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© The scientific sentence. 2010

Relativity: Newtonian gravitational field




1. Gravitational field of single mass

The weight of an object is the gravitational force of the the earth acting on it. More generally, the gravitational force F acting on an object of mass m, located at a point P, is due to the gravitational field g of another mass M.

The law of universal gravitation is:

F = - G m M r/r2
G is the gravitational constant, r is the distance between the two masses, and the unit vector r is directed from M to m. The force F is directed from m to M.

Therefore, we write:

g = G M /r2

So

F = - g m r

F = - g m r

The gravitational force acting on a particle is equal to its mass times the gravitational field created by another mass.

Hence the gravitational field g has then the following expression: (notice that rr = 1)

g = - (F/m)r = - (G M/r2) r

g = - (F/m)r = - (G M/r2) r

The gravitational field is the the gravitational force per unit of mass. It is function of the distance (latitude for earth) and mass of the field.



We have used the definition of g as the acceleration of an object due to the gravity of the earth of magnitude near the earth surface is - 9.80 m/s2. The sign minus is set because the unit vector is directed from the center of the earth and the acceleration is downward.



2. Gravitational field of many masses





For many masses m1, m2, m3, ..., mn of gravitational fields g1, g2, g3, ..., gn resprctively, the mass m undergoes a unique resultant net gravitational field g wich is:

g = - (G m1/r12) r1 - (G m2/r22) r2 - (G m3/r32) r3 ... - (G mn/rn2) rn
= - G Σ (mi/ri2) ri

g = - G Σ (mi/ri2) ri

3. Gravitational Gauss's law

3.1. single mass





Gauss's law is written for the gravitational field vector as:

s g . dS = ∫v ∇. g dV

At spherical distance r, we have g = g(r).

With g = - G m/r2 r, and g . dS = g dS n . r = g dS
Gauss's theorem becomes:

∫(s) g . dS = 4 π r2 g(r)
= - G m/r2 x 4 π r2 = - 4 πG m

Single mass

s ∇. g = - 4 πGm

3.2. Many masses inside a spherical contour S





m = Σ mi = ∫ dm

dm = ρ dV = ρ(r') d3r'

Gauss's theorem becomes:

v ∇ g dV = - 4 πG ∫v ρ dV
v∇ g dV = ∫v(- 4 πG ρ) dV

Therefore

∇ g = - 4 πG ρ

Many masses

∇. g = - 4 πGρ



3.3. Poisson's equation for gravity

Notice that by definition:
An acceleration field A(r) is equal to the negative derivative of its gravitational potential - ∂Φ/∂r

The Newtoniam (classical) accelerartion field g is:

g = g(r) = F/m = (GMm/r2)/m = GM/r2 = - ∂Φ/∂r,
where Φ = - GM/r is the Newtonian gravitational potential.

Its gradient has the expression:

∇. g = - 4 πGρ

The gravitational field g = g(r) is a vector, so is - ∂Φ/∂r.
We can then write:

g = - (∂Φ/∂r) er.
With del or gradient operator, it can be written:

g = - ∇Φ

Therefore:

∇. g = ∇.(- ∇Φ) = - ∇2Φ
That is:

- 4 πGρ = - ∇2Φ
or
2Φ = 4 πGρ

2Φ = 4 πGρ

That is Poisson's equation for gravity.




  


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