How long would it take for something to fall from one light year away into the sun

[Cheap Astronomy]

I'm imagining something without an existing orbital trajectory, but instead is very slowly commencing a straight-line trajectory towards the Sun. How long would it take to get there - or would it be compelled into an orbit and perhaps never get there?

I imagine we are talking hundreds of thousands of years so the proper motion of the Sun would be an issue, so I am also assuming there's an appreciable gravity well at one light year out that would not only keep a hold of the object but also accelerate it inwards? The object would be very small (grain of rice say). Thanks

[Jens]

One major issue would be the gravitational attraction of other bodies. But if you can hypothesize a universe with only two bodies, and you start them off with absolutely no velocity relative to each other, then yes, it would fall in. I don’t know how to do the math, but I’m pretty sure it’s just a fairly simple calculus problem.


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[The Backroad Astronomer]

If it were the sun it would take about 3.5 million years.
Here is the math.
https://twitter.com/DavidLPFairweat/...005632/photo/1
This is assuming just a simple gravitational attraction.

[grant hutchison]

If it were the sun it would take about 3.5 million years.
Here is the math.
https://twitter.com/DavidLPFairweat/...005632/photo/1
This is assuming just a simple gravitational attraction.

The maths doesn't look quite right - I think you have the wrong numerical constant in there. The time for free-fall under gravity from a distant stationary position to the centre of mass is

pi/[2*root(2)] * sqrt(D^3/GM)

Here D = 9.461e15 m (one light year), G = 6.67408e-11 m^3/kg.s^2 (universal gravitational constant), M = 1.989e30 kg (mass of the sun)

That comes out to 8.87e13 s = 2.8 million years.

You're getting a different answer because you've plugged in a numerical constant of root(2) rather than pi/[2*root(2)].

This formula famously figured as a plot element in Arthur C. Clarke's short story "Jupiter Five" - he pointed out that the free-fall time to the centre of mass is always equal to 1/[4*sqrt(2)] times the orbital period at the same distance.

Grant Hutchison

[grant hutchison]

Here's the derivation of the relationship, if anyone's interested. It's one of those integrals that explodes and then collapses in a satisfactory sort of way. Along the way, I derive the free-fall time to a fixed distance from the centre of mass (for instance, the surface of a planet).
scan0001.jpgscan0002.jpg

Grant Hutchison

[The Backroad Astronomer]

I was assuming a straight drop, no orbital mechanics.

[grant hutchison]

I was assuming a straight drop, no orbital mechanics.

My calculation is for a straight drop. It's all orbital mechanics - a straight drop is just a particularly simple example.

Grant Hutchison

[Jens]

But in any case, I would say that 2.8 million and 3.5 million years is pretty similar. In any case, it would take a few million years for the object to hit the sun,


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[grant hutchison]

But in any case, I would say that 2.8 million and 3.5 million years is pretty similar. In any case, it would take a few million years for the object to hit the sun

Yes, the order of magnitude was always going to be sqrt(D^3/GM).
The problem does have a pretty well-known exact solution, however, because it crops up as a worked example in astrophysics text-books, setting a lower limit to the length of time it would take a dust cloud to collapse. (Which, IIRC, is where Arthur C. Clarke ran into the formula.)
narlikar.jpg

Grant Hutchison

[grant hutchison]

Yes, the order of magnitude was always going to be sqrt(D^3/GM).
The problem does have a pretty well-known exact solution, however, because it crops up as a worked example in astrophysics text-books, setting a lower limit to the length of time it would take a dust cloud to collapse. (Which, IIRC, is where Arthur C. Clarke ran into the formula.)

I dug out "Jupiter Five". Here's the relevant text:

Now there's a well-known theorem stating that if a body falls from an orbit to the centre of attraction, it will take point one seven seven of a period to make the drop.

I'd misremembered Clarke as giving the exact value of 1/[4*root(2)].
Clarke's introduction (in The Collected Stories) states:

In 1962, I commented that I am by no means sure that I could write 'Jupiter Five' today; it involved twenty or thirty pages of orbital calculations and should by rights be dedicated to Professor G.C. McVittie, my erstwhile tutor in applied mathematics.

Grant Hutchison

[George]

Kepler offers a simple method by changing e.

iPad

[grant hutchison]

Kepler offers a simple method by changing e.

Not really by changing e, but by realizing that the time taken to fall is just the half-period of an orbit with a semi-major axis equal to half of the start distance. The straight fall is the special case of a degenerate ellipse with e=1. (So it really is all orbital mechanics.)

And that only works for the special situation of a fall to all the way to the centre of gravity - for other situations, you need my integrals I think.

Grant Hutchison

[grant hutchison]

I imagine we are talking hundreds of thousands of years so the proper motion of the Sun would be an issue, so I am also assuming there's an appreciable gravity well at one light year out that would not only keep a hold of the object but also accelerate it inwards?

You really need to control that by putting your test object at rest relative to the sun - otherwise it end up in an orbit that misses the sun, or indeed simply escapes outwards (because the solar escape velocity at one light year is pretty small).

Grant Hutchison

[grant hutchison]

Not really by changing e, but by realizing that the time taken to fall is just the half-period of an orbit with a semi-major axis equal to half of the start distance.

And there's the factor of 1/[4*sqrt(2)], popping out.
Since period squared varies with semi-major axis cubed, halving the SMA reduces the period by a factor of 1/sqrt(2^3), or 1/[2*sqrt(2)]. Half of that period = 1/[4*sqrt(2)].

The trouble with this derivation, I think, is that it's one of those that only makes sense once you know the right answer. Taking the limit of an elliptical orbit as e approaches 1, so as to come up with a "degenerate ellipse of e=1", which looks like a straight line connecting the starting point to the centre of gravity? That's likely to ellicit either howls of protest or blank looks. But once you know the answer you're trying to get, you say, "Oooooh, neat, I see how that works."

Grant Hutchison

[Ken G]

The trouble with this derivation, I think, is that it's one of those that only makes sense once you know the right answer. Taking the limit of an elliptical orbit as e approaches 1, so as to come up with a "degenerate ellipse of e=1", which looks like a straight line connecting the starting point to the centre of gravity? That's likely to ellicit either howls of protest or blank looks. But once you know the answer you're trying to get, you say, "Oooooh, neat, I see how that works."

Especially since it's kind of a singular situation where you don't really know if the right solution should look like a line that falls straight through the center and goes out an equal distance the other way. That would be the correct solution if the source of gravity is not infinitely concentrated into a point, yet allows the test particle to fall through it anyway. The Kepler solution does not fly out an equal distance the other side, it whips around and comes straight back out, like your degenerate ellipse, but that only happens if the test particle misses the Sun by a little bit. That wouldn't change the time it takes to fall in by any appreciable amount, so still works to get the time to fall straight in.

[George]

And there's the factor of 1/[4*sqrt(2)], popping out.
Since period squared varies with semi-major axis cubed, halving the SMA reduces the period by a factor of 1/sqrt(2^3), or 1/[2*sqrt(2)]. Half of that period = 1/[4*sqrt(2)].

The trouble with this derivation, I think, is that it's one of those that only makes sense once you know the right answer. Taking the limit of an elliptical orbit as e approaches 1, so as to come up with a "degenerate ellipse of e=1", which looks like a straight line connecting the starting point to the centre of gravity? That's likely to ellicit either howls of protest or blank looks. But once you know the answer you're trying to get, you say, "Oooooh, neat, I see how that works."

Sorry Grant, my wife was calling me and she was approaching reproach and I needed to lateral you the ball; I had to compress about 5 or 10 minutes into 20 seconds, so I attempted to maximize brevity. My attempt was to simply get the ball in your capable hands. Thanks.

[The Backroad Astronomer]

Well mine was down just using simple formulas from memory, and few values I looked up for verification at five in the morning in about 15 minutes or what some people call back of the envelope calculations.

[George]

The P² = T² = a³ method (2.81M) agrees with Grant's 2.8M year result. ["a" becomes 1/2 lightyear in value as e goes to 1]

I would assume the drop would fall towards (or is it "toward"?) the barycenter but shifting gently towards the Sun's center especially after arriving within the orbital distance of Saturn and more so when within Jupiter's orbit.

[Hornblower]

The P² = T² = a³ method (2.81M) agrees with Grant's 2.8M year result. ["a" becomes 1/2 lightyear in value as e goes to 1]

I would assume the drop would fall towards (or is it "toward"?) the barycenter but shifting gently towards the Sun's center especially after arriving within the orbital distance of Saturn and more so when within Jupiter's orbit.

In a perfectly straight-in fall in this thought experiment the Sun and the dropped object converge on the barycenter in a straight line, so the object is moving toward both points exactly. In this sort of thought exercise we usually treat the mass of the small object as negligible.

[DaveC426913]

... would it be compelled into an orbit and perhaps never get there?

If the object had any transverse motion greater than about 0.5 mm/h * then it would miss the Earth and fly back out to its apoapsis.

*the radius of the Earth (~6500km w/ atmo) per quarter period (1.4 million years)

[grant hutchison]

In a perfectly straight-in fall in this thought experiment the Sun and the dropped object converge on the barycenter in a straight line, so the object is moving toward both points exactly. In this sort of thought exercise we usually treat the mass of the small object as negligible.

I'm guessing George is talking about the solar system barycentre, rather than the sun/object barycentre. But of course if you introduce the messiness of the real solar system, there will be spacetime coordinates from which you simply can't drop an object straight into the sun starting from rest in the solar reference frame. But maybe there will always be some choice of initial (Sun centred) velocity that results in the object arriving at the sun on the "first pass" (that is, without ever having a perihelion outside the sun's surface). Perhaps there's some deep way of figuring that out, but I'm blank at present.

Grant Hutchison

[George]

In a perfectly straight-in fall in this thought experiment the Sun and the dropped object converge on the barycenter in a straight line, so the object is moving toward both points exactly. In this sort of thought exercise we usually treat the mass of the small object as negligible.

As Grant mentioned, it's the barycenter of the solar system as a whole that I was considering, which would be the point to which the object falls. Once it's inside the system, however, especially inside Jupiter's orbit, the path would be more and more directed toward the center of the Sun. [I first envisioned it falling through the barycenter and was almost puzzled with the thought.]

[Hornblower]

As Grant mentioned, it's the barycenter of the solar system as a whole that I was considering, which would be the point to which the object falls. Once it's inside the system, however, especially inside Jupiter's orbit, the path would be more and more directed toward the center of the Sun. [I first envisioned it falling through the barycenter and was almost puzzled with the thought.]

Yes indeed, I see what you meant. That has been analyzed with comets coming in from the Oort cloud. Not only will the trajectory drift as the comet gets close, but the orbit can change from an elongated ellipse to locally hyperbolic and back. Comet Kohoutek did just that in the 1973-74 flyby.

[George]

Yes indeed, I see what you meant. That has been analyzed with comets coming in from the Oort cloud. Not only will the trajectory drift as the comet gets close, but the orbit can change from an elongated ellipse to locally hyperbolic and back. Comet Kohoutek did just that in the 1973-74 flyby.

That's interesting but also logical since the approach of an interloper would very likely have an e value close to 1 as it entered, no doubt.

[Hornblower]

I found the Sky and Telescope report that included Brian Marsden's orbital determination. While still far from the solar system on the way in, it was in an extremely elongated ellipse with a period of about 5 million years. That is consistent with an aphelion of about a light year and e within a gnat's whisker of unity. At perihelion the orbit was slightly hyperbolic relative to the Sun. The comet had been accelerated by the combined forces of the Sun and the planets and thus was a bit too fast for the Sun alone when very close. During the flyby the comet lost a slight amount of orbital energy to the planets and ended up in a much less elongated ellipse with a period of some 75,000 years. That would be an aphelion of about 3,600 AU, or 0.06 light year.

[DaveC426913]

If the object had any transverse motion greater than about 0.5 mm/h * then it would miss the Earth and fly back out to its apoapsis.

*the radius of the Earth (~6500km w/ atmo) per quarter period (1.4 million years)

Oops. The Sun. Not the Earth.

[21st Century Schizoid Man]

Here's what I'm getting.

Idealised assumptions, object at rest relative to sun, sun is a point mass, Newtownian gravity, no other sources of gravitation, mass of object is negligible compared to sun.



is the initial distance (with no velocity), is the time at which object is at this initial distance, is gravitational constant, is mass of sun.

Formula then gives the time at which a given distance from centre of sun, , is achieved. Formula valid for .

This tells you how long to reach any given distance from the centre of the sun. Strictly speaking, it stops being valid once you hit the surface, because it assumes the sun is a point mass.

If we choose (how long to get to the centre of the sun), it simplifies to



But, I get something just a bit over the 2.8 million years already posted.

[grant hutchison]

Here's what I'm getting.

Idealised assumptions, object at rest relative to sun, sun is a point mass, Newtownian gravity, no other sources of gravitation, mass of object is negligible compared to sun.



is the initial distance (with no velocity), is the time at which object is at this initial distance, is gravitational constant, is mass of sun.

Formula then gives the time at which a given distance from centre of sun, , is achieved. Formula valid for .

This tells you how long to reach any given distance from the centre of the sun. Strictly speaking, it stops being valid once you hit the surface, because it assumes the sun is a point mass.

If we choose (how long to get to the centre of the sun), it simplifies to



But, I get something just a bit over the 2.8 million years already posted.

Given that (remarkably enough)

(acos[2x-1])/2 = pi/2 - asin(sqrt[x])

you and I seem to have exactly the same results.

Grant Hutchison

[grant hutchison]

Given that (remarkably enough)

(acos[2x-1])/2 = pi/2 - asin(sqrt[x])

you and I seem to have exactly the same results.

It's a mystery to me now why, when I did my original derivation twenty-odd years ago, I didn't simplify the (pi/2 - asin) to a simple acos.
But after that, we use the identity

2.acos(x) = acos(2x²-1)


to prove the equality.

Grant Hutchison

[21st Century Schizoid Man]

Given that (remarkably enough)

(acos[2x-1])/2 = pi/2 - asin(sqrt[x])

you and I seem to have exactly the same results.

Grant Hutchison

I saw the link with your derivation, but didn't realise there was a second page

That would have saved me a lot of time and trouble . . .

My method was rather different though.