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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, nonlinear differential equation with variables separable:
∂t = ∂r / √[2GM(1/r − 1/d)]
where G = 6.67384e11 m³ kg⁻¹ sec⁻²
t−t₀ = √[d/(2GM)] ∫ ∂r/√(d/r−1)
u = √(d/r−1)
∂u/∂r = −½ d r⁻²/√(d/r − 1)
t−t₀ = −2d √[d/(2GM)] ∫ ∂u/(u²+1)²
t−t₀ = −2d √[d/(2GM)] { ½ ∫ [∂u/(u²+1)] + ½ u/(u²+1) }
t−t₀ = −d √[d/(2GM)] { u/(u²+1) + ∫ [∂u/(u²+1)] }
∫ ∂u/(u²+1) = arctan u
t−t₀ = −d √[d/(2GM)] { u/(u²+1) + arctan u }
t−t₀ = −d √[d/(2GM)] { √(d/r−1)/[(d/r−1)+1] + arctan √(d/r−1) }
t−t₀ = −√[d/(2GM)] { √(rd−r²) + 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−r²) + 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 r«d then the above equation reduces to
t−t₀ ≈ π √[d³/(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 sun’s photosphere (r = 6.96e8 meters):
Sun’s 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 timetofall in a plunge orbit problem, I didn’t do the integration as I did it this time. Instead of getting
t−t₀ = √[d/(2GM)] { √(rd−r²) + d arctan √(d/r−1) }
directly, what I had was
t−t₀ = √[d/(2GM)] { √(rd−r²) − i d ln{[√(r/d) − i √(1−r/d)]} }
But through Euler’s theorem there’s 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−r²) + d arctan √(d/r−1) }

