San José State University

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Thayer Watkins
Silicon Valley
& Tornado Alley
U.S.A.

The Mass and Rate of Rotation
of a Protoplanet Sweeping Up
Material in a Ring Around a Star

Background

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

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

mv²/R = GmM/R²
the mass m cancels out
to give
v² = GM/R
and thus
v = γ/R½

where γ is equal to (GM)½.

The Conservation of Angular Momentum

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 Ro at a velocity vo and then moved to another orbit of radius R the conservation of angular momentum requires:

mvR = mvoRo
which reduces to
v/vo = Ro/R

The equation for the velocity of material in the Keplerian ring can be put into the form

vr/vo = (Ro/R)½

Thus if an object is originally in the ring at a radius Ro traveling with a tangential velocity of vo and then moves to a radius of R its tangential velocity will become vo(Ro/R) but the material in the ring at that radius will be travelling with a tangential velocity of vo(Ro/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 (vr−v) where vr 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−vr). 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.

Further Implications of the Model

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

dm/dt = σγΔv
[M/T] = [M/L³][L²][L/T]

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

m = (4/3)ρπr³
[M] = [M/L³][L³]

where ρ is the mass density of the protoplanet.

Angular Momentum

The rate at which the protoplanet is acquiring angular momentum L is

dL/dt = (dm/dt)Δvr
[ML²/T²] = [M/T][L/T][L]

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

r = (3m/(4ρπ))
[L] = [[M]/[M/L³]

Therefore

dL/dt = Δv(3/(4ρπ))m(dm/dt)
[ML²/T²] = [L/T][M/L³]−⅓[M][M/T]

An integration of the above equation yields

L = Δv(3/4)4/3(ρπ)−⅓m4/3
[ML²/T] = [L/T][M/L³]−⅓[M4/3]
or, more succintly as
L = βΔvm4/3 .

where β is equal to (3/4)4/3(ρπ)−⅓ and its dimensions are [M/L³]−⅓, which reduces to [M−⅓L]. The dimensions in the above equation are then
[ML²/T] = [M−⅓L][L/T][M4/3].

Moment of Inertia

The moment of inertia J of the protoplanet as a solid sphere is given by:

J = (2/5)mr²
[ML²] = [M][L²]

which can be expressed
in terms of the mass as
J = αm5/3
[ML²] = [M/L³]−2/3[M5/3

where α is equal to (2/5)(4/(3ρπ))2/3 and thus its dimensions are [M/L³]−2/3 which reduces to [M−2/3L²].

Rate of Rotation

The angular momentum L is equal to Jω, where ω is the rate of angular rotation. This means that

ω = L/J = Δvβm4/3/(αm5/3)
= Δv(β/(αm) = Δv(β/α)/m
[T]−1 = [L/T] [M−⅓L]/(M−2/3][L²][M]).

The ratio (β/α) evaluates to:

(β/α) = (3/4)4/3(ρπ)−⅓/(((2/5)(4/(3ρπ))2/3))
= (2/5)[(4/3)²/(ρ−⅓π) = 4.417865ρ

Therefore

ω = 4.417865Δvρ/m
[T−1] = [L/T][M/L³][M−⅓]

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

ω = KΔv/r
and hence
vt/Δv = K

where K is a constant and vt 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 trot is given by

trot = 2π/vt

The velocity Δv is roughly given by

Δv = vo[((Ro-r)/Ro − ((Ro-r)/Ro)½)

(To be continued.)

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