Everything becomes clear if you assign their proper meanings to words like "theory","law" etc. In a scientific context, "theory" is not an insult (as in the silly put-down "it's just a theory"). A theory is the most elaborate formof consistent scientific knowledge not yet disproved by experiment. In experimental sciences, a theory can never be proved, it can only be disproved by experiment. This is precisely was makes a theory scientific. A statement that cannot be disproved by experiment may still be highly respectablebut it's simply not part of any experimental science (it could be mathematics, philosophy or religion, but it's not physics). Now that we have the basic vocabulary straight, we may discuss gravity itself...
Gravity is a physical phenomenon which has obvious manifestations all around us. As such, it's begging for a scientific theory to describe it accurately and consistently. The rules within a theory are called laws and the inverse square law of the Newtonian theory of gravitationdoes describe gravity extremely well. Loosely stated:
Two things always attract in direct proportion to their masses and in inverse proportion to the square of the distance between them.
However, the Newtonian laws are not the ultimate laws of gravity. We do know thatGeneral Relativity (GR)provides more accurate experimental predictions in extreme conditions (e.g., a residual discrepancy in the motion of the perihelion of Mercuryis not explained by Newtonian theory but is accounted for by GR).
Does this mean Newtonian theory is wrong ? Of course not. Until we have atheory of everything (if such a thing exists) any physical theory has its own range of applicability where its predictions are correct at a stated level of accuracy (an experimental measurement is meaningless if it does not come with a margin for error). The Newtonian theory is darn good at predicting the motion of planets within the Solar System to many decimal places... That's all we ask of it.
Even General Relativity is certainly not the ultimate theory of gravitation. We know that much because GR is a classical theory,as opposed to aquantum theory. So, GR is not mathematically compatible with the quantum phenomenawhich become so obvious at very small scales...
Science is mostly a succession of better and better approximations. This is what makes it so nice and exciting. If you were to insist at all times on "the whole truth and nothing but the truth"in a scientific context, you'd never be able to make any meaningful statement (unless accompanied by the relevant "margin for error"). As a consistent body of knowledge, each theory allows you to make such statements freely, knowing simply that the validity of your discourse is only restricted bythe general conditions of applicability of a particular theory. Without such a framework, scientific discourse would be crippled into utter uselessness.
(2008-08-22)
The inverse square law of Newtonian gravity is also valid forelectrostatics: The force between two electric charges is proportional to thecharges and inversely proportional to the distance between them.
Consider two bodies with the same mass m carrying thesame electric charge q. If the following relation holds,there won't be any force between them, as their gravitational attractionis balanced by their electric repulsion, at any distance:
G m 2 = q 2/40
This happens when q/m = (40G) = 8.617350(44) 1011 C/kg.
In other words, two bodies carrying one elementary charge (1.602 1019 C) have no net force between them if their mass is about 1.86 g (which is roughlythe mass of a dust mite).
(2011-03-15) Motion of two isolated bodies gravitating around each other.
The two-body approximation discussed here applies with excellentprecision to the motion of two stars within abinary system, because all other bodies areeither too small or too distant to influence their relative motion. More importantly for the inhabitants of theSolar system, this simple discussion also gives themain motion of a planet around a star like the Sun. As long as other planets do not come too close, their influences can beeither ignored or treated as small perturbations.
Because they are assumed to be isolated, the two gravitating bodiesrevolve around their common center of mass.../...
(2016-06-08) In a two-body Keplerian motion, the celerity hodograph is a circle!
The term hodograph was coined in 1847 by Hamilton from the Greek words ' and , with the intended meaning of path written.
The hodograph of a vector is the trajectory of its extremity when its origin isfixed (more pedantically, it's the locus of a point whose coordinates are the same as a varying vector's coordinates).
(2009-08-18) Pendulums at the top and bottom of a mine give the mass of the Earth.
In 1826, the idea occurred to GeorgeBiddell Airy (1801-1892) that the periodof a pendulum in a mine depends on the mass of the rockabove it and below it. The latter, which is essentially the whole Earth,can be estimated from the former.
More precisely,let's assume that the Earth, of radius R, has a mass distribution whosedensity depends only on the distance r to its center. That density, which varies with the depth h = R-r, is near the surface andits average is o
o = 3 M / (4 R 3)
By Newton's theorem (thetheorem of Gauss applied to a sphericallysymmetric distribution) the gravitational field g at depth h is the same as that which would be due to the total mass M locatedat a greater depth if it was concentrated at the center of the Earth. We'll obtain the variation of g by differentiation:
g = G M / r 2 dg = -2 G M / r 3 dr + G / r 2 dM
For r slightly below R, we have dM = -4r 2 dh (since dh = -dr) and:
dg/dh = 2 GM / r 3 4 G = 2 GM / R 3 3 GM / R 3(o)
In the main, the gravitational field at a small depth h is thus:
g h = g 0 [ 1 + ( 2 3o) h / R ]
This spherical model ought to be a good approximation of reality if we let bethe average density of the local ground, since distant aspherical contribution to gravitywould yield nearly equal corrections at the top and bottom of the mine.
Airy had bad luck with two experimental attempts in 1826 and 1828 and he gave upon the idea for a while (he went on with his life and becameAstronomer Royal in 1835). In 1854 however, Sir Airy finally performed theexperiment conceived by his younger self, at the coal pit of West Harton, nearSouth Shields.
Precision timing revealed that a pendulum placed at the bottom of the pitwas faster by about 2.24 s per day (one part in 38572). At a depth of 383 m gravity was thusfound to exceed surface gravity by one part in 19286. Knowing that R = 6371000 m, the above equationthus means that the average density of the Earth is 2.6374 times that of the rock at West Harton.
The value Airy published for the mean density of the Earth was 6.566 g/cc (the currently accepted value is around 5.5153). This means that he estimated the density of the ground at West Harton to be around 2.49, a valuereportedly provided to him by the mineralogistWilliamH. Miller (1801-1880) of Millerindex fame. An estimate of 2.0912 would have given a perfect result.
The experimental details are delicate. Ideally, the pendulum should be locatednear the center of a spherical cavity. Local irregularities in the strata should be accounted for,but this is impractical... AlthoughRobert von Sterneck(1839-1910) used better technology in 1882 and 1885,he couldn't obtain consistent results from the similar experimentsthat he conducted at the St. Adalbert shaft of Pribram(Bohemia)and the Abraham shaft of Freiberg (Saxony).
In fact gravitational experiments involving large natural land masses are now mostlyused as evidence for the irregular distribution of superficial strata. Apparently, similar experiments have never been carried out at sea...
(2007-09-29) Arigid motion of three equidistant gravitating bodies,as they rotate around their common center of mass O.
The equilateral triangle at right tells the whole story:
If the bodies at A, B and C attract each other in direct proportion to their masses,the so-called paralellogram law forvector additiondoes indicate that each body is subjected to a centripetal acceleration toward O, whose magnitude is proporttional to its distance from the common center of mass O. (With a suitable scaling torepresent accelerations, the geometric construction of the center of massmatches the parallelograms involved in vector addition, as depicted above.)
This means that the triangle ABC rotatesrigidlyabout its center of mass O.
Quantitatively, the square of angular velocity is the scaling factor of the above diagram: To a distance R corresponds an acceleration 2 R.
This remark allows the value of that scale to be obtainedgeometrically in terms of Newton's universalgravitational constant (G) :
as a function of d = AB = AC = BC
2 d 3 = G M = G ( m A + m B + m C )
In the diagram, we observe that the arrow extremitiesdivide each side (of length d) into three segments whose lengths areproportional to the three masses (the coefficientof proportionality being d/M). Thus, an arrow toward B (from A or C) translates (byscaling lengths intoaccelerations) into the following component ofthe acceleration, which is equated to its gravitational counterpart(using Newton's inverse square law)to yield the advertised relation.
(2007-10-08) The 5 points where gravity balances the centrifugal force.
Theabove can be applied to the case of two bodiesin circular orbit around each other: A third body ofnegligible mass would follow their rotation rigidly if it'sin the plane of rotation and forms an equilateral triangle with those two bodies.
There are two such points (called L4 and L5). These are stable locations (in the sense that they seemto attract nearby test masses) provided the ratio of the larger mass to the smaller one exceeds 24.96 or,more precisely:
The Lagrange point L4 (the Greek triangular point) leads the smaller body in its orbit around the larger one, whilethe Lagrange point L5 (the Trojan or trailing triangular point) lags behind.
In addition,there are three unstable Lagrangian points (aligned with the two orbiting bodies) where the centrifugal forceexactly balances gravity.
L1 (the inner Lagrangian point) is located between the two orbiting bodies. L2 is outside those twobodies, on the side of the lighter one, while L3 is on the side of the heavier one.
(2010-12-31) Satellites whose positions in the sky remain nearly fixed.
(2019-07-21) Simplest way to go from one circular orbit to another.
(2019-08-08) Newtonian motion around an [oblate] central body.
(2013-03-27) Negative, proportional to the mass squared and the inverse of the radius.
In a spherically-symmetric distribution of mass, the Newtonian gravitationalfield at any point is the same as what would be created at that pointby whatever mass is located at a lesser distance from the center if that mass wasconcentrated at the center (the field created by any homogeneous sphericalshell at a greater distance is exactly zero). This property of the inverse-square law is often called Newton's theorem (students of electrostatics may recognize it as a straight applicationof thetheorem of Gauss).
In an homogeneous sphere of mass M and radius R, the mass within a distance r from the center varies as r3 which makes the field directlyproportional to r. All told, the radial field at distance r is:
- GM r / R3 when r < R and - GM / r2 when r > R
Multiplying this into an infinitesimal mass dM gives the gravitational force exerted on that mass at distance r. The gravitational energy dU of dM within thismass distribution is obtained byintegrating that force from r to infinity (making the convenient convention that attributes zero gravitationalenergy to very distant stuff, everything has negative gravitational energy):
dU
= - G M dM
R
xdx
+
dx
r
R3
R
x2
= - G M dM
R2-r2
+
1
2 R3
R
Now, the mass dM located in the shell between r and r+dr is equal to the shell's volume 4r2dr multiplied into the mass density 3M/(4R3 ).
dM = ( 3 M r2 / R3 ) dr
Therefore, the ball's total self-energy is given by the following integral:
U =
R
- G M
R2-r2
+
1
( 3 M r2 / R3 ) dr
0
2 R3
R
Gravitational self-energy of an homogeneous sphere of mass M and radius R :
U = - 6 G M2 / 5R = - 1.2 GM2/R
Something from nothing :
The above Newtonian expression of gravitational energy is sufficient toexplain qualitatively how the entire Universe can have zero total energy. Discarding the (interesting)effect of temperature,the total energy is:
M c2 6 GM2 / 5R
At face value, this would mean that a (cold) uniform ball of zero energyhas a mass proportional to its radius. In the presence of gravity, a ball of radius R created from nothing would thus have the following mass:
M = (5 c2 / 6G) R
Numerically, the constant is 1.122241027 kg/m. This entails a density so large that Newtonian theory is no longer applicable;the above constant is 66.67% more than what's required tocreate black holes, according toGeneral Relativity...
Thus, the problem is not so much to explain how gravity can create somethingfrom nothing, which turns out to be conceptually simple, but to realize that themechanism is so powerful quantitatively that it cannot run raw. Somehow, the above calculation might qualitatively describethe creation of mass locked in particles of very small sizes (essentially, tiny black holes whose ultimateHawking evaporationis forbidden by the conservation of some quantized number, like electric charge).
Douglas G. (2010-11-25) How the braking effect of earthly tides makes the Moon drift away.
To investigate the causes and magnitudes of Earthly tides, we shall usea simplified model where a perfectly spherical Moon moves along a circlearound the Earth which itself moves along a circular orbit around aperfectly spherical Sun. We'll also make the drastic assumption thatthe Earth is a solid sphere completely covered by a single ocean ofseawater (without any continents, islands or other irregularitiesin the sea floor).
Consider two solid bodies (of masses m and M) with perfect spherical symmetry. Their outer radii are respectively r and R. They are both gravitating aroundtheir combined center of gravity (O) in a perfect circle. The centers of the two spheres are at a distance D from each other:
The distance from O to the center of m is MD / (M+m)
The distance from O to the center of M is mD / (M+m)
The whole system rotatesrigidly with angular velocity
According to Newton's theorem (i.e.,Gauss's theoremapplied to Newtonian gravity) the two spheres attract each other as would two masses concentrated at the centerof each sphere. The gravitational force between them is equal to:
F = G m M / D 2 = m 2 MD / (M+m) = M 2 mD / (M+m)
Dividing the above by mM we obtain the following, in two different ways:
2 D 3 = G (m+M)
Let's consider a cartesian frame of reference rotating about O at the angular speed . The x-axis goes throughboth of the centers of the spheres and the y-axis is parallel to the axis of rotation. The z-axis is, of course, perpendicular to both of the above.
Let's study the apparent gravitational field at a distance R+z from the center of the sphereof mass M (assuming that the sphere does not spin at all) at a point of latitude (with respect to the Ox axis) and longitude (where is defined to be zero for the half-meridian at thesurface of the M sphere in the xOy plane which is nearest to the m sphere.
It is the sum of three terms:
The centrifugal field.
The gravitational attraction of the sphere of mass M.
The gravitational attraction of the sphere of mass m.
The vectorial sum of those three component, as computed below,involves a main tidal term (inversely proportional to the cube of the distance D) whose values on the Ox axis yield the size of the tidalbulge of a liquid ocean coating the sphere of mass M.
(2019-08-10) If a body had spherical symmetry, its field would be that of a point-mass.
Newton's shell theorem says that an homogeneousspherical shell produces a zero gravitational field inside it, while the outside field is exactly what would be produced byits entire mass if it was concentrated at the center.
By symmetry, a point-particle exerts no torque on an homogeneous spherical shell. Neither does any distribution of mass. Conversely, a spherically-symmetric body cannot exert any gravitationaltorque on any mass distribution, symmetric or not, rigid or not. ( The total angular momentum of the wholesystem would be conserved if it was isolated.)
Figure of the Earth :
In the main, the Earth is an oblate spheroid (which is to say that weassume rotational symmetry about the polar axis but not necessarily north-southsymmetry with respect to the equatorial plane).
The perceived gravity derives from an apparent potential U equal to thesum of the purely gravitational potential V (which verifies Laplace's equation V = 0) and a term due to rotation... Using to denote the Earth's angular rate of rotation, that apparent potential U for an observer at a latitude and a distance r from the center of the Earth is:
U = V ½2 r2 cos2
is the geocentric latitude of the observer, namely the angle berween the equatorialplane and the line going from the Earth's center to the observer's location. It's not the geodetic latitude (also called astronomical latitude) normaly used in geography (always called , in Numericana) which is the angle between the equatorial plane and a plumb line orthogonal to the local horizon.
In his Principia(1687) Newton proposed thatthe Earth assumes the shape of an homogeneous incompressible fluid subjected only tocentrifugal forces and self-gravitation (U is constant over the whole surface). In the main, that's an ellipsoid of revolution of equatorial radius a and polar radius b.
The flattening f depends on two dimensionless ratiostraditionally dubbed J2 and m, through an approximative formula due to Clairaut (1743) :
f = (a b )/ a = ½ ( 3 J2 + m ) where m = a32/ GM
In this, J2 is called the second dynamic form factor (more about that soon) whereas m is defined as the ratioof the normal acceleration at the equator ( a2 ) to the standardized gravitational field GM / a 2 (which would be the purely gravitational field at the equator of a spinlessspherical Earth).
J2 is the coefficient of the second zonal harmonic, per the following expansion of the gravitational potential of a body with axial symmetry:
V (r,) =
-GM
a n
JnPn(sin )
r
r n
The above only holds for bodies with axial symmetry. Otherwise a more generalexpression is needed which entails longitude as well.
P0(x) = 1 ; P1(x) = x ; P2(x) = ½ (3x2 - 1) ; P3(x) = ½ (5x3 - 3x)
The series is convergent and it's equivalent to the term of order zerofor large values of r. The same flux is thus produced through a sphere of large radiusas would be observed with just a central mass M. Therefore, J0 = 1.
We have J1 = 0 by virtue of the fact that the center of mass isthe origin of our coordinate system. This makes J2 the first nontrivial coefficient.
For the Moon, a / r is less than 1/60 making lunar precessions depend almost entirely on the second zonal harmonic of ther Earth gravitational field. For artificial satellites in low Earth orbits, the higher harmonics become significant.
Dynamic form factors of the Earth (Yoshihide KOZAI, 1964-09-22)
2n
J2n (ppm)
J2n
2n+1
J2n+1 (ppm)
J2n+1
0
1000000.000
0
1
0.000
0
2
1082.645
0.006
3
-2.546
0.020
4
-1.649
0.016
5
-0.210
0.025
6
0.646
0.030
7
-0.333
0.039
8
-0.270
0.050
9
-0.053
0.060
10
-0.054
0.050
11
0.302
0.035
12
-0.357
0.044
13
-0.114
0.084
14
-0.179
0.063
15
It's impossible to determine the internal massdistribution solely from the external gravitational field, because two distributions whose difference is spherically symmetric produce the same field (by the shell theorem). Conversely, two mass distributions with the same external fieldmust differ by such a symmetric distribution (possibly allowing negative masses).
(2010-12-31) Wake-up call: Detecting large asteroids on a collision course with Earth.
The 400 m asteroid dubbed 2004 MN4 was discovered in June 2004. Initially, that Aten asteroid (now called 99942 Apophis) was given a 2.7% chance of impacting Earth on April 13, 2029. Refined data shows that Apophis will come no closerthan 30000 km above Earth's surface on that date. It will again come extremely close to Earth in 2036 and 2068.
On2013-01-13,NASA officially ruled out the possibility of an impact in 2036 (based on fresh data) as it also announced therecord-settingflyby of 2012DA14 (an asteroid about 40 m in size) for Feb. 15, 2013.
Amazingly, 16 hours before that actual event, anotherunrelated meteor (about half the diameter of 2012DA14) made front-page news as it exploded at an altitude of 23.3 km in the atmosphere near the Russian town ofChelyabinsk, releasing more than 20 times the energy of the Hiroshima bomb! Nobody had ever detected that one before it entered the atmosphere (it's thought to have been an Apollo asteroid).
(2013-01-21) , 1933) What makes galactical rims rotate so fast?
More than 99% of the mass of the Solar system is concentrated at its center (within the Sun itself). That's why orbital speeds around the Sun decrease fairly rapidly withdistance. (The orbital speed is the size of the orbit divided by theorbital period. So,Kepler's third lawimplies that it varies inversely as the square root of the orbital radius.)
On the other hand, within a spherically symmetric distribution of mass of uniformdensity, the centripetal force of gravity is proportional to the radius (this is a simple application of thetheorem of Gauss,applicable to any inverse-square force field). For thecentripetal acceleration v2/R to be proportional to R, the orbital speed (v) must be proportional to R as well.
The orbital speed v wouldn't depend on the radius R ifthe centripetal force was proportional to 1/R, which corresponds to a densityof matter inversely proportional to the square of the radius (by the theorem of Gauss).
In 1932, Jan Oort...
The picture which might be emerging is that of a universe where thesupermassive black holes formed first and pulled around themstuff which either gave birth to ordinary matter or didn't.
(2023-02-28) ) vector. A quantity that's conserved in the case of
(2007-09-29) 
