## The gravitational constant

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.

Other interesting things in a gravitational field are:
• The escape velocity v=2 r/t so that the kinetic energy of any escaping object, opposite to its gravitational potential energy, is its mass multiplied by the gravitational potential v2/2=(r/t)2, (which is proportional to 1/r outside the planet since the density in 1/t2 is then proportional to 1/r3) while r/t is the circular orbital velocity.
• The tidal effect, directly equal to 2T tearing apart in the vertical direction, but an opposite component equal to T in any horizontal direction.
It is this quantity T that measures the curvature of space-time in the time direction, as this is how gravitation is described according to General relativity : the proportionality of this curvature to the density of mass is one of the components of the equation of this theory. It says for any two nearby objects separated by a given distance, how much speed towards each other or away from each other, they gain per second in proportion to the distance between them (a speed per time and per distance, in s-2) = . The tidal effect of the Moon on the Earth causes the tides on the ocean (in a complex way as it forms big waves whose shape depends on the shapes of the oceans). Let us compute its amplitude from other data : at the surface of the moon, the gravity is Tm rm = 1.622 m/s2, where rm = 1,737 km, thus Tm=9.338 ×10−7 s-2 = (17 minutes)-2=  (smaller than that of Earth, Te=1,542×10−6 s-2 = (13.4 minutes)-2 because the density of the moon, 3.346, is lower than that of Earth, 5.515). The earth-moon distance varies between 362,570 km and 405,410 km. Let us take a value d= 370,000 km. The amplitude of the tidal effect on Earth is T= (rm/d)3 Tm . To find out the range of order of the height of tides it would bring on Earth, we have to compare it with the Earth's Te . This relative effect on the Earth's gravity gives the range of magnitude of the relative effect on the Earth's radius re=6,371 km, that is a variation of height with magnitude order re T/Te = re (rm/d)3 Tm/Te =  6,378000*(1,737/370,000)^3* (3.346/5.515) = 0.40 m.

The tidal effect that the Earth exerts on the Moon causes its tidal locking, that is its always showing the same face to the Earth : it is stretched in the direction of the Earth. This does not exactly make it like a baseball but an ellipsoid with different sizes on all 3 axis, due to the centrifugal force: the global values of tidal stretching by (Earth + centrifugal force) on the Moon are proportional to : 3 towards the Earth (a pull) ; 0 on the orbital direction, and -1 (a push) on the polar direction. Not to speak about the variable components due to the excentricity of the Moon's orbit.
Its amplitude can be computed by exchanging the roles of Earth and Moon in the above formula : rm (re/d)3 Te/Tm= 1737000*(6,378/370,000)^3* (5.515/3.346) =14.6 m.
The tidal effect is also responsible for the fact that Saturn's rings stay at ring and do not form satellites closer to a distance from Saturn than the Roche limit, that is a proportion of Saturn's radius determined by the ratio of densities between Saturn and the satellites that may be formed.
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### Effects of General Relativity

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