The Time to Fall in a Plunge Orbit

So, after simplifying the equation,v = r₁ √[GM(2/r − 1/a)]...we can solve for the time elapsed:t = r₁ √[GM(2/r − 1/a)] − 1/2 − π/4...which is about 0.14132295 seconds.f
  • #1

Homework Statement

Two masses, M and m, are initially (at time t=0) at rest with respect to each other and separated by a distance d. They are affected only by their mutual gravitational attraction. Find the amount of time elapsed beginning at the instant the separation is equal to d/2 and ending at the instant the separation is equal to d/3. (Ignore relativity.)

Homework Equations

The Vis Viva equation is relevant as a starting point. Solve it for the speed in orbit, and, since the orbit is a plunge orbit, change v into the differential dr/dt, creating an ordinary non-linear differential equation. This ODE is to be solved to give time elapsed from t=0 as a function of r, such that 0 ≤ r ≤ d.

The Attempt at a Solution

This was a homework problem that I was given as an undergraduate, along with those who were my classmates at the time. We solved it, but it was somewhat more challenging that we'd at first expected. The solution is

t(r) = √{d/[2G(M+m)]} {√(rd−r²) + d Arctan[√(d/r−1)]}

t(d/2) = (1/2 + π/4) √{d³/[2G(M+m)]}

t(d/3) = {√(2)/3 + Arctan[√(2)]} √{d³/[2G(M+m)]}

Δt = {√(2)/3 + Arctan[√(2)] − 1/2 − π/4} √{d³/[2G(M+m)]}

Δt ≈ 0.141322975518 √{d³/[2G(M+m)]}

From t(r), you may notice that when r«d then, effectively, r ≈ 0, and

t ≈ π √[d³/(8GM)]

This approximate solution can be used to estimate the amount of time needed for a cloud of interstellar gas to collapse into a proto-star, where d is the initial radius of the cloud and M is the cloud's total mass.

Check my answer,

t(r) = √{d/[2G(M+m)]} {√(rd−r²) + d Arctan[√(d/r−1)]}
  • #2
To check - try another approach:
ie. figure out the equation if ##M>>m## and ##\delta r \to 0## ... see if you get the uniform gravity (time to fall h).

Since they are initially at rest, ##M## and ##m## just fall directly towards each other ... so this is a 1D problem: choose coordinates so the x-axis connects both centers of mass and put ##M## at ##x_M## and ##m## at ##x_m## so that ##x_M>x_m##, then, for ##\dot x_M,\dot x_m <\!< c## they should obey Newton's laws as:

##-Gm = (x_M-x_m)^2\ddot x_M##
##GM = (x_M-x_m)^2\ddot x_m##
... using the +x direction for +ve on vectors, and the direction is indicated by the sign.

Change variable to ##x=x_M-x_m## (the separation of the masses) we can combine these equations giving the IVP:
##G(M+m) = x^2\ddot x : x(0)=d, \dot x(0)=0##
... which is awkward sure, but not impossible.
You can solve this to see if it looks like your result... or just see if this can be made to look like your initial DE.

The vis-viva approach is basically conservation of energy... so how about trying without using the official vis-viva equation?
$$mv_m^2 + Mv_M^2 = 2GMm\left(\frac{1}{x}-\frac{1}{d}\right)$$ ... defining (by conservation of momentum): ##mv_m - Mv_M = 0##

...but you will prefer to use the speed at which the separation is closing instead: ##v=-\dot x = v_M+v_m##
manipulate the simultaneous equations to get a DE.

It seems you can get an equation for v(x) fairly easily... and it is easier to solve.
Is there a way to use this to get the time as a function of x as t(x)?
  • #3
You can check my result numerically. For example, suppose that Earth halted in its orbit around the sun, so that initially (at time t=0) Earth and Sun were at rest and separated by d=1.495978707e11 meters. Find the time until Earth falls through the sun's photosphere (r=6.96e8 m).

GM = 1.3271284e20
dt = 0.01 (one-hundredth second increments in time)
t = 0
v = 0
r = 1.4959787e11
t = t + dt
a = −GM/r²
r = r + v dt + ½ a (dt)²
v = v + a dt
Until r ≤ 6.96e8

When you run that, you find out that the elapsed time is 5577992.68 seconds.

Now, solving the equation:

t(r) = √{d/[2G(M+m)]} {√(rd−r²) + d Arctan[√(d/r−1)]}

d = 1.4959787e11 m
r = 6.96e8 m
G(M+m) = 1.3271284e20 m³ sec⁻²

...the result is:

t = 5577992.74 sec

From the limiting equation

t = π √[d³/(8GM)]

we find that the time for Earth to fall from 1 AU to the center of the sun would be

t = 5578753.56 sec

So, neglecting forces that dissipate kinetic energy and regarding the sun as a point mass, it would take only 760.82 seconds for Earth to move from the sun's photosphere to its core.
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  • #4
Here is the derivation of the equation for the time to fall in a plunge orbit.

Two bodies having a total mass M (i.e., M = M₁ + M₂) are initially at rest, separated by a distance d, in vacuum, and isolated from all forces except their mutual gravitational attraction. Find the time elapsed from the initial moment to the moment at which the separation is r, such that 0<r<d.

The problem involves the conservation of energy in a gravity dominated physical system, so we begin with the Vis Viva equation, solved for the speed in orbit:

v = √[GM(2/r − 1/a)]

The eccentricity of a plunge orbit is one, just as it is for a parabolic orbit. The apoapsis separation of M₁ and M₂ is therefore twice the semimajor axis.

e = 1
d = a(1+e)
d = 2a
a = d/2

Therefore, the Vis Viva equation can be rewritten as

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 = 6.6743e-11 m³ kg⁻¹ sec⁻²

t−t₀ = √[d/(2GM)] ∫ ∂r/√(d/r−1)

We make this substitution:

u = √(d/r−1)
∂u/∂r = −½ d r⁻²/√(d/r−1)
r = d/(u²+1)
∂r = (−2d) u [∂u/(u²+1)²]

t−t₀ = −2d √[d/(2GM)] ∫ ∂u/(u²+1)²

Integrating by parts,

t−t₀ = −2d √[d/(2GM)] { ½ ∫ [∂u/(u²+1)] + ½ u/(u²+1) }
t−t₀ = −d √[d/(2GM)] { u/(u²+1) + ∫ [∂u/(u²+1)] }

We notice an integral-trigonometric identity.

∫ ∂u/(u²+1) = arctan u


t−t₀ = −d √[d/(2GM)] { u/(u²+1) + arctan u }

Reversing the substitution,

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

After simplification, we get

t−t₀ = −√[d/(2GM)] { √(rd−r²) + d arctan √(d/r−1) }

The minus sign indicates that the distance decreases with time, so we just remove the minus sign (which represents flipping the limits on the original integral), and

t−t₀ = √[d/(2GM)] { √(rd−r²) + d arctan √(d/r−1) }

If r«d then r≈0 and

t−t₀ ≈ π √[d³/(8GM)]

When determining the time to fall from r₁ to r₂, such that r₁>r₂>0,

t₂−t₁ = (t₂−t₀) − (t₁−t₀)

For example, how much time is required for the separation to decrease from r₁=d/2 to r₂=d/3?

t₁−t₀ = √[d/(2GM)] { √(r₁d−r₁²) + d arctan √(d/r₁−1) }
t₁−t₀ = √[d/(2GM)] { √(d²/2−d²/4) + d arctan √(2d/d−1) }
t₁−t₀ = √[d/(2GM)] { √(d²/4) + d arctan √(2−1) }
t₁−t₀ = √[d/(2GM)] { d/2 + dπ/4 }
t₁−t₀ = {1/2 + π/4} √[d³/(2GM)]

t₂−t₀ = √[d/(2GM)] { √(r₂d−r₂²) + d arctan √(d/r₂−1) }
t₂−t₀ = √[d/(2GM)] { √(d²/3−d²/9) + d arctan √(3d/d−1) }
t₂−t₀ = √[d/(2GM)] { √(2d²/9) + d arctan √(3−1) }
t₂−t₀ = √[d/(2GM)] { √(2d²/9) + d arctan √(2) }
t₂−t₀ = {√(2/9) + arctan √(2)} √[d³/(2GM)]

t₂−t₁ = {√(2/9) + arctan √(2) − 1/2 − π/4} √[d³/(2GM)]
t₂−t₁ ≈ 0.141322975518 √[d³/(2GM)]
  • #5
This equation,

t−t₀ = √[d/(2GM)] { √(rd−r²) + d arctan √(d/r−1) }

is the classically exact solution to the time-to-fall problem. It does not depend on the acceleration of gravity being constant throughout the fall. Nor does it approximate the resulting time difference by an approximate method of numerical integration.

This equation (used twice) can be used in place of the approximation taught in high school physical science class,

t₂−t₁ = (t₂−t₀) − (t₁−t₀)


t₀ = the time when the test mass is let go, at distance d from the center of the planet

t₁ = the time, during the fall, when the object is at a distance r₁ from the center of the planet
(It is permitted that r₁ can be equal to d, but this isn't required.)

t₂ = the time, during the fall, when the object is at a distance r₂ from the center of the planet
(It is permitted that r₂ can be equal to the radius of the planet, but this isn't required.)

Suggested for: The Time to Fall in a Plunge Orbit