How to mathematically calculate a fall through the Earth 

"Your fragile face scraping against the rock surface will cause friction, which is frankly inconvenient for the maths."

“Earths density does not change” well that just triggered all the geologists. The center of the earth is much more dense than the exterior. In fact the core is a mars scized chunk of iron! Now if you were to just fall through the iron bit the time to fall through would less than the 42 minutes. This means that going through the core would take LESS TIME! Right? Or am I missing something?

The re-capping of the sharpie at 4:32 was done so professionally.

0:13 "a apple"

Walking through Fremantle today and my son grabs my sleeve - "Dad, Dad, that's the guy who does those cool RU-vid videos!" I learned two things today: - The world is smaller then you think. - My son is a bigger nerd than I realised.

2:07 Until you said "Oh grow up" I never noticed. THANKS MATT.

Repeatedly falling through the hole, emerging at the other end of the sphere and plummeting once again is technically orbiting (mathematically).

Matt, what on EARTH are you doing (@9:28)?! You're using a dot as both a decimal and thousands separator!!!

It is actually more fun with air resistence. You will have damping that willl give you some decaying oscillation that will likely end you sitting susupended in the middle of the earth.

2:13 how did I not see that? What's happened to me? Did I accidentally remove my mind from the gutter?

"Oh grow up!" What, were you saying it was silly to think we'd just bounce back and forth? (Thirty entire seconds pass) Wait...

"assuming there is no air resistance" me, a highschool student: "air resistance? What air resistance?"

I actually had this problem as an exercise demonstrating harmonic oscillation in my high school physics textbook. Good times.

Wait... in the 'London to New York' part of the video, wouldn't you need a non linear path/tunnel? Gravity would not pull you straight from London to NY but instead would pull you along the side of the tunnel towards the lopsided mass side... So how did we manage to get the same answer for time? Did we take a parabolic route sling-shot-ing around the core of earth similar to a gravity assist in space?

this had to involve the apple, the iconic symbol of gravity.

I have been watching these math channels for years and they are so amazing. I started watching these videos when i was a freshman in high school taking algebra 1, now i am in 12 grade AP Calculus and AP Physics. The math has gone from complete gibberish to things I understand. Its so cool. So to all of the math channels, thanks for being a part of my life and thanks for being one of the few intelligent corners left on youtube. ❤️ love you guys

This question came up as a free response question on my physics final and I had no idea how to do the math but I remembered this video, wrote 42 minutes and got full credit 😎

As though 42 minutes wasn't a good enough connection, "There is an art to flying, or rather a knack. The knack lies in learning how to throw yourself at the ground and miss. ... Clearly, it is this second part, the missing, that presents the difficulties." The hitchhikers guide to the galaxy is the best trilogy ever and if you don't think so I'll fight you

Isn't this assuming that the Earth's mass is even distributed? Most of the Earth's mass comes from the very center.

I love how he says "oh, grow up!" before it even occurred to me what was going on. Thanks for putting your ideas into my mind Matt. ;)

3:13 that seems a bit counterintuitive. I've always thought of it as if I'm in a basement, I'm closer to the center of the earth, so that gravity has more of a pull on me. But I guess his logic makes sense, because if you continue this until you reach the center, you are "essentially" weightless. But if you go higher than the surface, your weight will also decrease, so the surface is where you weigh the most?

"This is not going to be exact... I'm using a food product."

The Answer to Life, the Universe and Everything, plus 10.5 sec! Is that a coincidence??

"oh, grow up" you got me 😂

For falling through a hole not on a central axis, you'd end up scraping against the walls I assume

well, you take 42 minutes (and a bunch of seconds) to fall from one spot to any other spot on earth (the math going on is absolutely amazing btw.) and you take this time for any planet no matter which size but with the same density as earth, may this is the thing that they wanted to point out, when they said “forty-two is the answer of all questions“ in the movie “the Hichhiker's guide to the galaxy“!! I'm absolutely amazed!! mind's blown!!

I don't know if in the 1,619 comments this was addressed, but if you take a simple pendulum of a certain length, and you take a hula hoop of a diameter equal to that of the length of the simple pendulum (ignoring the lack of isochronism, of course), your Periods would be identical! I noticed that during this excellent discussion. Great work on the video!!! Thanks a million!

Why didn't you cut the onion the other way? Then the layers are much more concentric and you can peel them easier.

At 10:50, my first thought was "Gee, if only we had a planet with an earth-like chemical makeup and a diameter of at least 299792.458 * 60 * 42.18 Kilometers!"

I probably laughed more than I should have at 2:12 !

I have been a fan for years and was pleasantly surprised today when I came across you on a tv science show talking about rainbows! What a delight. You are awesome.

9:45 just thinking he should mention it doesn't matter if your going through the center or not, then saw video was'nt even half over, good on ya Matt

I wish there was a love button because this is an absolutely awesome video. I loved it to bits.

"Will cause friction which is, frankly, inconvenient for the maths." It's also inconvenient for the face that's scraping along the rock.

This takes me back. Did this during introductory theoretical physics back in the day. :)

this was so entretaining. i saw years ago minutephysics video about the same subject. but the math perspective on this one is a great bonus.

Hey Matt, I just finished this, as always fantastic, video and the saw the sponsoring at the end. Which I think is great to keep your channel up and running. But then I was wondering whether there was another way to support you? Since I'm from Germany, I was only able to buy a digital copy of your "festival" and couldn't show up in person - btw, love it! Something like flattr or patreon or at least a PayPal account to donate something. I love your channel and your videos and I understand that you're doing this in your spare time, so give us a chance to support you even more! Lovely greetings from Aachen, Germany.

Your enthusiasm is totally catching!

the orbit at 0 and the time taken to fall through are linked despite what you say! Orbits are basically just falling (with style) except you miss the planet in the process. From the side your corenaut and the orbit will be identical.

I'm not sure if I missed it during the explanation, but wouldn't it be more of a circular horizontal slice than just the sphere around the center that determines your gravitational pull force? Like, some of the spheres "above" you as you fall have a portion of the sphere below you as well, so instead of an onion wouldn't it be like a line perpendicular to the falling direction? Because (In your scenario in which the earth is all the same density, which it isn't actually) it doesn't matter that it's a part of the outer sphere(s), it's still pulling on you with the same mass as before and that pull would have a vertical and horizontal portion of its pull vector, but the horizontal portions would get counteracted by the same portion from the opposite direction, so just the vertical part would remain Do the numbers work out to the same results with this line of reasoning?

2:07 I like that a lot...................

Great connection between the hole in the Earth problem and simple harmonic oscillators. One correction at around 12:24. Your first equation of motion should read s=ut+1/2at^2. The other two are fine though.

And the period of a simple pendulum of length R as well! Pendulum, orbit and drop all the same! when I discovered this in high school I was amazed!

8:00 Lost me at 'density of the Earth doesn't change', especially after all that earlier mention of the core.

Another Parker Square: drawing on the inside of an apple with a Sharpie.

We finally have the question to the ultimate answer of life, the universe and everything. 42 = how long to orbit or fall through a planet? Simply stunning.

Matt parker! thanks for the videos! always looking forward to them when i get home, after a long day of work, only to turn into mush what ever chunks of brain i have left. I appreciate it.

11:50 What is bugging me the most is that in reality, the density of the sphere below you is not constant, it's a function of x!

Not even kidding my eyes started watering when he cut the onion 😮

So I have two big problems with this video. The first is when you mentioned the situation in which one falls through a tunnel that doesn't go through the center of the Earth. If one were to jump into such a hole/tunnel, gravity will still pull them directly towards the center of mass of the Earth, so they wouldn't fall perfectly centered through the hole, and would eventually hit the wall of the tunnel, making it impossible to fall through such a hole/tunnel without an external force to keep them centered within that hole. I guess if you can create a device that fights against your off-centered acceleration towards the wall of the tunnel in order to keep you centered, making it only affect that and nothing else, then you'd fall through the off-centered tunnel in the same amount of time, but if it's a system where only gravitational force is applied, then it's only possible to fall through the center of mass. The other problem has to deal with time dilation when gravity changes, which affects how you would calculate the time it takes to cross the planet depending on which point-of-view you're calculating from, but that's on a whole new level of complication, so I don't fault you for not taking that into account.

Actually that bit at 20:40 made me actually understand it for the first time. Of course it has to be the same time. Because if you throw a ball sidewise and drop a ball at the same time, they will fall the same distance and land on the ground at the same time. And your Satellite is basically just a ball that starts with sideways movement and falls straight down just like if you fall through earth - just that it also gets pulled to the side aswell.

The video begins: Stick man falling through onions and apples... The video ends: SO, we just integrated multiple derived equations of motion to derive harmonic oscillators that describe the gravitational influence on objects falling in and around a sphere...

00:13 Matt: "a apple" Me thinking: "Were my kindergarten books wrong?!"

"It's very hard to draw on an onion." Ah, yes. The kind of quality information I came here to learn.

My train of thought as Matt is demonstrating the ideal motion through the planet given a perfect system: "I need to gif this." Matt: "Oh, grow up!"

Can you explain the side hole more in depth? my understanding of this problem up to now was that your path must pass straight through the center of the planet so all side gravitational forces cancel out.

24:06 So, if the Earth fell through the sun, surviving intact, it would reach the opposite side of its normal orbit in the same amount of time? whoa

"The reason I'm using [the onion]... it's made of concentric spheres." Yeah, onions have layers. Like ogres.

by making that hole, would nt it decrease the mass of earth

Hi Matt, I'm taking a numerical analysis course in university right now, and... your "cheating" method is Euler's Method, and it's actually what's used most often in the real world! Especially for difficult differential equations. This comment isn't meant to correct you, but rather to say thank you for letting me instantly understand what my professor was talking about when I went "wait... this sounds familiar!" in a lecture.

This prompted me to _finally_ look up the shell theorem. I've been wondering how to prove that literally for years, but i've never found it before because... well, maybe i just didn't look hard enough.

So, it would take me 42 mins to fall to me neighbors house? What would that even be like?

The figures are a fair way out due to the approximation of uniform density. If you use an accurate density model of the Earth you find that acceleration remains high all the way down to surface of the outer core, where it actually peaks at around 10.5 m/s² before then reducing down through the core to zero at the centre. This is because the core is very dense, holding up the mass beneath you as you fall relative to the uniform density model. This effect in turn means that you reach a much higher maximum speed at the centre - around 9.9 km/sec. The time to the centre at these greater speeds is reduced to just over 19 mins, so a total of 38 mins surface to surface.

This also assumes that you don't start accelerating to the point where your speed approaches light speed. In a large enough body in which to travel across through the body required going faster than the speed of light in order to travel in the given time, obviously there's a problem (which comes about in the case of many different formula in which the effect of relative mass is removed from the proper equation due to insignificance in a practical sense but in which that change in mass proportional to velocity becomes a substantive effect at higher speeds).

"is someone cutting onions in here?" hahahaha

Would, when falling from London to new York, the gravity of the center of the earth be great enough to pull you along the wall of the tunnel causing friction?

The answer to everything is 42!

Rho does change, because it is defined as the density of the mass in the sphere below you, which becomes more hotter and dense as you get closer to the center of mass

instant sub after the leapyear episode. love being precise

I feel like you glossed over the point that the gravity above you is cancelled out by the gravity of the shell outside your altitude below you. Is that something to do with the way the inverse square law for gravity falling off over distance cancels out the volume of a hollowed sphere?

Matt, the velocity at the centre 7,909 Ms-1 exceeds the detonation velocity of many explosives. So your 'fallanaut' could set off a bomb as he passes and outrun the explosion!

That was actually a well-put-together demonstration... Taking it apart was another story!

but the density of the earth does change. I'm pretty sure that the molten / solid iron core (we're not sure what state it's in under such great temprature and pressure) has a higher density than that of the mantle.

I didn't even think about it untill he said: oh grow up.

The movie total recall (new 2012 one) has its flaws, but one set piece of the movie is a way of transporting people and cargo by this exact method of falling through the earth. I remember the travel time quoted to be around 40 minutes so credits to the scriptwriters for getting that math correct.

This answers a question that was on my 1st year mechanics exam!

2:13 was specifically at me :(

Not like any other youtubers have done this ........

The implication here being that, once constructed, a trans-global tunnel could operate very efficiently, requiring only enough energy to counter wind resistance (which the boffins will have reduced significantly with clever aerodynamics) so long as the two platforms are //exactly// the same height above sea-level. So, under no circumstance is anyone allowed to use Imperial Scale when constructing this.

I have an idea for the material and/or cosmology analysis. When you are on the planet of unknown mass, make a tunnel through the core, fill it with the vacuum (unless it's already) and measure the travel time. Congratulations - now you know the density of that planet and thus, assuming you know the radius, its mass.

I'm disappointed. That's not even remotely a coincidence. The ball's motion along the "y" axis is the same. It just also happens to be moving along the x axis as well.

I remember once i was watching an episode of doctor who and in the episode there was a hole dug to the core of the earth where there was some evil spider nest. Later in the episode the doctor flooded the hole with water from the river Thames, killing the nest at the bottom and the spider queen. I thought "that doesnt seem realistic" so i calculated the actual volume of the cylindrical hole to find out how much water could fit in that hole. I believe i was watching the show with my mother at the time too.

14:20 I would think that splitting it into 1m chunks would actually be _more_ effort than the integration even using a computer.

Note also: the period of a pendulum with length = r of the earth would have the same ½ period as the a/r fall through ½ orbit of the earth at the surface. (ref: Schuler).

I dont know why i watched this whole video about a subject i dont understand and still really dont,and honestly really dont care about..but i was quite entertained and enjoyed the vid.

You know what's even more amazing? It will ALWAYS take the SAME amount of time no matter from where the hole starts, and wherever it ends (assuming ideal conditions). It could literally be 2m away from the start point, as long as that end point is also on the circumference of the Earth. EDIT: Nevermind got too excited. He did say it later on.

18:00 Assuming you can throw a ball at 40 m/s (= 144 km/h = 89 mph), then you could do this on an Earth-density spherical rock with radius 32 km. With a more realistic density for an asteroid-sized object (~1000 kg/m^3), then you could throw a ball at 40 m/s and catch it on the other side of a spherical asteroid with radius 75 km. Check for yourself: R = (v^2 / (4/3 * pi * G * density))^0.5 R = radius v = throw speed

The density of the earth is not equally distributed. The heaviest elements are concentrated in the center. So the decrease of acceleration towards the center is actually smaller than presumed.

Of course, rho isn't actually constant in this equation, since inner layers of the Earth are much denser than outer layers. In fact, as you descend into the Earth, for a while the acceleration actually increases as the effect of getting closer to the dense core outweighs the effect of neglecting the mass above you. Present models show that the freefall acceleration would continue to increase until you reach the edge of the outer core, at which point it would begin to decrease roughly linearly. Wikipedia has a cool graph here: upload.wikimedia.org/wikipedia/commons/5/50/EarthGravityPREM.svg.

One thing I noticed that may have been deliberately overlooked is the fact that a sphere with a hole though the center is actually a toroid (though not a torus). With an inner radius just large enough for a person to fall though the difference is probably negligible but the presence of such a tunnel could still be seen as affecting the average density of the earth. From what I can imagine, this probably adds too much extra work to be worth the difference in the answer but I still thought I should point it out.

At 18:52, he claims that a satellite orbiting at the surface of the earth would go halfway around in the same time and at the same speed as a person falling through the earth. That claim doesn't hold up however, when you consider that the length fallen by the person (2r) is not the same as the distance around the planet (pi * r). The ball somehow moves at the same speed for the same time but travels further?

(correct me if I'm wrong, I'm interested in learning) assuming we have friction by air and therefore the falling / up-and-down motion is slowed down (dampened): we'd sooner or later end up with zero velocity - in the very core of the earth - floating at zero G ... cool 😎

Back in the late Seventies and early Eighties our F104 Starfighters has the inertial navigation system LN3 which had Schuler pendulums. If I recall correctly they had the same period of roughly 84 minutes.

42 minutes, so 42 is really answer to everything, all mighty number.

I thought the coincidence would be that 42 (ignoring those couple of seconds) is The Answer to the Ultimate Question of Life, the Universe, and Everything, thereby suggesting that The Ultimate Question of Life, the Universe, and Everything is how long it would take to fall trough the earth (or any earth-dense object) in a perfect vacuum with no resistence

24:06 How can you possibly "fall" in a straight line that does not go through the center ("London to New York")? That would have to be an ellipsis (for a periodic fall) and in order to avoid the center and make it an ellipsis you would need some non-zero tangetial initial velocity.

I was screaming "no, no, no, that can't be!" when I saw the number for the maximum speed and time, because it reminded me somehow of the orbital speed/period(/2) of the ISS (at least +- a little bit). Yes, that *is* nice, although in hindsight I would not call it a coincidence.

Since inside the planet you have more planet under your head than under your feet, you get the opposite of tidal forces. In a really dense planet you will get the opposite of spaghettified (pizzafied?)

Sorry for only reading thru 3months of comments, so I don't know if this is already said. But the gravity of the moon would pull you to the side of the tunnel. That's quite obvious so let's forget the moon. Also gravity of the Sun would pull you to the side of the tunnel. Because your center of mass is not at the same place where earths CoM is. That's why your orbit would be different than the earth's orbit... But does it cancel out when you are oscillating, I cannot do the math...? Also if you're interested in these problems, check out Scott Manleys video in RU-vid: Could an astronaut orbit a space station. Thanks for the good video!

What an interesting question! I'm sure that my teenage self could have done this calculation, but having forgotten most of the tools I'd need, I thought I'd try to work out a very rough figure, and then see how closely it compared to Matt's answer. Avoiding any calculus, and without looking up any formulæ I could not remember. I had no idea what the answer would be, but I was expecting it to be several hours. I looked up only the radius of the earth = 6731km approx. I remembered that the acceleration due to gravity at the surface of the earth is 9.81 m/s². I realised that the acceleration at the earth's centre would be zero, and that the time to reach the centre would represent half the time to go the whole way - say London to Auckland-ish. I remembered one formula (one that Matt said was useless), namely s = ut + ½at² where s = distance travelled, t = time taken, a = acceleration; u = initial velocity (which in this case is zero: we're assuming a standing start). In this case, we want T, being twice the distance to get to the centre, so T = 2t. Setting s = 6371km = 6731k m I got 6371k = 0 + ½at² So t² = 2 × 6371k ÷ a = 12,742,000 ÷ a Then I made my huge approximation: knowing that a varies from 9.81m/s² (or approx 10) at the start to zero at the centre of the earth, I simply took a figure of 5m/s² as a very rough value across the whole radius. So t = √ (12,742,000 ÷ 5) = √ (2,548,400) So t = 1596 sec So T = 2t = 3193 sec, or about 53 minutes 13 seconds. My "gut feeling" was that this was a bit on the low side ... but I was very happy to have got a figure within 25% of Matt's answer.. By the way ... "a apple" .. "a earth-orbiting satellite" ? Oh - and what if the earth doesn't have constant density ...?