Governing the gravitational interaction, its value is G= 6.6738×10−11 m3 kg-1 s-2.
It relates a mass (in kg) with some space-time quantities.
In practice, the most usual situation to describe is that of one
spherical massive body (a planet or a star) with mass M, and the
situation of its gravitational field at a given position with
respect to this body; namely at a distance r from its center,
either on its surface, or near its surface, or more far away.
So, if we give to the gravitational constant some specific mass M and some length r, what quantity does it give back in reply ?
It gives back a quantity T = GM/r3 and says that this
quantity must be counted in s-2. In other words, that
this is the inversed square of a time t, so that the period of the
circular orbit with radius r around this body is 2πt.
First let us look at M/r3 : as a quotient of the mass
by the cube of the distance to the center, it describes the
average density ρ of mass inside the sphere centered on the planet
and with radius r.
More precisely, as the volume of this sphere is 4π r3/3, we have ρ =
(3/4π)M/r3.
Thus, we get T = (4πG/3)ρ
= t-2, in other words t=√3/4πG ρ-1/2.
Now for convenience we need to express ρ in g/cm3. Thus
the constant we need is the amount of time
√(3/4πG) (1cm3/g)
= 1891.3 s = 0.5254 hours = 31.5 minutes, or 2π × 0.5254 = 3.301 hours = if we want
the orbital period.
The density of the Earth is 5.515g/cm3. Thus we get
for the surface of the Earth, a value Te =5.515* (1891.3)−2
= 1.542×10−6
s−2
= (805.35 s)−2 = (13.42 min)−2 and a low
earth orbit period of 3.301/√5.515=2π
× 13.42 min =5060 s = 1.406 hours, or rather a little more
because these orbits have some altitude above the Earth.
For example the orbit of the International Space Station has
period 92 minutes 36 seconds = 1.54 hours.
For the Earth we have a theoretical low orbital speed a sea
level, 6,371 km / 805.35
s = 7,911 m/s, while the real orbits being at some higher altitude
are a bit slower.
The rotation period (day) of any planet must be significantly
longer than its low orbital period, as it would otherwise break
apart by the centrifugal force. The flattening of a planet is
roughly proportional to the square of the ratio between them : in
the case of the Earth, the radius at the equator, re=6,378.1 km,
is 21.3 km larger than the polar radius, 6,356.8 km, that is
a 0.33% difference, close to the squared ratio of orbital to day
periods (1 sidereal day = 86164 s),
(3.3/24)2= 0.0034.
The acceleration of gravitation is g = rT = r/t2 (the one that would send an object initially at rest to a distance r/2 in the time t=1/√T)
On the Earth, it varies from 9.780 m·s−2
at the equator to 9.832 m·s−2 at the poles,
that is a difference of 0.53%. In this difference, the centrifugal
force is directly responsible for a decrease of re(2π/sidereal day)2 =
6,378.1km(2π/86164s)2= 0.0339 m·s−2,
that is a 34.5% difference.
The rest of the difference to account for, from an abstractly
corrected 9.814 m·s−2 at the equator to
9.832 m·s−2 at the poles, is a mere 0.18%.
(We can check that taking the mean Earth radius, 6,371 km, and
dividing it by (5060 s)2, gives 9.823 m·s−2,
that is, up to computation errors (?) the average between polar
and equatorial values ((2×9.814
+9.832)/3=9.820).
This difference is smaller than would be computed by taking the
mass of the Earth and the difference of distance to the center,
because of the distortion of the Earth's gravitational field by
the flatening itself.