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Old October 15th, 2016 #4
Jerry Abbott
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Jerry Abbott
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If two bodies having total mass M are initially at rest and separated by a distance d, the time to fall until the separation is r, such that r<d, is found by integrating a differential form of the Vis Viva equation:

v = √[GM(2/r − 1/a)]

Since the apoapsis of a plunge orbit is twice its semimajor axis,

a = d/2

v = √[2GM(1/r − 1/d)]

Since all the motion in a plunge orbit is radial (i.e., there is no transverse component),

∂r/∂t = √[2GM(1/r − 1/d)]

We derive an ordinary, non-linear differential equation with variables separable:

∂t = ∂r / √[2GM(1/r − 1/d)]

where G = 6.67384e-11 m kg⁻ sec⁻

t−t₀ = √[d/(2GM)] ∫ ∂r/√(d/r−1)

u = √(d/r−1)

∂u/∂r = − d r⁻/√(d/r − 1)

t−t₀ = −2d √[d/(2GM)] ∫ ∂u/(u+1)

t−t₀ = −2d √[d/(2GM)] { ∫ [∂u/(u+1)] + u/(u+1) }

t−t₀ = −d √[d/(2GM)] { u/(u+1) + ∫ [∂u/(u+1)] }

∫ ∂u/(u+1) = arctan u

t−t₀ = −d √[d/(2GM)] { u/(u+1) + arctan u }

t−t₀ = −d √[d/(2GM)] { √(d/r−1)/[(d/r−1)+1] + arctan √(d/r−1) }

t−t₀ = −√[d/(2GM)] { √(rd−r) + d arctan √(d/r−1) }

The minus sign indicates that the distance decreases with time. We can remove the minus sign by reversing the limits on the integral, and so we will, since we prefer our times positive.

t−t₀ = √[d/(2GM)] { √(rd−r) + d arctan √(d/r−1) }

This answer has the advantage of being applicable when r is an appreciable fraction of d; i.e., you can still use it when r is not very much less than d.

If rd then the above equation reduces to

t−t₀ ≈ π √[d/(8GM)]

If we take the initial distance, d, of each planet in the solar system to be equal to the semimajor axis of its actual orbit, then we get these times for the planet to fall to the suns photosphere (r = 6.96e8 meters):

Suns mass: 1.98855e30 kilograms

Mercury, d = 5.79090e+10 meters, Δt = 15.5421 days

Venus, d = 1.07477e+11 meters, Δt = 39.3108 days

Earth, d = 1.4959787e+11 meters, Δt = 64.5601 days

Mars, d = 2.2792e+11 meters, Δt = 121.416 days

Ceres, d = 3.8262e+11 meters, Δt = 264.102 days

Jupiter, d = 7.78e+11 meters, Δt = 765.407 days

Saturn, d = 1.42939e+12 meters, Δt = 1906.76 days

Uranus, d = 2.87504e+12 meters, Δt = 5439.9 days

Neptune, d = 4.50445e+12 meters, Δt = 10668.1 days

Pluto, d = 5.90638e+12 meters, Δt = 16018.3 days

Note: The last time I solved the time-to-fall in a plunge orbit problem, I didnt do the integration as I did it this time. Instead of getting

t−t₀ = √[d/(2GM)] { √(rd−r) + d arctan √(d/r−1) }

directly, what I had was

t−t₀ = √[d/(2GM)] { √(rd−r) − i d ln{[√(r/d) − i √(1−r/d)]} }

But through Eulers theorem theres a relationship between the trigonometric functions (and their inverses) and the natural logarithms of complex numbers. In this case, the identity was

arctan{√[(1−x)/x]} = i ln[√x − i √(1−x)]

And, making the substitution and doing the algebra, we get, again, the answer

t−t₀ = √[d/(2GM)] { √(rd−r) + d arctan √(d/r−1) }