Formula
The Orbital Velocity Formula
and three force formulas:
F = m * a
F = GmM / R 2
F = m v 2 / R
The Momentum Force Formula
The Gravitation Force Formula
The Centrifual Force Formula
From the geosynchronous formula, R^3 = GMT^2 / 4 pi^2,
G = 4 pi^2 R^3 / T^2 M
And since
G = K / M,
Then also,
K = 4 pi^2 R^3 / T^2
In order to calculate the mass of the planets, it was necessary to obtain a unit value for
gravity. How much gravitational force does one gram of matter exert? This was answered
when the gravitation constant was directly measured in the Cavendish experiment.