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Energy Transitions:
Electrostatic and Gravitational

The elecrostatic and gravitational forces obey inverse distance squared laws. This means that the kinetic energy in bound system is the negative of one half the potential energy. This is easily shown using the Virial Theorem or other methods. At the atomic level when an electron changes states the change in potential energy is twice the magnitude of the change in kinetic energy. The loss in potential energy that does not go into kinetic energy goes into the emission of a photon. The loss in potential energy in a gravitational bound systems such as a planet revolving around a star is also twice the magnitude of the gain in tangential kinetic energy. Since a photon cannot be emitted for such a transition the question is what happens to the excess energy.

Equilibrium in a Solar System

Let M be the mass of the star and m the mass of the planet. The attractive force between them is

F = −GMm/r²

where G is the graviational constant and r is the distance separating them. For simplicity now it is assumed that r is also the radius of the orbit of the planet. The centrifugal acceleration of a body traveling in an orbit of radius r is v²/r where v is the tangential velocity. A dynamic balance requires

mv²/r = GMm/r²
which reduces to
mv² = GMm/r
and thus the kinetic
energy of the planet is
K = ½mv² = ½GMm/r

On the other hand the potential energy V of the system is

V(r) = −∫r(−GMm/z)dz
which reduces to
V(r) = −GMm/r

Thus it is easily seen that

K = ½|V|


Suppose the system is in equilibrium and then something happens which disrupts that equilibrium. For a solar system that could be a planetary collision but the effects of the collision would mask the effect being sought. Suppose instead that an exoplanet the size of Jupiter crashes into the Sun. That would raise the mass of the Sun by 0.1 of 1 percent and thereby create a disequilibrium. The dynamics of the system would be given by the equation

m(d²r/dt²) = mv²/r − GMm/r²

The planet would be drawn in toward the Sun. The angular momentum L of the planet would be preserved so

L = mvr
and hence
v = L/(mr)

When this relation is substituted for v in the dynamic equation the result is

m(d²r/dt²) = L²/(mr³) − GMm/r²
or, equivalently
(d²r/dt²) = L²/(m²r³) − GM/r²

This equation can be solved analytically, as is done in the Appendix, but all that is needed here is an example of its numerical solution, as shown below.

The radial distance of the planet from the Sun would oscillate about the new equilibrium distance.

At the atomic level such an oscillation of a charged particle like an electron produces an electromagnetic perturbation; i.e., a photon. Thus the mechanism for the generation of a photon is revealed and the examination of the effect of a change of state for a solar system produces an insight into the mechanism involved in a change of state at the atomic level.


The dynamics of orbital distance r is given by the equation

(d²r/dt²) = L²/(m²r³) − GM/r²

This equation can be multiplied by (dr/dt) and the result integrated to yield

½m(dr/dt)² = E −½L²/(mr²) + GMm/r

where E is a constant of integration. The quantity ½L²/(mr²) is equivalent to the tangential kinetic energy K. Thus the equation can be expressed as

½m(dr/dt)² + ½mv² − GMm/r = E

The first two terms on the left are, respectively, the radial kinetic energy and the tangential kinetic energy. The third term is the potential energy. Thus E is the total energy, which is constant.

To get an analytical solution the equation

½m(dr/dt)² = E −½L²/(mr²) + GMm/r
can be solved for (dr/dt) as
(dr/dt)² = (2E/m) −L²/(m²r²) + 2GM/r
and hence
(dr/dt) = [(2E/m) −L²/(m²r²) + 2GM/r]½
and multiplication by r gives
r(dr/dt) = [(2E/m)r² + 2GMr −L²/m²]½
or, equivalently
r(dr/dt) = (2E/m)½[r² + (GMm/E)r − L²/(Em)]½

In differential form this last equation is

rdr/[r² + (GMm/E)r − L²/(Em)]½ = (2E/m)½dt

The integration of this equation gives

(To be continued.)

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