whitehorse456
(2007-08-07)
Is [Newtonian] gravity a theory or a law?
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 form
of 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 respectable
but 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 gravitation
does 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 that General Relativity (GR)
provides more accurate experimental predictions in extreme conditions
(e.g., a residual discrepancy in the motion of the perihelion of Mercury
is not explained by Newtonian theory but is accounted for by GR).
Does this mean Newtonian theory is wrong ?
Of course not.
Until we have a
theory 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 a quantum theory.
So, GR is not mathematically compatible with the quantum phenomena
which 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 by
the general conditions of applicability of a particular theory.
Without such a framework, scientific discourse would be crippled into utter uselessness.
(2008-08-22) Comparing Gravity and Electrostatics
The inverse square law of Newtonian gravity is also valid for
electrostatics:
The force between two electric charges is proportional to the
charges and inversely proportional to the distance between them.
Consider two bodies with the same mass m carrying the
same electric charge q. If the following relation holds,
there won't be any force between them, as their gravitational attraction
is balanced by their electric repulsion, at
any distance:
G m 2 = q 2/
4p e0
This happens when q/m =
Ö(4p e0 G)
=
8.617350(44) 10-11 C/kg.
In other words, two bodies carrying one elementary charge
(1.602 10-19 C)
have no net force between them if their mass is about
1.86 mg
(which is roughly the mass of a dust mite).
(2011-03-15) Binet's Formulas: Deriving Kepler's Laws
Motion of two isolated bodies gravitating around each other.
The two-body approximation discussed here applies with excellent
precision to the motion of two stars within a
binary system, because all other bodies are
either too small or too distant to influence their relative motion.
More importantly for the inhabitants of the
Solar system, this simple discussion also gives the
main motion of a planet around a star like the Sun.
As long as other planets do not come too close, their influences can be
either ignored or treated as small perturbations.
Historically, the two-body approximation
(with circular orbits) was first applied by
Isaac Newton to the
motion of the Earth and the Moon around each other.
Although the result is only a terse approximation of a very complicated
three-body motion (involving the Sun) the good numerical
agreement in this particular case
is what convinced Newton of the validity of his inverse-square law
of Universal Grabitation.
Because they are assumed to be isolated, the two gravitating bodies
revolve around their common center of mass
.../...
Jacques Binet
Jacques Binet
(1786-1856; X1804)
is also known for another set of equations by the same name,
popular with recreational mathematicians and others:
Binet's formulas (1843)
give explicit expressions for the successive terms of the
Fibonacci sequence and other recursively defined sequences.
In practice, there's very little risk of confusion between the two sets of formulas.
Binet is also remembered for first describing the general rule
for matrix multiplication, in 1812.
He succeeded Poisson to the chair of
mechanics at Polytechnique in 1815.
(2016-06-08) Hodograph: The curve traced by a (celerity) vector.
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 'odoV and grajw,
with the intended meaning of path written.
The hodograph of a vector is the trajectory of its extremity when its origin is
fixed (more pedantically, it's the locus of a point whose coordinates are the same as a varying vector's coordinates).
"The Hodograph, or a New Method of Expressing in Symbolic Language the Newtonian Law of Attraction"
by William R. Hamilton.
Proceeding of the Royal Irish Academy,
3 (1847) pp. 344-353.
(2009-08-18) Weighing the Earth (Airy's method, 1826 & 1854)
Pendulums at the top and bottom of a mine give the mass of the Earth.
In 1826, the idea occurred to
George
Biddell Airy (1801-1892) that the period
of a pendulum in a mine depends on the mass of the rock
above 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 whose
density depends only on the distance r to its center.
That density, which varies with the depth
h = R-r, is r
near the surface and
its average is ro
ro =
3 M / (4p R 3 )
By Newton's theorem (the
theorem of Gauss applied to a spherically
symmetric distribution) the gravitational field g
at depth h
is the same as that which would be due to the total mass M located
at 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 = -4pr 2 r dh (since dh = -dr) and:
dg/dh = 2 GM / r 3
- 4 G p r
=
2 GM / R 3
- 3 GM / R 3
(r/ro )
In the main, the gravitational field at a small depth h is thus:
g h = g 0 [ 1 +
( 2 - 3 r/ro )
h / R ]
This spherical model ought to be a good approximation of reality if we let
r be
the average density of the local
ground, since distant aspherical contribution to gravity
would 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 up
on the idea for a while (he went on with his life and became
Astronomer Royal in 1835).
In 1854 however, Sir Airy finally performed the
experiment conceived by his younger self, at the coal pit of
West Harton, near
South Shields.
Precision timing revealed that a pendulum placed at the bottom of the pit
was faster by about 2.24 s per day
(one part in 38572).
At a depth of 383 m gravity was thus
found to exceed surface gravity by one part in 19286.
Knowing that R = 6371000 m, the above equation
thus means that the average density of the Earth is
2.6374 times that of the rock at West Harton.
2 - 3 r/ro
= 6371000 / ( 19286 . 383)
so:
ro / r
= 2.63739747...
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 value
reportedly provided to him by the mineralogist
William
H. Miller (1801-1880)
of Miller
index fame.
An estimate of 2.0912 would have given a perfect result.
The experimental details are delicate. Ideally, the pendulum should be located
near the center of a spherical cavity.
Local irregularities in the strata should be accounted for,
but this is impractical... Although
Robert von Sterneck
(1839-1910)
used better technology in 1882 and 1885,
he couldn't obtain consistent results from the similar experiments
that 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 mostly
used as evidence for the irregular distribution of superficial strata.
Apparently, similar experiments have never been carried out at sea...
(2007-09-29) Rigid Motion of a Rotating Triangle
A rigid 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 for
vector addition
does 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 to
represent accelerations, the geometric construction of the center of mass
matches the parallelograms involved in vector addition, as depicted above.)
This means that the triangle ABC rotates rigidly
about its center of mass O.
Note that this much is true regardless of the dependence of forces on distance,
since the 3 bodies are at the same distance from each other.
Quantitatively, the square of angular velocity
w is the scaling factor of the above diagram:
To a distance R corresponds an acceleration
w2 R.
This remark allows the value of that scale to be obtained
geometrically in terms of Newton's
universal
gravitational constant (G) :
w
as a function of d = AB = AC = BC
w2 d 3
= G M
= G
( m A + m B + m C )
Proof :
In the diagram, we observe that the arrow extremities
divide each side (of length d) into three segments whose lengths are
proportional to the three masses (the coefficient
of proportionality being d/M).
Thus, an arrow toward B (from
A or C) translates (by scaling lengths into
accelerations) into the following component of
the acceleration, which is equated to its gravitational counterpart
(using Newton's inverse square law)
to yield the advertised relation.
(2007-10-08) Lagrange points of two bodies in circular orbit
The 5 points where gravity balances the centrifugal force.
The above can be applied to the case of two bodies
in circular orbit around each other: A third body of
negligible mass would follow their rotation rigidly if it's
in 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 seem
to attract nearby test masses)
provided the ratio of the larger mass to the smaller one exceeds 24.96
or, more precisely:
½
( 25 + 3 Ö69 )
= 24.959935794377112278876394117361238...
The Lagrange point L4 (the Greek
triangular point) leads the
smaller body in its orbit around the larger one, while
the Lagrange point L5 (the Trojan or
trailing triangular point) lags behind.
L4 and L5 are sometimes collectively known as the "Trojan points".
Several asteroids which reside there in the Sun-Jupiter system are
named after heroes of the
Trojan war.
The leading triangular point L4 is home to the
Greek
camp led by
588 Achilles
(discovered in 1906 by Max Wolf)
with 659 Nestor,
911 Agamemnon,
1143 Odysseus,
1404 Ajax,
1583 Antilochus,
1437 Diomedes
and 1647 Menelaus.
The trailing Trojan point L5 marks the
Trojan
camp where
884 Priamus,
1172 Aeneas,
1173 Anchises
and 1208 Troilus
reside.
Early naming has left only two so-called "spies"
(both discovered in 1907 by August Kopff)...
617 Patroclus
is the lone Greek in the Trojan camp.
624 Hector
is the lone Trojan among the Greeks.
In addition,
there are three unstable Lagrangian points
(aligned with the two orbiting bodies) where the centrifugal force
exactly balances gravity.
L1 (the inner Lagrangian point) is located
between the two orbiting bodies. L2 is outside those two
bodies, on the side of the lighter one, while
L3 is on the side of the heavier one.
(2013-03-27) Gravitational self-energy of a uniform sphere
Negative, proportional to the mass squared and the inverse of the radius.
In a spherically-symmetric distribution of mass, the Newtonian gravitational
field at any point is the same as what would be created at that point
by whatever mass is located at a lesser distance from the center if that mass was
concentrated at the center (the field created by any homogeneous spherical
shell 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 application
of the theorem 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 directly
proportional 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 this
mass distribution is obtained by
integrating that force from r
to infinity (making the convenient convention that attributes zero gravitational
energy to very distant stuff, everything has negative gravitational energy):
dU
= - G M dM
[
ò
R
xdx
+
ò
¥
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 4pr2dr
multiplied into the mass density 3M/(4pR3 ).
dM = ( 3 M r2 / R3 ) dr
Therefore, the ball's total self-energy is given by the following integral:
U =
ò
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 to
explain 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 energy
has 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
1.12224(14) 1027 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 to
create black holes, according to General Relativity...
Thus, the problem is not so much to explain how gravity can create something
from nothing, which turns out to be conceptually simple, but to realize that the
mechanism is so powerful quantitatively that it cannot run raw.
Somehow, the above calculation might qualitatively describe
the creation of mass locked in particles of very small sizes
(essentially, tiny black holes whose ultimate Hawking evaporation
is forbidden by the conservation of some quantized number, like electric charge).
Douglas G. (2010-11-25)
Tidal forces from the Sun and the Moon
How the braking effect of earthly tides makes the Moon drift away.
To investigate the causes and magnitudes of Earthly tides, we shall use
a simplified model where a perfectly spherical Moon moves along a circle
around the Earth which itself moves along a circular orbit around a
perfectly spherical Sun. We'll also make the drastic assumption that
the Earth is a solid sphere completely covered by a single ocean of
seawater (without any continents, islands or other irregularities
in 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 around
their 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 rotates rigidly with angular velocity
w
According to Newton's theorem
(i.e., Gauss's theorem
applied to Newtonian gravity)
the two spheres attract each other as would two masses concentrated at the center
of each sphere. The gravitational force between them is equal to:
F = G m M / D 2
= m w 2 MD / (M+m)
= M w 2 mD / (M+m)
Dividing the above by mM we obtain the following, in two different ways:
w 2 D 3
= G (m+M)
Let's consider a cartesian frame of reference rotating about O
at the angular speed w. The x-axis goes through
both 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 sphere
of mass M (assuming that the sphere does not spin at all) at a point of latitude
q (with respect to the Ox axis) and longitude
j (where
j is defined to be zero for the half-meridian at the
surface 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 tidal
bulge of a liquid ocean coating the sphere of mass M.
(2019-08-10) Gravitational Field around a Rigid Body
If a body had spherical symmetry, its field would be that of a point-mass.
Newton's shell theorem says that an homogeneous
spherical shell produces a zero gravitational field inside it,
while the outside field is exactly what would be produced by
its 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 gravitational
torque on any mass distribution, symmetric or not, rigid or not.
(HINT: The total angular momentum of the whole
system 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 we
assume rotational symmetry about the polar axis but not necessarily north-south
symmetry with respect to the equatorial plane).
The perceived gravity derives from an apparent potential U equal to the
sum of the purely gravitational potential V (which verifies Laplace's equation
DV = 0) and a term due to rotation...
Using w to denote the Earth's angular rate of rotation,
that apparent potential U for an observer at a latitude q
and a distance r from the center of the Earth is:
U = V -
½ w2 r2 cos2 q
In this, q is the geocentric
latitude of the observer, namely the angle berween the equatorial
plane 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 j, 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 that
the Earth assumes the shape of an homogeneous incompressible fluid subjected only to
centrifugal 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 ratios
traditionally dubbed J2 and m, through an approximative
formula due to Clairaut (1743) :
f = ( a - b ) / a
= ½ ( 3 J2 + m ) where
m = a3 w2/ GM
In this, J2 is called the second dynamic form factor
(more about that soon) whereas m is defined as the ratio
of the normal acceleration at the equator
( a w2 )
to the standardized gravitational field
GM / a 2
(which would be the purely gravitational field at the equator of a spinless
spherical 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,q) =
-GM
¥
å
n = 0
a n
JnPn (sin q)
r
r n
The above only holds for bodies with axial symmetry. Otherwise a more general
expression 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 zero
for large values of r.
The same flux is thus produced through a sphere of large radius
as 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 is
the 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)
DJ2n
2n+1
J2n+1 (ppm)
DJ2n+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 mass
distribution 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 field
must differ by such a symmetric distribution
(possibly allowing negative masses).
(2010-12-31)
99942-Apophis and gravitational keyholes
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 closer
than 30000 km above Earth's surface on that date.
It will again come extremely close to Earth in 2036 and 2068.
On 2013-01-13,
NASA officially ruled out the possibility of an impact in 2036
(based on fresh data) as it also announced the
record-setting
flyby 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 of
Chelyabinsk, 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) Dark Matter (Oort, 1932.
Zwicky, 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 with
distance.
(The orbital speed is the size of the orbit divided by the
orbital period. So, Kepler's third law
implies that it varies inversely as the square root of the orbital radius.)
On the other hand, within a spherically symmetric distribution of mass of uniform
density, the centripetal force of gravity is proportional to the radius
(this is a simple application of the theorem of Gauss,
applicable to any inverse-square force field).
For the centripetal 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 if
the centripetal force was proportional to 1/R, which corresponds to a density
of 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 the
supermassive black holes formed first and pulled around them
stuff which either gave birth to ordinary matter or didn't.
(2007-09-29) Rigid Motion of a Rotating Triangle
