**Newton’s law of gravity**

Fig 1

(1) F = –(GmM)/R

^{2}

In vector form, this reads as:

(2)

**F**= – [(GmM)/R

^{2}]

**e**

_{r}

Consider a system of particles. We label the particles 1, 2, 3... i...j...

Fig 2

(3)

**F**= – ∑

_{i≠j}[(Gm

_{i}M

_{j})/R

_{ij}

^{2}]

**e**

_{r}

Define Gravitational Field:

(4)

**A**(x) =

**F**(x)/m

_{i}

Substituting (3),

(5)

**A**= = – ∑

_{i≠j}[(GM

_{j})/R

_{ij}

^{2}]

**e**

_{r}

Gauss Theorem

(6) ∫

_{v}

**∇**•

**A**dV = ∫

_{s}

**A**•d

**S**

Substitute (5) into the right hand side of (6),

(7) ∫

_{s}

**A**•d

**S**= – ∫

_{s}[(GM)/R

^{2}]

**e**

_{r}•d

**S**

However, the magnitude of the infinitesimal area element dA is just the area of the infinitesimal solid angle dΩ, given by,

(8) d

**S**= R

^{2}dΩ

**e**

_{r}

Substitute (8) into (7),

(9) ∫

_{s}

**A**•d

**S**= – ∫

_{s}[(GM)/R

^{2}]

**e**

_{r}•R

^{2}dΩ

**e**

_{r}

= – GM ∫

_{s}

**e**

_{r}•

**e**

_{r}dΩ

= – GM ∫

_{s}dΩ ,(

**e**

_{r}

^{2}=1)

Therefore,

(10) ∫

_{s}

**A**•d

**S**= –4πGM

But the mass can be written in terms of the density,

(11) M = ∫

_{v}ρdV

Therefore equation (10) becomes,

(12) ∫

_{s}

**A**•d

**S**= –4πG∫

_{v}ρdV

Using Gauss Law , equation (6),

(13) ∫

_{s}

**A**•d

**S**= ∫

_{v}

**∇**•

**A**dV = –4πG∫

_{v}ρdV

Therefore,

(14) ∇•A = – 4πGρ

Define the Gravitational Potential:

(15) A = –∇φ

We get,

(16) ∇

^{2}φ = 4πGρ

Note that Gauss theorm allows us to say that a particle is attracted as if all the other matter was concentrated at the center of a sphere.

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