VbvAstE-001

where C Ω is a constant.
Thus
and
E x = 2 C Ω x, E y = 2 C Ω y, E z = −4 C Ω z (8.10)
and we obtain important equations:
div E Ω = 0 (8.11)
ρ Ω = 0; (8.12)
γg Ω = ρE Ω. (8.13)
Since centrifugal force must be contra-balanced by electric force
and
γ 2Ω 2 x = ρ 2C Ω x (8.14)
C Ω = γ Ω2
ρ
The potential of a positive uniform charged ball is
ϕ(r) = Q ( ) 3
R 2 − r2
2R 2
= Ω2
√
G
(8.15)
(8.16)
The negative charge on the surface of a sphere induces inside the sphere the potential
ϕ(R) = − Q R
(8.17)
where according to Eq.(8.4) Q = √ GM, and M is the mass of the star.
Thus the total potential inside the considered star is
√ )
GM
ϕ Σ =
(1 − r2
+ √ Ω2
r 2 (3cos 2 θ − 1) (8.18)
2R R 2 G
Since the electric potential must be equal to zero on the surface of the star, at
r = a and r = c
ϕ Σ = 0 (8.19)
and we obtain the equation which describes the equilibrium form of the core of a
rotating star (at a2 −c 2
a 2 ≪ 1)
a 2 − c 2
≈ 9 Ω 2
a 2 2π Gγ . (8.20)
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