There's Always Something Else to Calculate

May 23, 2018 12:09 PM

The first XKCD "What If" to appear in more than a year answer's a 5-year-old's question: "If there were a kind of a fireman's pole from the Moon down to the Earth, how long would it take to slide all the way from the Moon to the Earth?"
posted by mr_bovis (55 comments total)

Very cool.
posted by 4ster at 12:22 PM on May 23, 2018

It's a fun question, but is there an actual answer, as in a period of time, in the answer?*

* Did I just Stack Overflow this?
posted by maxwelton at 12:23 PM on May 23, 2018 [6 favorites]

Years. Because of the climbing.
posted by drdanger at 12:42 PM on May 23, 2018 [5 favorites]

Plus "another few weeks" for the falling.
posted by Greg_Ace at 12:50 PM on May 23, 2018 [2 favorites]

Would any of this be simpler if you used a polar pole (or Arctic Circle pole) instead of an equatorial pole?
posted by clawsoon at 12:51 PM on May 23, 2018

The moon doesn’t go over the poles
posted by aubilenon at 12:52 PM on May 23, 2018 [6 favorites]

The pole would be more or less perpendicular to the Pole; once you're close enough you can let go of the pole and drift downward. I'm sure the snow would cushion your fall.
posted by Greg_Ace at 12:54 PM on May 23, 2018 [1 favorite]

Or, y'know, at the North Pole in a few years it'll be water, which is even softer than snow, right?
posted by Greg_Ace at 12:55 PM on May 23, 2018

So what you're saying is it might be a good idea to install a really long pole at the L1 point
posted by vibratory manner of working at 1:03 PM on May 23, 2018 [3 favorites]

the rest of the pole is a bad idea, obviously
posted by vibratory manner of working at 1:04 PM on May 23, 2018 [1 favorite]

Coincidentally, something pretty similar is described in Greg Egan's new novella "Phoresis". Except that it becomes a lot more feasible when you're constructing your "pole" between a pair of low-gravity dwarf planets whose surfaces are tidally locked to each other.
posted by teraflop at 1:04 PM on May 23, 2018 [2 favorites]

My takeaway from that is that it would make a lot more sense to have a ladder instead of just a pole.
posted by Naberius at 1:04 PM on May 23, 2018 [6 favorites]

You could make a profit by cutting the ladder in half and selling the two poles.
posted by justsomebodythatyouusedtoknow at 1:07 PM on May 23, 2018 [8 favorites]

I know he already handwaved off the "magic pole," but I was just thinking about a couple other pole problems:

1) The majority of the pole is on the Earth side of L1, which means the overall pull on the pole is going to be towards Earth. It'll come crashing down pretty quickly.

2) Ignoring the first point (assume a massless pole...), as soon as you start climbing the pole you're going to pull it towards the moon, causing it to crash in to the moon.
posted by backseatpilot at 1:11 PM on May 23, 2018 [1 favorite]

So what you're saying is it might be a good idea to install a really long pole at the L1 point

I mean, that's literally what the Space Elevator thing is all about.
posted by tobascodagama at 1:12 PM on May 23, 2018 [3 favorites]

This kind of shit is why the sea levels are rising.
posted by goatdog at 1:15 PM on May 23, 2018 [2 favorites]

If we ignore the moon, the orbital motions, the rotation of the bodies, atmosphere, etc, and just consider how long it would take to fall from the height of the moon to a stationary Earth, my quick calculation is 1915 days. Most of this time would be going nowhere fast in very weak Earth gravity at an altitudes over 350000 kilometers.

Or maybe I made a mistake.
posted by SemiSalt at 1:16 PM on May 23, 2018 [3 favorites]

This kind of shit is why the sea levels are rising.

Wait! How much string would it take to move all that extra ocean back to the Moon? It'd be nice if we could get it to Mars instead but that would be logistically silly, obvs
posted by FatherDagon at 1:20 PM on May 23, 2018 [1 favorite]

If we ignore the moon

Ah yes, the Star Wars school of physics - first, assume that's no moon.
posted by backseatpilot at 1:20 PM on May 23, 2018 [19 favorites]

backseatpilot: The majority of the pole is on the Earth side of L1, which means the overall pull on the pole is going to be towards Earth. It'll come crashing down pretty quickly.

What if you extended the pole past the moon, so that centrifugal forces pulled it equally in both directions? Rest it on the top of the moon and the top of the earth (i.e., tangent to both).
posted by clawsoon at 1:21 PM on May 23, 2018 [2 favorites]

If we ignore the moon, the orbital motions, the rotation of the bodies, atmosphere, etc, and just consider how long it would take to fall from the height of the moon to a stationary Earth, my quick calculation is 1915 days.

I think it would take much less time than that.

A vertical fall is just the extreme limiting case of an extremely long and narrow elliptical orbit, with an apogee near the moon and a perigee near the center of the earth. If we take the limit as the orbital eccentricity approaches 1, this orbit has a semi-major axis equal to half the moon's orbital radius. And the elapsed time between apogee and perigee is half an orbital period.

So using Kepler's third law, the fall would take 1/√32 = 17.7% of the moon's orbital period, or about 116 hours. (Actually a tiny bit less, because we want the time taken to reach the earth's surface, not its center.)
posted by teraflop at 1:30 PM on May 23, 2018

I mean, that's literally what the Space Elevator thing is all about.

The Space Elevator isn't at L1. It's attached to a satellite in a geostationary orbit (with a long tether for a counterweight). That's low enough that the gravity of the moon can be ignored (it's not like there aren't enough other problems to contend with).
posted by It's Never Lurgi at 1:42 PM on May 23, 2018

teraflop: "Coincidentally, something pretty similar is described in Greg Egan's new novella "Phoresis". Except that it becomes a lot more feasible when you're constructing your "pole" between a pair of low-gravity dwarf planets whose surfaces are tidally locked to each other."

Or you could just sail over via hot air balloon, like in Bob Shaw's The Ragged Astronauts.
posted by Chrysostom at 1:47 PM on May 23, 2018 [1 favorite]

It took Apollo 11 a little under 60 hours to fall back to Earth from Moon orbit according to this timeline (from Transearth injection burn at 135:23:42 to Splashdown at 195:18:35).
posted by rhamphorhynchus at 1:48 PM on May 23, 2018

This perfectly explains the popularity of that excruciating "novel" The Martian.

(Fabulous post, though).
posted by Jody Tresidder at 1:50 PM on May 23, 2018 [1 favorite]

But they don't actually answer the question, how long does it take? Years for sure (as stated), but that's a pretty awful approximation. With all the info given it wouldn't take too much to cut the distance into segments and calculate a reasonable approximation (climbing from moon, accelerating to 100 mph, distance travelled while at 100 mph, acceleration to 11 km/s, decelerating from 11km/s to desired re-entry speed, decent into atmosphere, parachute).

I am at work so cannot do the math now, but I am surprised that XKCD didn't solve it already.
posted by Vindaloo at 1:55 PM on May 23, 2018

The correct answer is to have a pole that rotates around its center such that the end near the earth's surface is traveling at the same speed as the earth's surface. You stand at the designated point, the pole descends and very nearly touches the surface of the earth, you grab on, and then hold on as the pole rotates up again out of the earth's atmosphere. You can then either hang on until you reach the point where letting go flings you in the appropriate direction, hang on until your pole lines up with another rotating pole to transfer, or you climb to the center of the pole and drift away.

I leave it as an exercise for the reader to determine how harrowing an experience it would be to jump on a moving pole at the earth's surface that will lift you up out of the earth's atmosphere.
posted by Mr.Encyclopedia at 1:55 PM on May 23, 2018 [3 favorites]

Granted, I'm answering in the opposite direction, so instead imagine a series of rotating poles that take you from the moon's surface up to the L1 point then down to the earth's surface. Maybe going down to the earth's surface on a rotating pole will be less harrowing than going up.
posted by Mr.Encyclopedia at 1:58 PM on May 23, 2018

The moon doesn’t go over the poles

That's no reason not to anchor the earth end of the pole there. You just need a fitting that lets it rotate, and a pole flexible enough to curve around over the horizon when the moon is below it, and not so flexible that it goes through the earth instead and chops off the top of the planet.
posted by sfenders at 1:58 PM on May 23, 2018 [1 favorite]

Or, y'know, at the North Pole in a few years it'll be water, which is even softer than snow, right?

At these speeds you'd rather hit snow that water. You're going fast enough that you'd want the structures in the snow to collapse to break some of your fall.

You're going fast enough that water cannot get out of the way quickly, and in liquid form is by far water's densest form (water is weird that way). Hitting open water is basically the same as hitting solid granite.
posted by jmauro at 1:58 PM on May 23, 2018 [3 favorites]

I found my mistake. I was restarted the fall at zero velocity every Km.

New estimate for simple free fall from moon altitude: 4.85 days.
posted by SemiSalt at 2:05 PM on May 23, 2018

New estimate for simple free fall from moon altitude: 4.85 days.

Why isn't it a quarter of a [sidereal] month, less the radius of the Earth?
posted by aubilenon at 2:08 PM on May 23, 2018 [1 favorite]

You're going fast enough that water cannot get out of the way quickly
You're going fast enough that air cannot get out of the way quickly. See also.
posted by rhamphorhynchus at 2:10 PM on May 23, 2018

At these speeds you'd rather hit snow that water.

Oh I know, I was being facetious in both cases.
posted by Greg_Ace at 2:14 PM on May 23, 2018

Why isn't it a quarter of a [sidereal] month, less the radius of the Earth?

It's surprisingly close to that, if the calculation is correct. It's less because gravity.
posted by sfenders at 2:21 PM on May 23, 2018 [1 favorite]

what if it's a moving escalator and you have small planes on your feet
posted by poffin boffin at 2:48 PM on May 23, 2018 [2 favorites]

Airplanes or like woodworking planes?
posted by aubilenon at 2:53 PM on May 23, 2018 [7 favorites]

the flying kind.
posted by poffin boffin at 3:23 PM on May 23, 2018

pfft you serve me the right kind of mushroom soup and I will do it in 5 hours and throw in the second half of Inception as a bonus
posted by seanmpuckett at 3:34 PM on May 23, 2018 [1 favorite]

"Coincidentally, something pretty similar is described in Greg Egan's new novella "Phoresis". Except that it becomes a lot more feasible when you're constructing your "pole" between a pair of low-gravity dwarf planets whose surfaces are tidally locked to each other."

When does a pole become a bridge in this scenario?
posted by GoblinHoney at 3:50 PM on May 23, 2018

You are all missing the point that this is a brilliant way to get a 5yo to never ask another question for fear of the answer.

Alternatively, the 5yo might just keep asking "why?"
posted by GenjiandProust at 3:54 PM on May 23, 2018 [2 favorites]

People often say, "It's not the fall that kills you, it's the sudden stop at the end." Unfortunately, in this case, it's probably going to be both.

Did not know there was an 18.5 year cycle that the moon makes, astronomy is really mind boggling, just how do you get the data to calculate an 18.5 (and it's probably a crapload more decimal places) cycle?
posted by sammyo at 4:23 PM on May 23, 2018

Depending on you you ask, it's either "Lasers pointed at reflectors that the Apollo astronauts set up that give very precise data about how far away the moon is." or "It's all made up by NASA/FEMA to confuse and ensnare people by obscuring the truth of the Flat Earth."
posted by Mr.Encyclopedia at 4:29 PM on May 23, 2018 [3 favorites]

Footnote 7:

At the distance of the Moon's orbit and the speed it's traveling, centrifugal force pushing away is exactly balanced by the Earth's gravity—which is why the Moon orbits there.

That seems pretty obvious, but I never saw it stated in this particular way. This gave me a new way to think intuitively about orbits.
posted by mbrubeck at 4:41 PM on May 23, 2018 [1 favorite]

Alternatively, the 5yo might just keep asking "why?"

In my experience that is almost always the likeliest outcome.
posted by Greg_Ace at 5:16 PM on May 23, 2018 [2 favorites]

I feel like someone could solve this in Kerbal Space Program with less words and more cute animations. But I'm not complaining!
posted by RolandOfEld at 5:55 PM on May 23, 2018

What if you extended the pole past the moon, so that centrifugal forces pulled it equally in both directions? Rest it on the top of the moon and the top of the earth (i.e., tangent to both).

I believe you will have created a very expensive process for separating fire poles into two pieces, one of which comes crashing down to earth, and the other one goes flying out into the solar system.

Sarcasm aside, because the gravity of Earth is, as mentioned in the article, much greater than the centrifugal force throughout the majority of this distance, and the centrifugal force only increases linearly with distance from the Earth, your pole would be mostly on the other side of the moon.

If I hadn't had such a long day I would figure out the integral calculus to find the actual distance, but let's just say "well past the moon"?
posted by thegears at 5:59 PM on May 23, 2018 [1 favorite]

I'm still waiting for the answer to "What if you stopped the earth's rotation momentarily?"
posted by ckape at 6:02 PM on May 23, 2018 [1 favorite]

My knowledge of astrophysics is limited, but I have been to both Chimborazo and Cayambe.

I suspect that means I'm less the narrator and closer to the stick figure in this scenario.
posted by nickmark at 7:22 PM on May 23, 2018

You are all missing the point that this is a brilliant way to get a 5yo to never ask another question for fear of the answer.

Have you ever met a five year old? (Joke, I'm aware that you have been a five year old.)

My five year old recently wanted to know why some kids smoke cigarettes if smoking makes you die. My attempt at an age-appropriate explanation was to say that there are bad people who sell cigarettes so they can make money, and they want kids to buy cigarettes to get even more money, so they try to trick kids into smoking. We're now discussing plans to escape from the bad guys if they corner us and try to get us to smoke cigarettes. Looks like we'll jump off the cliff with parachutes in our backpacks. I don't know why this solution isn't more often discussed in response to manipulative marketing, and it could also help trips back to earth from the moon.
posted by medusa at 10:28 PM on May 23, 2018 [11 favorites]

Do I get to be the first to say that there’s no such thing as centrifugal force?

Yay! I’m a physics pedant.
posted by Segundus at 1:09 AM on May 24, 2018 [1 favorite]

Have you ever met a five year old? (Joke, I'm aware that you have been a five year old.)

I have, although I had to surrender my own 4th-7th years in return for certain.... considerations that are... inconvenient to describe.

I do agree that parachutes could be deployed to solve many personal dilemmas on a regular basis. I keep a few in my office for this very purpose. They have not helped with lunar problems. Yet.
posted by GenjiandProust at 1:44 AM on May 24, 2018

> estimate for simple free fall from moon altitude

It's simple enough if you don't want it to be exactly right. Simply use Newton/Kepler and calculate the period P - the time to fall will be approximately 1/4 of this. In this case "a" will be the distance from the earth to the moon, and you can use just M as the mass of the earth rather than M1 + M2. It's not quite right because you'll end up with the time it takes to reach the centre of the earth on that, but it'll be good enough for an approximation.

If you want to do it exactly right, it's harder - either you can use Kepler's position as a function of time (you'll have to convert to polar coordinates and it may help if you set the semi-minor axis to something like a metre as otherwise you're dealing with a degenerate ellipse and may confuse you). There's some detail here but there's multiple steps to getting the answer.

Alternatively, for this problem (essentially a degenerate ellipse), you can use some horrid calculus to get the right answer, which I ended up having Maxima do for me.
posted by BigCalm at 2:51 AM on May 24, 2018

Do I get to be the first to say that there’s no such thing as centrifugal force?

Footnote 6: “As usual, anyone arguing about ‘centrifugal’ versus ‘centripetal’ force will be put in a centrifuge.”
posted by hwyengr at 4:25 AM on May 24, 2018 [2 favorites]

Mr.Encyclopedia: ...imagine a series of rotating poles that take you from the moon's surface up to the L1 point then down to the earth's surface. Maybe going down to the earth's surface on a rotating pole will be less harrowing than going up.

Mr.Encyclopedia, You are Neal Stephenson and I claim my £5.
posted by wenestvedt at 8:20 AM on May 24, 2018 [1 favorite]