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The Mass and Rate of Rotation of a Protoplanet Sweeping Up Material in a Ring Around a Star |
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Consider our own solar system with respect to the matter of the direction of revolution and rotation.
All of these point to the formation of the planets from a ring or rings of material circling the Sun.
A Keplerian stellar ring is one in which all of the particles are in balance with the gravitataional attraction to the star and the centrifugal force in the orbit. Let m be the mass of a bit of material and M the mass of the Sun. The gravitational attraction of the Sun for this bit is equal to GmM/R², where G is the gravitation constant and R is the distance separating the bit from the center of the star;. The centrifugal force on the bit in an orbit of radius R is mv²/R.
Equating these two gives
where γ is equal to (GM)^{½}.
Let m be the mass of a bit traveling in a circular orbit of radius R with a velocity of v. Its angular momentum is mRV. If the bit were originally traveling in an orbit of radius R_{o} at a velocity v_{o} and then moved to another orbit of radius R the conservation of angular momentum requires:
The equation for the velocity of material in the Keplerian ring can be put into the form
Thus if an object is originally in the ring at a radius R_{o} traveling with a tangential velocity of v_{o} and then moves to a radius of R its tangential velocity will become v_{o}(R_{o}/R) but the material in the ring at that radius will be travelling with a tangential velocity of v_{o}(R_{o}/R)^{½}.
Thus if the object moves toward the star it will be moving faster than the surrounding ring material and if moves away from the star it will be moving slower than the surrounding ring material. That means that an object moving to a smaller orbit will crash into the ring material and roll up mass like a snowball. On the other hand an object to a larger orbit will find the ring material colliding into it. In either case the object will not only accumulate mass it will accumulate angular momentum and rotate in the same direction the ring is revolving around the star. The following diagram illustrates the effects.
In the diagram the white area represents ring material moving from the southeast to the northwest. The center of the star is toward the southwest in the diagram. The brown circle represents an object moving to a larger orbit. It is moving slower that fing material and hence ring material crashes into it. It crashes into the spherical object roughly parallel to its surface at a distance of roughly the radius r of the spherical object. The velocity at which it crashes into the object is equal to the (v_{r}−v) where v_{r} is the velocity of the ring material and v is the velocity of the object. This would start the process of the accumulation of larger objects.
The green circle represents an object moving to a smaller orbit. It acquires ring material at its surface traveling at a velocity of (v−v_{r}). This causes the object to rotate in the same direction the ring is revolving about the star.
There are two factors which could cause objects in a Keplerian ring to move out of their original orbits. One is radiation pressure. If the Keplerian ring was established before the central material ignited into a star. the radiation pressure would push all material outward but it would push the smaller material relatively more strongly.
Once any larger object is developed there would be orbital resonance which would move material out of orbits which have periods of revolution which are one half or a similar simple fraction of the period of the larger object. This phenomenon is displayed by the Kirkwood Gaps in the orbits of asteroids in the asteroid belt between Mars and Jupiter.
The material does not have to move very far to break the resonance. Thus planets are formed near the orbits of resonance. This is illustrated in the Bode-Titius relation.
Let m and r be the mass and radius of a protoplanet in a Keplerian stellar ring. The protoplanet is moved out of its original orbit and accumulates mass. Whether it moves toward or away from the star it acquires rotation in the same direction as the rotation of the stellar ring.
Let σ be the volume mass density of the material being acquired at the equator of the protoplanet and γ the area of impact. Let Δv be the velocity by which the material impacts the protoplanet's equatorial edge. Therefore
In the above [M/T] = [M/L³][L²][L/T] represents a check on the dimensionality of the equation above it. [MLT] denotes mass, length and time, respectively.
From geometry
where ρ is the mass density of the protoplanet.
The rate at which the protoplanet is acquiring angular momentum L is
In the relation dL/dt = (dm/dt)Δvr, Δv changes very slowly and can be taken to be constant over an extended interval of time.
Since m=ρ(4/3)πr³, the radius r can be expressed as
Therefore
An integration of the above equation yields
where β is equal to (3/4)^{4/3}(ρπ)^{−⅓} and
its dimensions are [M/L³]^{−⅓}, which reduces
to [M^{−⅓}L].
The dimensions in the above equation are then
[ML²/T] = [M^{−⅓}L][L/T][M^{4/3}].
The moment of inertia J of the protoplanet as a solid sphere is given by:
where α is equal to (2/5)(4/(3ρπ))^{2/3} and thus its dimensions are [M/L³]^{−2/3} which reduces to [M^{−2/3}L²].
The angular momentum L is equal to Jω, where ω is the rate of angular rotation. This means that
The ratio (β/α) evaluates to:
Therefore
The dimensions of Δv are [L/T] and those of ρ are [M/L³]
The ratio (m/ρ) is equal to the volume of the protoplanet and its cube root is proportional to its radius. Therefore
where K is a constant and v_{t} is the tangential velocity of surface of the protoplanet at its equator.
If r is the radius of the protoplanet then its period of rotation t_{rot} is given by
The velocity Δv is roughly given by
(To be continued.)
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