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 The Time Path of a Body Falling Under the Force of a Central Gravitational Attraction

The dynamics of a body falling toward a mass M a distance r away from it is given by the equation

#### d²r/dt² = −GM/r²

where G is the gravitational constant.

Suppose the body is initially at rest at a distance of r0. If the above equation is multiplied by v=(dr/dt) and integrated, taking into account the initial conditions the result is

#### ½v² = GM[(1/r)−(1/r0)] thus dr/dt = (2GM)1/2[(1/r)−(1/r0)]1/2

If this equation is divided by r0 on the left and equivalently muliplied by (r0)1/2/(r0)3/2) on the right and y substituted for r/r0 the result is

#### dy/dt = (2GM/(r0)3))1/2[1/y - 1]1/2

A further substitution of 1/y-1=z and y=1/(z+1) gives

#### dy/dt=-(1/(z+1)2)(dz/dt) = (2GM/(r0)3))1/2z1/2which leads to (1/(z+1)2)(1/z1/2)(dz/dt) = −(2GM/(r0)3))1/2and finally (2/(z+1)2)(d(z1/2)/dt) = −(2GM/(r0)3))1/2

Now let z1/2=q, and hence z=q². The above equation then takes the form

#### (2/(q²+1)²)dq/dt = −(2GM/(r0)3))1/2

Indefinite integration of this equation gives

#### q/(q²+1) + tan-1(q) = −(2GM/(r0)3))1/2 + C

where C is an arbitrary constant.

The variable r ranged for r0 to 0 so y ranged from 1 to 0. The variable z=1/y-1 then ranged from 0 to ∞ and likewise for q=z1/2.

For an integration with respect to q from q=0 to q=∞ the first term, q/(q²+1), vanishes at both the upper and lower limits. The second term, tan-1(q) has the value π/2 for q=∞ and 0 for q=0. The value of the integral is −π/2 because of the inversion of the limits in going from y to q. Let tf be the time at which r goes to 0. Then

#### π/2 = (2GM/(r0)3))1/2tfand hence tf = (π/2)((r0)3/(2GM))1/2

For a comparison consider a particle moving under a constant acceleration g. In time t it travels a distance ½gt². Thus the time tc required to travel a distance r0 under constant acceleration is

#### tc = (2r0/g)1/2

The value of g is GM/r0² so the above expression reduces to

#### tc = (2r0³/GM)1/2

This leads to the interesting result that

#### tf/tc = π/4

That is to say, the time which it takes a particle to reach the center of the gravitational attraction is 78.54% of the time it would take to reach it under constant acceleration equal to its initial acceleration.

### A Spherical Gas Cloud Without Any Interaction of Particles

If the mass is distributed uniformly throughout a spherical cloud the value of M depends upon r0. In fact

#### M = (4/3)πρr03

where ρ is the mass density.

When this expression is substituted for M in the equation for tf the result is

#### tf = (π/2)(1/(2G(4/3)ρπ))1/2

Thus tf is independent of r0. This means all elements of the spherical cloud collapse to the center at exactly the same time! Of course this would be prevented by the build up of material and pressure in the process of collapse.