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**INTRODUCTION:**

This web page provides insight as to how energy and forces arise from electric, magnetic and gravitational fields.

**FIELDS:**

An important concept in physics is that the total energy of a particle is finite but the particle's energy distribution in space stretches out to infinity. Hence a high vacuum contains energy due to nuclear, electric, magnetic and gravitational fields originating from distant particles.

The concept that nuclear, electric, magnetic and gravitational vector fields contain energy leads to well known equations for apparent nuclear, electric, magnetic and gravitational forces as well as to a better understanding of physical phenomena.

**ISOLATED PARTICLE:**

Each isolated particle has corresponding surrounding nuclear, electric, magnetic and gravitational vector fields that extend out to infinity but which contain only a finite amount of distributed energy. The effective range of nuclear fields is very short.

**INTERACTING PARTICLES:**

When multiple particles co-exist their vector fields overlap everywhere in the universe. At every point in the universe there is a net nuclear field vector, a net electric field vector, a net magnetic field vector and a net gravitational field vector. The field overlap causes a change in the field energy density of every element of volume. The law of conservation of energy simultaneously causes an equal but opposite change in the kinetic energy content of the same element of volume and/or the emission/absorption of a photon. The apparent force between particle i and all the other particles in a cluster is really just the change in total field energy with respect to a change in the position of particle i in the cluster.

**ASSUMPTION:**

Assume that at every point in space the electric, magnetic and gravitational field vectors from different particles linearly add to give position dependent net electric, net magnetic and net gravitational field vectors.

The net nuclear field vector **Fn** at position **X** is given by:

**Fn** = Sum of all **Fni**

Fni denotes the nuclear field vector at **X** due to the ith particle. Note that the nuclear fields contain an Exp(- R / Rn) term which restricts their effective radius of action.

The net magnetic field vector **Fm** at position **X** is given by:

**Fm** = Sum of all **Fmi**

where **Fmi** denotes the magnetic field vector at **X** due to motion of the ith particle.

The net electric field vector **Fe** at position **X** is given by:

**Fe** = Sum of all **Fei**

where **Fei** denotes the electric field vector at **X** due to the ith particle.

The net gravitational field vector **Fg** at position **X** is given by:

**Fg** = Sum of all **Fgi**

where **Fgi** denotes the gravitational field vector at **X** due to the energy of the ith particle.

**ASSUMPTION:**

Assume that the nuclear, magnetic, electric and gravitational field vectors of a particle are all mutually mathematically orthogonal. In effect this assumption means that field unit vectors have six dimensions. In addition to the normal **X, Y, Z** cartesian co-ordinate unit vectors there are imaginary co-ordinate **jX, jY, jZ** unit vectors, where **j = (-1)^0.5**. The additional imaginary unit vectors allow gravitational field energy to be negative.

**ASSUMPTION:**

Assume that for each field type **the rest energy density of the field is proportional to the net field vector squared**.

**TOTAL FIELD ENERGY DENSITY:**

Then the element of total field energy **dEo** contained in an element of volume **dV = (dX dY dZ)** at **X** can be expressed as:

**dEo = [(Kn / 2) |Fn|^2 + (Km / 2) |Fm|^2 + (Ke / 2) |Fe|^2 - (Kg / 2) |Fg|^2] dV**

where:

**Fn** = net nuclear field vector at **X**

**Fm** = net magnetic field vector at **X**

**Fe** = net electric field vector at **X**

**Fg** = net gravitational field vector at **X**

**Kn** = positivee natural constant

**Ke** = positive natural constant

(Ke = Epsilon = electric permittivity of free space)

**Km** = positive natural constant

(Km = Mu = magnetic permeability of free space)

**Kg** = positive natural constant = 1 / (4 Pi G)

where:

G = Newton gravitational constant

Note that the gravitational field energy of normal matter is negative. When the field energy density at position **(X - Xo)** becomes more negative the kinetic energy density at position
**(X - Xo)** instantaneously becomes more positive by the same amount so that the total energy density at position X is unchanged.

Note that the above formulation of energy density implicitly assumes that the energy content of a universe with no energy and no charge is zero.

**RELATIVE SIZE OF ENERGY COMPONENTS:**

In most practical situations:

Nuclear Field Energy >> Electromagnetic Field Energy >> |Gravity Field Energy|

Thus changes in gravity field energy are only important when both the nuclear energy and the electromagnetic energy components are stable. Similarly changes in electromagnetic energy are only important when the nuclear energy component is stable.

The nuclear field energy density decays exponentially with increasing radius from the nominal particle position. The magnetic field energy density component decreases less quickly with increasing radius. The electric and gravitational field energy density components both decrease slower and in proportion to:

(1 / R)^4

**PARTICLE CHARACTERIZATION:**

Each particle **i** can be characterized as having a charge **Qi**, a magnetic moment **Mi**, a radius **Ri** and a rest energy **Eoi**. The rest energy **Eoi** includes both non-field energy and the energy content of the particle's electric, magnetic and gravitational fields.

**FIELD ENERGIES:**

The field energies are integrals over spacial volume of the field energy densities.

**MULTIPLE PARTICLES:**

Most real systems involve multiple particles. A real system has a Centre of Momentum (CM), which serves as a system position reference point.

**CONSERVATION OF ENERGY FOR AN ISOLATED CLUSTER OF INTERACTING PARTICLES:**

The law of conservation of energy requires that for any fully isolated system the total system energy with respect to an inertial observer is constant. Hence, if progressive overlap of vector fields causes a change in field energy the law of conservation of energy requires a corresponding opposite sign change in kinetic energy and photon energy to keep the total energy constant.

**CONSERVATION OF ENERGY FOR TWO ISOLATED INTERACTING PARTICLES:**

If two particles forming an isolated system interact the individual particle energies can remain unchanged or an element of energy **dE** can be transferred from one particle to the other or a third particle can be formed but the total system energy remains unchanged. Note that a process involving creation of a third particle from an interaction between two particles is usually not reversible because such reversal requires a three body interaction. The probability of a random three body interaction is usually extremely small compared to the probability of a random two body interaction.

A common example is two bodies interacting and liberating a photon in circumstances where the probability of photon capture is very small. Thus conservation of energy in combination with photon emission leads to formation of assemblies of particles mutually bound in potential wells and determines the direction of evolution of many processes in the local universe. eg Condensation of water vapor to form liquid water.

**CLUSTER OF PARTICLES:**

A nearly isolated cluster of particles can absorb or emit photons, which increase or decrease the total cluster energy. Hence the total cluster energy will gradually change until the rate of energy absorption equals the rate of energy emission. The Earth, in its orbit in space around the sun, is an example of a nearly isolated cluster of particles.

**CHANGE IN FIELD ENERGY DUE TO CHANGE IN FIELD OVERLAP:**

Vector fields from different particles add linearly. Hence, overlap of vector fields from multiple particles changes the net local vector field strength linearly. However, the local field energy density is proportional to the net local vector field strength squared. Hence the local field energy density changes nonlinearly as the vector field overlap changes.

**UNDERSTANDING FIELD OVERLAP:**

Consider an electric field of the form:

(field) = Q K / R^2

For an isolated particle with charge Q:

(field)^2 = Q^2 K^2 / R^4 = electric field energy density

For an isolated particle with charge dQ:

(field)^2 = (dQ)^2 K^2 / R^4 = field energy density

For an isolated particle with charge (Q + dQ):

(field)^2 = (Q + dQ)^2 K^2 / R^4

= [Q^2 K^2 / R^4] + [(dQ)^2 K^2 / R^4] + [2 Q dQ K^2 / R^4]

= field energy density

Thus bringing the isolated particles Q and dQ together introduced an additional field energy density component of:

[2 Q dQ K^2 / R^4]

The total additional field energy introduced by this component is:

Integral from R = Ro to R = infinity of:

[2 Q dQ K^2 / R^4][4 Pi R^2] dR

= Integral from R = Ro to R = infinity of:

[8 Pi Q dQ K^2 / R^2] dR

= [- 8 Pi Q dQ K^2 / R]|R = infinity -[- 8 Pi Q dQ K^2 / R]|R = infinity|R = Ro

= [8 Pi Q dQ K^2 / Ro]

= (additional field energy)

The change in additional field energy with respect to distance is:

d((additional field energy) / dRo = d[8 Pi Q dQ K^2 / Ro] / dRo

= - [8 Pi Q dQ K^2 / Ro^2]

= electric force

Make the substitution:

(8 Pi K^2) = 1 / (4 Pi Epsilon)

or

K^2 = 1 / (32 Pi^2 Epsilon)

Then:

(electric force) = Q dQ / (4 Pi Epsilon R^2)

as expected.

(electric field energy density) = Q^2 K^2 / R^4

= Q^2 / (R^4 32 Pi^2 Epsilon)

= [Q / (4 Pi Epsilon R^2)]^2 [Epsilon / 2]

= conventional expression for electric field energy density

A similar derivation can be made for magnetic and gravitational fields. Thus the concept of force at a distance is hogwash. Forces at a distance are really changes in potential energy that arise from changes in particle field overlap.

**FORCES:**

A change in the total cluster field energy dEo with respect to a change in a position d**Xi** of a particular particle is expressed as a force on that particle. If the velocity of a particle affects its potential energy (eg an electrically charged particle moving in a magnetic field which creates a secondary magnetic field) then there is an additional dynamic force component.

A force on a particle due to a potential energy gradient causes a corresponding change in energy motion. Energy motion is also known as momentum.

Let **Xo** be an observers position vector. The force **Fi** on particle **i** causes a change in kinetic energy dEki during a change in particle position vector:

d(**Xi - Xo**).

Hence:

dEki = **Fi** * d(**(Xi - Xo)**)

**FORCE SIGNS:**

Electric and magnetic vector fields have real unit vectors.

Overlap of electric vector fields from charges of opposite sign causes net far field vector cancellation and hence a reduction in total positive field energy, leading to an increase in kinetic energy which appears to be the result of an attractive force. Similarly, overlap of electric vector fields from charges of the same sign causes net far field vector addition and hence an increase in total positive field energy, leading to a reduction in kinetic energy which appears to be the result of a repulsive force.

Magnetic fields, when viewed as originating from small electric current loops, also diminish in proportion to (1 / |**X - Xi**|^2) but are orientation dependent with respect to sign. As with electric charges, opposite signs lead to an attractive force whereas same signs lead to a repulsive force.

Gravitational fields from normal matter have unit vectors that are proportional to (-1)^0.5, which when squared causes negative potential energy / unit volume. Overlap of gravitational vector fields from mutual proximity of normal matter masses causes far field vector addition and hence an increase in magnitude of the negative field energy. Hence as normal masses approach each other they gain positive kinetic energy to keep the total energy constant, leading to the appearance of an attractive force.

Hence forces are simply a result of the law of conservation of energy.

Pair production lifts energy from below the field free vacuum state to above the field free vacuum state. Hence an anti-matter particle is in effect an energy hole with negative energy. The gravitational field related to this hole should reverse direction. When an electron-positron pair annihilate each other the change in rest mass energy is twice the electron rest mass energy. Thus the energy of the positron is negative with respect to the field free vacuum reference.

The form of the vector field equations suggests that overlap of the gravitational vector field from ordinary matter with the gravitational vector field from antimatter results in the far field vector cancellation which makes the total potential energy less negative and hence causes a force that is repulsive. Hence normal matter and antimatter will gravitationally repel each other. Hence we do not expect to find any free antimatter in our solar system.

The deduced repulsive gravitational force between normal matter and antimatter may explain certain aspects of the intergalactic expansion of the universe that have been observed by astronomers. Two galaxies, one composed of normal matter and the other composed of anti-matter will likely repel each other. Some parties attribute this repulsion to "dark energy". The gravitational interaction between matter and antimatter is presently impossible to resolve in the laboratory because the electric and magnetic forces affecting single particles are many orders of magnitude larger than the gravitational force.

**FINITE CONSTANT CHARGE:**

An isolated particle has a finite constant electric charge Q. This charge is the same regardless of its distance from the observer. The electric field E is a result of the contained charge. The surface area at radius R from the nominal position of a particle is (4 Pi R^2). Hence the electric field E at radius R is given by:

E = Q / (4 Pi Epsilon R^2)

where:

Epsilon = natural constant

**ELECTRIC FIELD RELATED FORCE:**

For an electric field:

**Ke = (Epsilon)**

where:

**Epsilon** = permittivity of free space

= **8.85 X 10^-12 coulomb^2 newton^-1 m^-2**

Epsilon is one of a handfull of independent natural constants that can only be determined by experimental measurement.

Consider a particle with charge Qi with radius Ri. For R > Ri the electric field **Fe** around isolated charged particle i is given by:

**Fe = Qi / (4 Pi R^2 Epsilon)**

where:

**Pi** = 3.14159

**R** = radius from the particle center

The corresponding electric field energy density is given by:

**(Ke / 2) Fe^2 = (Epsilon / 2)[Qi / (4 Pi R^2 Epsilon)]^2**

Then the electric field energy **Ee** surrounding the charge Qi is given by:

**Ee = Integral from Ri to infinity of:
[Epsilon / 2][Qi / 4 Pi R^2 Epsilon]^2] 4 Pi R^2 dR**

=

This

Thus an isolated charge **Qi** has an electric field rest energy **Eei** of:

**Eei = (Qi^2 / 8 Pi Epsilon Ri)**

Similarly an isolated charge **Qj** has an electric field rest energy **Eej** of:

**Eej = (Qj^2 / 8 Pi Epsilon Rj)**

If the two charges are both within radius Ro the total electric field rest energy **Eet** is given by:

**Eet = ((Qi + Qj)^2 / 8 Pi Epsilon Ro)
= ((Qi^2 + Qj^2 + 2 Qi Qj) / (8 Pi Epsilon Ro)**

Hence the change in total electric field potential energy required to bring two isolated charges together is:

**Eet - Eei - Eej
= (2 Qi Qj) / (8 Pi Epsilon Ro)**

Differentiating this expression with respect to Ro gives:

**F = d(Et - Ei - Ej) / dRo
= - (1 / 4 Pi Epsilon)( Qi Qj / Ro^2)**

Recall that the electrostatic force **Fe** attracting two charges **Qi** and **Qj** separated by distance **Ro** is given by:

**Fe = - (1 / 4 Pi Epsilon)(Qi Qj / Ro^2)**

Hence we have shown that the electric force is simply the change in electric field energy with respect to position that results from overlap of electric fields.

**MAGNETIC FIELD ENERGY:**

For a magnetic field the constant **Km** is given by:
**Km = (1 / Mu)**

where:

**Mu** = permiability of free space

= 4 Pi X 10^-7 webers / amp-m

= 4 Pi X 10^-7 T-m / amp

1 Tesla (T) = 1 weber / m^-2

Consider a toroidal solenoid. The magnetic field inside the solenoid is given by:

**B = (Mu N I) / L**

where:

**B** = magnetic field strength
**N** = number of turns

**I** = current through each turn

**L** = average magnetic path length

The magnetic field volume within the toroidal solenoid is:

**L A**

where **A** is the cross sectional area of the magnetic field.

The solenoid self inductance is:

**N B A / I
= (N A / I) (Mu N I / L)
= (Mu N^2 A) / L**

The magnetic field energy Em is:

**Em = (inductance) I^2 / 2
= (Mu N^2 I^2 A) / 2 L**

The magnetic field energy density is:

**(Mu N^2 I^2 A) / (2 L (L A))
= (Mu N^2 I^2) / 2 L^2
= (Mu / 2)(N I / L)^2
=(1 / 2 Mu) B^2
= (Mu / 2) H^2
= (Km / 2) H^2**

Hence the magnetic field energy density is given by:

**(Km / 2) Fm^2 = (Mu / 2) H^2**

where:

**Fm = H = (B / Mu)**

Thus we have shown the relationship between toroidal solenoid parameters and magnetic field energy density.

**MAGNETIC FORCE:**

In an electrical contactor the closing force results from reducing the length of a magnetic circuit of approximately uniform cross-section. The stored magnetic energy is:

**Em = (Mu N^2 I^2 A) / 2 L**

The contactor closing force is given by:

**dEm / dL = - (Mu N^2 I^2 A) / 2 L^2**

**GRAVITATIONAL FIELD:**

Every element of volume has associated with it a gravitational vector field component that contains gravitational field energy. The gravitational field vector is mathematically orthogonal to both electric and magnetic field vectors. Like the electric and magnetic fields the gravitational field has a net vector flux per unit area that when squared is proportional to gravitational field energy density.

However, because gravitational fields contain energy and gravitational fields are themselves the result of energy the resulting forces are not exactly proportional to **(1 / R^2)**.

Assume that each particle is surrounded by a radial gravitational vector field.

For gravity the total external radial vector flux from a particle is proportional to the contained energy.

The total gravitational vector flux due to contained energy Ec is given by:

Flux = + j Ec / Kg C^2

where:

j = (-1)^0.5

Kg = natural constant

C = speed of light

Ec = total energy contained in sphere with radius **R**.

The surface area of a sphere of radius **R** is:

**(4 Pi R^2)**.

Hence, for a single isolated particle the external local vector field strength diminishes approximately in proportion to:

**1 / (R^2)**.

The gravitational field flux per unit area at radius **R** due to energy **E** contained inside radius **R** is given by:

**Gravitational Field Flux / area = Fg
= j Ec / (C^2 Kg 4 Pi R^2)**

Assume:

Ec = M C^2

so that M = mass contained within radius R.

**Assume that the gravitational field energy density is proportional to the square of the vector field flux per unit area.**
Then the gravitational field energy density at **R** is given by:

**(Gravitational Field Energy / unit volume) = (Kg / 2) (Gravitational Field Flux / unit area)^2**

= (Kg / 2) [j Ec / (C^2 Kg 4 Pi R^2)]^2

= (Kg / 2) [j M / Kg 4 Pi R^2]^2

= - [1 / (32 Kg)] [(M^2) / (Pi^2 R^4)]

Note that the gravitational field energy density is negative.

Let R = Ro at the surface of a solid sphere of mass Mo. The gravitational field energy between radius R = Ro and R = infinity is:

Integral from R = Ro to R = infinity of:

- [1 / (32 Kg)] [(M^2) / (Pi^2 R^4)] 4 Pi R^2 dR

= Integral from R = Ro to R = infinity of:

- [1 / (8 Kg Pi)] [(M^2) / (R^2)] dR

Define:

Mi = mass inside a sphere of radius Ri where Ri > Ro

and

Mo = same mass inside radius Ro

However: Mi = Mo + Integral from R = Ro to R = Ri of (dM / dR) dR.

Mass density = (energy density / C^2). Then:

dM / dR = (Field energy density) 4 Pi R^2 / C^2

= - [1 / (32 Kg)] [(M^2) / (Pi^2 R^4)] [4 Pi R^2 / C^2]

= - [1 / (8 Kg)] [(M^2) / (Pi R^2 C^2)]

Thus:

[dM / M^2] = - [1 / (8 Kg Pi C^2)] [dR / R^2]

Integrate from R = Ro, M = Mo to R = Ri, M = Mi giving:

{[- 1 / Mi] - [- 1 / Mo]} = - [1 / (8 Kg Pi C^2)]{[- 1 / Ri] - [- 1 / Ro]}

or

{[1 / Mi] - [1 / Mo]} = [1 / (8 Kg Pi C^2)][(1 / Ro) - (1 / Ri)]

or

{(Mo - Mi) / Mo Mi} = [1 / (8 Kg Pi C^2)] [(1 / Ro) - (1 / Ri)]

or

(Mo - Mi) = [(Mo Mi) / (8 Kg Pi C^2)][(1 / Ro) - (1 / Ri)]

or

(Eo - Ei) / C^2 = [(Mo Mi) / (8 Kg Pi C^2)][(1 / Ro) - (1 / Ri)]

or

(Eo - Ei) = [(Mo Mi) / (8 Kg Pi)] [(1 / Ro) - (1 / Ri)]

Assume:

**Kg = 1 / (G 4 Pi)**

Then:

(Eo - Ei) = [(G Mo Mi) / (2)][(1 / Ro) - (1 / Ri)]

or

**(Ei - Eo) = - [(G Mo Mi) / 2][(1 / Ro) - (1 / Ri)]**

which is the gravitational field energy between R = Ri and R = Ro.

In order to evaluate the full gravitational field energy we let Ri go to infinity. Then:

**(Ei - Eo) = - [(G Mo Mi) / 2 Ro]**

Remember that in this formula Mo and Mi refer to the same object except that Mi includes its gravity field energy which is negative. Hence (Ei - Eo) is the gravity field energy formed by bringing bits of matter from R = infinity to R = Ro. Note that there is simultaneous creation of an equal amount of positive kinetic energy.

Mi = Mo + (Ei - Eo) / C^2

Hence:

(Ei - Eo) = - [(G Mo Mo) / 2 Ro] - [(G Mo (Mi - Mo) / 2 Ro]

= - [(G Mo Mo) / 2 Ro] - [(G Mo (Ei - Eo) / 2 C^2 Ro]

Thus:

(Ei - Eo)[1 + (G Mo / 2 C^2 Ro)] = - [(G Mo Mo) / 2 Ro]

or

**(Ei - Eo) = - [(G Mo^2) / 2 Ro] / [1 + (G Mo / 2 C^2 Ro)]**

**GRAVITATIONAL FORCE:**

Gravitational force

= (change in gravitational energy with respect to a change in Ro)

= **d(Ei - Eo) / dRo** = + {[(G Mo^2) / 2 Ro^2] / [1 + (G Mo / 2 C^2 Ro)]}

- {[(G Mo^2) / 2 Ro](G Mo / 2 C^2 Ro^2) / [1 + (G Mo / 2 C^2 Ro)]^2}

= + {[(G Mo^2) / 2 Ro^2] / [1 + (G Mo / 2 C^2 Ro)]}

- {[(G^2 Mo^3) / 4 C^2 Ro^3] / [1 + (G Mo / 2 C^2 Ro)]^2}

=+ {[(G Mo^2) / 2 Ro^2] / [1 + (G Mo / 2 C^2 Ro)]}

- {[G Mo / 2 C^2 Ro][(G Mo^2) / 2 Ro^2] / [1 + (G Mo / 2 C^2 Ro)]^2}

= **{[(G Mo^2) / 2 Ro^2] / [1 + (G Mo / 2 C^2 Ro)]} {1 - ([G Mo / 2 C^2 Ro] / [1 + (G Mo / 2 C^2 Ro)])}**

For:

**[(G Mo) / (2 C^2 Ro)] << 1**

this gravitational force expression simplifies to:

**d(Ei - Eo) / dRo ~ [(G Mo^2) / 2 Ro^2] [1 - (G Mo / C^2 Ro)]**

Note that the gravitational force is not exactly proportional to 1 / R^2 due to the energy contained within the gravitational field itself. However, this effect is very small and is difficult to detect with nearly circular planetary orbits. Precise measurement of the advance of the perihelion of the elliptical orbit of Mercury was required to experimentally observe this effect.

**NEWTONIAN METHOD:**

Newtons gravitational force equation is:

Force = G Ma Mb / R^2

In Newtons method there are forces at a distance but there is no concept of a field.

According to Newton the potential energy released in bringing mass dM from infinity to Ro is:

Integral from R = infinity to R = Ro of:
dE = (G M dM) dR / R^2
= - G M dM / Ro

Then the potential energy released in building up M at Ro from M = 0 to M = Mo is:

DeltaE = - G M^2 / 2 Ro

Thus according to Newton moving mass Mo from distributed particles at infinity to a spherical shell at R = Ro causes a change in potential energy :

**DeltaE = - G Mo^2 / 2 Ro**

Thus the field method and Newton's method give similar results if:

**(G Mo / C^2 Ro) << 1**

or if

**(G Mo / Ro) << C^2**

Note that (2 G Mo / Ro)^0.5 is the theoretical escape velocity from a Newtonian system. This parameter becomes significant when the escape velocity is a significant fraction of the speed of light C as is the case in the proximity of a large black hole.

**ESCAPE VELOCITY:**

The escape velocity is the minimum radial velocity that an object must have to escape from a gravitational potential well.
In Newtonian gravitation:

Gravitational acceleration is:

dV / dT = - G Mo / R^2

or

dV = - (G Mo) dT / R^2

= - (G Mo) (dT / dR) dR / R^2

= - (G Mo) (1 / V) dR / R^2

Hence:

V dV = - (G Mo) dR / R^2

Integrating from V = Vo, R = Ro to V = Vf, R = Rf gives:

(Vf^2 / 2) - (Vo^2 / 2) = (G Mo) [(1 / Rf) - (1 / Ro)]

Escape is prevented if Vf = 0 at Rf = infinity

or

- (Vo^2 / 2) = (G Mo) [ - (1 / Ro)]

or

Vo^2 = (2 G Mo / Ro)

Thus a particle cannot directly escape from a gravitational potential well if its initial velocity Vo is:

Vo < (2 G Mo / Ro)^0.5

Hence if:

(2 G Mo / Ro) > C^2

then nothing can escape and there is a black hole.

Consider the value of (G Mo / C^2 Ro) near the surface of the sun:

**G** = 6.67384 X 10^-11 m^3 kg^-1 s^-2

**Ms** = 1.98892 X 10^30 kg

**Rs** = 6.955 X 10^8 m

**C** = 3 X 10^8 m / s

Substituting in the above formula gives:

[G Mo / (Ro C^2)]

= [6.67384 X 10^-11 m^3 kg^-1 s^-2 X 1.98892 X 10^30 kg] / [6.955 X 10^8 m X 9 X 10^16 m^2 / s^2]

= .212 X 10^-5

= **2.12 X 10^-6**

Under favorable circumstances this error fraction of two parts per million in apparent gravitational force can be detected by precise astronomical observations.

This web page last updated July 4, 2018.

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