Game Theory: Mario's LUNAR APOCALYPSE!! (Super Mario Odyssey)

Mr. MatPat, how many moons does it take to get to New Donk City? Let's find out. a! *crisp noises* two-hoo! *more crunching noises* a three! Twenty minutes later... 46 47 48! 48 sugar moons. Now, if you'll excuse me, I'm not feeling very well. BBLLEEEAARGH Hello Internet! Welcome to Game Theory, the original way to tackle all the toughest topics in gaming. Today we'll strap in and jump into Super Mario Odyssey, one of the most satisfying and confusing Mario worlds I've ever had to experience. I mean, you have a Rainbow Forks kingdom right next to Nintendo's version of Dark Souls. Then, due to Mario's cannon height of 5 feet 1 inch, the people of New Donk City are over 10 feet tall, but of course Wario isn't in the

game

so I can't prove it for sure.

Once and for all, my calculations from four years ago weren't as crazy as they initially seemed, okay? But you know, arguably the most fascinating part of this

game

has to do with the last place you go: the moon. Not because of the Dark Side stages or the onslaught of bosses, but because when you really sit down and do the math, it turns out that the moon is the biggest threat to Mario. No, I mean the entire Mushroom Kingdom! Forget about Bowser and his petty little plans to marry Princess Peach... That's nothing. This thing. This right here is a threat more terrifying than anything Nintendo has ever invented.

We're talking Majora's Mask levels of planet-wide

lunar

catastrophe, people. But to really understand why this thing, the moon, is such a threat to Mario in this game, we'll have to do some math, so get ready, theorists. Let's open up some formulas and get out the TI-84s. Yes, 84 (we have to bring out the big guns today) because here we go, off the rails to explore Mario's big, crazy world of astrophysics. To begin, we must first establish how big Mario's Earth and Moon are and how far these celestial bodies are from each other. Now, by now you know how many headaches the Mario franchise causes me on the pixel measurement side.

From changing heights to crazy gravity, Mario doesn't so much fight Bowser as he constantly fights my math. But determine the size of the planet in this game, this one. This was the last straw. I was sure there would be a way to do it between all the maps and globes that are present in the game's marketing and in the game itself. And let me tell you: I tried. I tried measuring pixels, but we didn't have reference points of consistent size. I tried measuring steps, but none of the stages allow you to actually see an entire continent.

We tried Al-Biruni's method. The Iranian scholar who determined the radius of the Earth by observing the heights of mountains. But without being able to reach a point on the horizon, we couldn't really establish a distance between anything. I looked at the balloons, the brochures, the splash screen, the opening menu, the promotional materials for this damn game, so many different things, but damn, Nintendo, you always have to make it hard for them. Then you know what I gave up. You heard right. I just... threw in the towel. To my knowledge, there is no definitive way in this game to demonstrate the actual size of Mario's planet and his moon.

I invite you all to try it. Consider this the Mario Odyssey challenge. If you figure it out, we can do an episode dedicated to you and your math and no, let me cut this one in your ass right now. You can't assume that the Hat ship on the world map is the actual size that the proportions would make. of this globe is absolutely absurd and wouldn't actually match what we see in the game with the actual ship or Bowser's ship being much smaller relative to the actual size of the continents, just throwing that in seriously, I feel like it I tried everything.

Here, however, just because I abandoned the measurement doesn't mean the

theory

itself was dead in the water. Doing this for as long as I have, do I know there's more than one way to skin a Mario? Oh boy. That is a horrifying image. Sorry, it didn't work out anyway. The key to this

theory

is the Roche limit. A theoretical concept first proposed by game theorist Reddit user JohnsonAndJohnCena. For those of you who don't know there is a magical distance between a moon and a planet to remain a moon. Now, what do I mean by that? Well, think about it this way: any planet or moon is held together by its own gravity, right?

That's why all these things are balls. All the material on the planet is being drawn towards the center, turning it into an orb IN SPACE. But remember that everything has gravity. Big things like the Earth, smaller things like the Moon, even really small things like you or me. We all have those gravitational forces that act on us. For example, the Earth, under the influence of the Moon's gravity, has entire continents elevated by about 50 centimeters due to the Moon's attraction on the Earth and, conversely, the Moon's surface moves about five meters towards the Earth in response to the Earth's gravitational pull and to make Those facts even more incredible.

Remember that right now the Moon orbits the Earth at an average distance of about three hundred eighty-four thousand kilometers or two hundred thirty-eight thousand miles. That's how strong the force of gravity is. So that begs the question. What happens when you bring those two objects together? Where the large object like the Earth with a lot of gravity is able to pull even harder on the small object with less gravity like the Moon? Well, something unexpected actually. The moon would shatter because Earth's gravity is stronger than the gravitational forces holding the moon together. Have you ever wondered how Saturn got its rings?

This is believed to be the reason. A small moon orbited too close to Saturn and was separated by gravity, with the pieces now orbiting the planet as rings and this, loyal theorists, defines the Roche limit: the distance from a planet that an orbiting body must be so as not to be destroyed by the planet's gravity. The Roche limit of Earth is eighteen thousand four hundred seventy kilometers or about eleven thousand four hundred seventy miles. That's about one-twentieth the distance the moon orbits now, so if it ever ventured within that eighteen thousand kilometer limit, it would break apart. and the Earth would suddenly have some rings and that, ladies and gentlemen, is how humanity could finally reach its final form by making our own planet shine.

We liked it and decided to put some moon rings on it. Which brings us back to Odessy. Have you ever wondered why we collect moons, instead of stars, in this game? Could it be that what we're seeing here are early signs that Mario's moon is being torn apart because it's orbiting within Mushroom Planet's Roche boundary? You probably haven't stopped to ask yourself that, and as absurd as it may seem, it seems to be the case here. Look at those huge chunks of space rocks scattered around the Mushroom Planet. We know for sure that they come from the moon.

Not only by name, but also by seeing them on the moon itself. But to be really sure we need to determine the Roche boundary of Mario's Earth in the game and then see where the moon falls in relation to it. Now, as you can imagine, that involves a lot of math. So I'll try to remove anything that's not too interesting. Scientists are the small sponsor of LUNA food day, but hey, that's just a theory. Sorry, maybe I deleted too much there. In all seriousness, the Roche limit of a planet is equal to 2.4 times the radius of the largest object multiplied by the cube root of the density of the largest object divided by the density of the smallest object and, as you can probably see , there's a lot. of things in that equation that we don't know.

I mean, I just admitted two paragraphs ago that I can't find the size of the planet, so without that, how do we have even a slim chance of calculating the radius of the planet, much less the density of the planet? And then we have to do the same with the moon. Well, it turns out that we don't actually need exact distances as long as a common unit is used in all of these calculations. So stay with me here and let's base everything on the radius of the moon. Give it a variable. Let's call it small 'r'.

Now when you zoom in on the world map, there are two frames where Mario's moon and the planet are next to each other and that's important so that perspective doesn't distort their proportions. Taking pixel measurements from both frames, we can see that the radius of the moon is approximately 30% of the radius of the planet. Now, earlier when I presented the Mario Odyssey challenge to you, I said that the proportions on the screen might not be one hundred percent true, but fortunately we can verify this as well. In New Donk City, there is a model of the Earth and the Moon on the ground in front of City Hall and wouldn't you know it?

But once again, mathematics proves that in this model it is determined that the radius of the Earth is 3.3 radii of the moon and with that we can say that the radius of the moon is equal to the small 'r' and the radius of Mario's planet is equal to 3.3 times the small 'r'. Eventually we are working on finding the density. The next step is to find the mass of both Mario's Earth and Mario's Moon. Which again seems like it should be impossible, but luckily we have another secret backdoor here. Gravity. You see, when it comes to finding the properties of planets and stars, there are many different formulas we can use, since you know there is no real way for scientists to weigh a planet without using some kind of math, so here we can use gravity.

To calculate mass using the following equation mass equals acceleration multiplied by the radius squared over the gravitational constant. That acceleration right there. It's as easy as finding out how fast Mario falls after jumping. First on the Mushroom Planet and then again on the Moon, something we've done many times here on the channel, especially in my episode Covering Mario's Performance at the Olympic Games, where we discovered that the Mushroom Kingdom's gravity is more extreme than the of Jupiter. . Anyway, for that all we need is the distance Mario is falling and the time it takes for him to fall.

Plug those numbers into this equation and we get OMG! On his home planet, Mario's normal jump has a downward acceleration of 70.16 m/s^2. That's more than seven times Earth's normal gravity. It's three times the gravity of Jupiter and is more than double what it's been in any other 3D Mario game. If you're wondering why Mario feels heavier and falls faster in this game, then there you have it. It's not Mario packing the cannoli. It's his planet gaining weight. We do the same on the moon and obtain an acceleration of 11.74 m/s^2. 1.19 g forces. Which is pretty incredible if you think about it.

This right here? This is what Mario could jump if you were subject to our miserable gravity. Now, with those accelerations in hand, we return to the mass for both the planet and the moon. These are just some basic calculations. Let me review. This quickly speeds things up. Checking the acceleration numbers, making sure to leave 3.3 R for the planet radius when we calculate for Mario's Earth and just the old little R when we calculate for the Moon and get a mass of 1.14*10^13 r^. 2 kilograms for the planet and 1.75*10^11 r^2 kilograms for the moon. Now, with the mass and radius of each body determined, determining the density, the final part of the Roche limit equation is very simple.

Density, remember, is just a trip to the DMV. D is equal to M over V Mass of the object divided by the volume of the object The volume of the sphere, which applies to both the Moon and the Earth, is calculated as 4/3*π*raduis^3. I won't bore you with math. Basically it is 150.53 r^3 for the Earth and 4.19 r^3 for the Moon. Which leads to densities of 7.6*10^10 divided by the radius of the Moon to the Earth and 4.18*10^. 10 divided by the radius of the moon for the moon and that ultimately means we have all the variables we need for the Roche limit equation.

As a reminder, the Roche limit of a planet is equal to 2.4 times the radius of the largest object multiplied by the cube root of the density of the largest object divided by the density of the small object. Therefore, the Roche limit of Mario's plane is 2.4 times the radius of the largest planet, which is 3.3 'r. As you remember, it's 3.3 times the radius of the moon multiplied by the cube root of that mess of densities we just calculated. Don't forget that when you divide a complex fraction, you multiply it by the inverse, so those little r's cancel out there, just like those 10^10.

Do you have all that? No? Perfect. Don't worry, you didn't miss anything. Going through all those numbers, we see that the Roche limit is equal to 9.66

lunar

radii. In other words, thanks to all the calculations we just performed, we show that if Mario's moon is within 9.6 lunar radii of the center of his Earth, the Earth's gravity will slowly tear it apart, giving us the which is why we have hundreds of minimoons and moon rocks scattered across the surface of the planet.Mario's planet and you know, that's exactly what's happening here. You know, in the Moon Kingdom there is a place where you can see the Earth not so far in the distance.

Well, using that point of view it seems pretty obvious that the Moon is very close to Mario's planet, so we could assume it's inside. that Roche limit, but that's not good enough for me or the standards I have for this program, so I use that point of view in conjunction with the game's compass. You heard right. Using the in-game compass to measure real angles may seem unbelievable, but it's actually possible to determine exact distance with trigonometry, string lengths, and good old days.SohCahToa But since this is the kind of math I personally didn't want to see come back to appear in my life, I can only assume that three of you really care about this kind of essential detail.

I'll do it as a mini theory if that's enough. of you and the comments are really excited about that sort of thing, suffice it to say that running all those numbers, the moon is about 5.5 lunar radii from Earth, confirming that it is within the Roche limit and that they are eating her alive. by the planet's gravity and remember this is not hypothetical, it's actually not just a theory, this is actually how Mario's planet would behave in relation to Mario's moon. It would be eating him alive. That's crazy, and while that goes a long way toward explaining why the moon rocks and multiple moons are scattered everywhere, it also explains some dire consequences heading straight for Mario's planet.

To give you an example, consider this. When our Sun begins to burn up and become a red giant, it is predicted that our moon will be knocked out of its orbit and approach Earth, crossing the Roche limit. At that moment it will be destroyed. On the plus side, it will give our planet a cool temporary ring. On the downside, it will shower the planet with huge chunks of lunar debris, destroying all types of land, causing tides of chaos, and sending the planet into another wave of extinction from space rocks crashing onto the surface all over the world just because. , while some of those pieces will orbit as small rings, most will be pulled by gravity toward the Earth's surface.

And that's the fate Mario Odyssey has already begun to allude to for the Mushroom Kingdom. And that is the fate that is determined for is based on science. And it certainly doesn't help that in the final battle Mario speeds up this process by destroying huge chunks of the moon's core. So enjoy running around to collect those moon rocks while you can Mario very soon. You won't have a choice as they charge towards you maybe that's why in Kirby and the Crystal Shards Shiver Star is a cold and dead land Maybe that's why in Kirby

super

star you see Mario traveling the galaxy He's looking for a new home, who knows , but all that needs will unfold in another episode.

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