plot twist- the moon must pull the same(negligible) amount and be to scale by mass to the real moon. find x, the distance between the moon and planet plot twist part 2- the organisation offers you an alternative, have the moon have the same period as earths moon instead of mass, find distance.
The moon has a mass equal to 1.2% that of Earth's approximately, so you could have a few. Although Saturn's rings compared to Saturn's own mass is smaller by a magnitude of like six. I'd throw that in as well.
I love this. Perfect for people who love the presentation and style of Vihart and wished that there was something with slightly higher-level maths with a similar vibe. Fully deserves the praise Grant has given it.
To be fair, earth 2.0 would be almost the correct gravity at all points given how close it is to a sphere. I would take 7.4% off of an expensive purchase like that any day.
i think, given that they're legally required to make sure your planet has the gravity force you're used to, that realistically, this implies you are not allowed to move from the single point. but yes, assuming that they do are not liable for gravitational inconsistencies after the purchase, then you should definitely take the deal and trick the gravity police
I'm honestly surprised this video doesn't have more views. It looks like an incredible amount of work went into it with all the claymation animations. Great work.
I recently read a paper ("The Cartan-Hadamard conjecture and the little prince") that used in a key way the solution to this problem in a non-Euclidean geometry (which surprisingly is not significantly harder than the Euclidean case). Your exposition made the solution much clearer for me, thanks!
Very fun entry, glad to see this. As an extra note, the coordinates for the surface of this shape in 3D space seem to have the equation y / (x² + y² + z²)^(3/2) = C, which offers a nice way to see that it's symmetrical along 2 of the axes.
This is an amazing video with a unique claymation + Vihart style! Keep up the great work, and soon enough, you'll make it to the upper echelon of RU-vid educational channels :)
Great job, this is really cool! I haven't taken physics classes in a long while, but I paused the video to do some calculations before the big reveal, and I was actually surprised to see that both the "long spaghetti earth" and "flat earth" were terrible options. For some reason my intuition told me that the "long spaghetti" would be best since all the matter is directly under your feet, maximizing its contribution - but of course it sends the matter way to far, defeating the purpose. I had fun and the explanation was very clear and engaging! Keep up the good work :)
When I saw the title of this video and its thumbnail, I thought for sure I was about to watch a wild fever-dream video. But I was pleasantly surprised with an enlightening video instead. Thank you!
Pretty cool video! I'm wondering... Your dewdrop shape has a point with max gravity equal to Earth's, but what is the average gravity on its surface? Also, what's the lowest gravity point and where is it located? Thanks a lot for the video 😊
Something I would be interested in is the gravity at every other point. What is the lowest downward force you experience on that shape? (By intuition I would guess that every other point you could stand on would have lower gravity than earth gravity, but it would be cool to find out for sure!)
holy moly this was such a treat to watch from start to end. I can't imagine the work that went into it but it turned out fantastic! Excited to see more in the future if you have stuff planned.
Great video about a really fun puzzle! My intuition was that the optimum would actually end up being a sphere and there was no money to save, so I'm pleasantly surprised that I wasn't *too* far off, but the answer still ended up being more interesting :)
Amazing, bravo! Another cool thing that I remember is that if your planet is a "thick sphere", by which I mean it is like an empty coconut, then the gravitational force inside of it is zero. So you could buy a hollow planet with a cork and a zero G playground inside of it.
This is such a fantastic video! Great production, interesting problem, super well-explained, it's just missing unicycles. My small nitpick is that a symmetry in a problem doesn't necessarily imply a symmetry in a solution--it might be that the problem has a whole symmetric family of asymmetric soultions. But I loved how you really crunched the problem down with clever argumentation!
what a great video!! i think the topic was a little too advanced for me, but it was still entertaining to go through, and it made me want to learn more so i can one day understand it better :))
This is a very neat video! I wonder what the gravity is at other points on the optimal shape? Since its so close to a sphere, it must be pretty close, only slightly lower and pointing at slight angles anywhere else I imagine. I suppose that makes the planet more interesting than a plain sphere!
Astronomical bodies above a certain size tend to mould themselves into a more and more perfect sphere shape the higher their surface gravity is relative to the materials they're consisted of (assuming no rotation, which pushes them out slightly toward the equator and squashes slightly toward the poles). The point of Earthlike gravity on the optimized planet would also be the lowest point on the planet topographically. With no mountain ranges to isolate it or lower points to drain to, it would quickly be buried and/or drowned by water (if your planet is not a desert). So, your earthlike gravity point would be short lived and/or at the bottom of an exceptionally deep ocean.
Even intergalactic companies cannot resist the lure of putting a catchy and marketable name to anything they can get a hold of. In this case, naming as "ThickBrick" what is essentially neutron star paste.
This video is wonderful. For anyone who is interested in a deeper dive into what actually causes gravity, I recommend Vsauce's old video called "Which was is down"
I suppose the natural next question would be where on this planet is gravity *minimized*. Or if I wanted to have a lot of fun, what shape maximizes the difference between its lowest and highest gravities? Thanks for giving me this idea!
@@inQEDible I haven’t done anything formal yet, but it did dawn on me that a hollow spherical shell will have no gravity inside, and a giant plane will have very little gravity on the flat sides with noticeable gravity on the edges. With that realization, I think I may have to impose some restrictions about concavity to prevent hollow shells (or shapes that are, in the limit as a hole approaches zero diameter, a hollow shell), and possibly some limit on volume to prevent ridiculous cases like incredibly large disks.
@@SeanBlantonPhD I believe I had used a different coordinate system than the video, so even though I had arrived at a different formula the shape may very well be the same.
This works on paper, wonderfully in fact, but it doesn't take into account hydrostatic equilibrium. Hydrostatic equilibrium is solid matter acting like a liquid when enough mass is present, and is the physics reason (or result, depending on how you slice it) for why planets Earth's (or even Mars') size are always round.
The word 'with' instead of 'using' in the title confused me completely. I thought this video was about affording a planet that had geometry (implying there are non-geometric planets, which sounded silly and interesting).
Let's say you're still short on galactic bucks (it's hard to do fine in this interstellar economy, with all this cosmic inflation and whatnot). You know this other guy who wants their own planet, you're not really friends so the two of you don't want to be neighbors but you both would be fine having if you lived on opposite sides. Then you wouldn't really see each other and each one might as well pretend they have the planetoid for themselves. A sphere would work, sure, but we could probably use our funds more efficiently, right? Also having gravity g at just two points will ensure each planetoidmate stays on their end of the spacerock. The reasoning that follows is very similar to that in the video but this time we care about two points where the little co-princes will live. Let's say that those points lie at x = -1 and x = + 1 on the x axis. Bear in mind that this means that 'down' will mean "towards positive x" for the inhabitant at (x, y, z) = (-1, 0, 0) while 'down' will mean "towards negative x" for the inhabitant at (x, y, z) = (+1, 0, 0). Another way of thinking about this is that each asteroidmate will consider the other one to be directly below them, the same as two people at antipodal points on Earth (they could 'dig down' through the planet and reach the other). Many of the arguments in the video still hold: we'll want our shared planet to have rotational symmetry along that axis and it will have no holes (so finding the surface will be enough to fully characterize its shape). The whole mass of the planet will have to lie between x = -1 and x = 1, the scale of what "distance 1" means will have to be adjusted later to make everything work. Also we'll treat both co-owners equally, so the whole planet will have mirror symmetry across the x = 0 plane. In the video, Daria discussed a surface where the 'downwards' gravitational pull was constant for the observer. Since we have two observers we care about, we'll now consider a surface where the sum of the 'downwards' pull P for each observer is constant, so P₁(x, y, z) + P₂(x, y, z) = k For this problem we’ll be better off using cylindrical coordinates along the x axis: (x, r, ɸ) where r = √(y²+z²) is the distance to the the x axis and ɸ is an angle defining the direction of the y and z components. Since we have rotational symmetry along the x axis, we don’t have to worry about ɸ; we’ll only be working on a plane defined by the x axis and a perpendicular r axis where P₁(x, r) + P₂(x, r) = k P₁(x, r will be the force F₁ a mass at point (x, r, ɸ) exerts over (-1, 0, 0), projected onto the x direction. The gravitational force is inversely proportional to the square distance, which can be calculated as: F₁ = 1 / d² d = √((x+1)² + r²) -> F₁ = 1 / ((x+1)² + r²) Using some trigonometry, we can calculate that P₁ = F₁ · cos(θ₁), where θ₁ can be calculated as θ₁ = atan(y/(1+x)) This gives us a formula P₁ = cos(atan(y/(1+x))) / ((x+1)² + r²) In a similar way, we can compute that P₂ = cos(atan(y/(1 - x))) / ((x - 1)² + r²) so our surface will be one where P₁ + P₂ = cos(atan(y/(1+x))) / ((x+1)² + r²) + cos(atan(y/(1 - x))) / ((x - 1)² + r²) = k The value of k and the scaling factor for the real distance between the two little co-princes will depend on the density of the material we’re using. The resulting overall shape depends on the value we get for k. One way to visualize it is to use the 3D graph function in Google, just search up the following equation: z=cos(atan(y/(1+x)))/((x+1)²+y²)+cos(atan(y/(1-x)))/((x-1)²+y²) and set the scales on the bottom right to graph the equation for x between -1 and 1, y between -1 and 1 and z between 0 and a positive number k. The actual shape the planetoidmates get can vary considerably. For k > 2, we actually get two separate planets, if k is 3 or larger the result looks a lot as if each co-owner had bought one of the planets in the video. For k between 1 and 2 we get a funny ‘peanut’ shape, with each asteroidmate living on the far ends of the peanut. For k = 1 we get something barrel-like, with each person living on the lids. For smaller values of k, the barrel gets wider and wider. By k = 1 - (1/e) = 0.632 the barrel gets as wide as its tall, giving us the most sphere-like shape we’ll get. For even smaller values we get a barrel-like shape that is wider than it’s tall, we might call it pumpkin-like. Still, I suspect that for very small values of k the resulting shape is no longer cheaper than a sphere.
It's relatively easy to extend this to a scenario with 4 people sharing the planet, located at points (1, 0, 0), (0, 1, 0), (-1, 0, 0) and (0, -1, 0). This would give us contours like those of: z=cos(atan(x/(1+y)))/((y+1)²+x²)+cos(atan(x/(1-y)))/((y-1)²+x²)+cos(atan(y/(1+x)))/((x+1)²+y²)+cos(atan(y/(1-x)))/((x-1)²+y²) It should be noted that things are no longer rotationally symmetric along the x axis, although there is still a mirror symmetry along the x, y plane. The shape we get from the contours is just a slice through the middle of the planet along that plane. Once again, for large values of k we get separate small planets roughly of the same shape as the one in the video. For smaller values of k those shapes join up, first into a clover shape , then a cross, eventually an octagon (where I suppose there should also be space for 4 other people) and finally something roughly circular. Still, I have a strong suspicion that these arrangements wouldn't be optimal and that four people could get a better deal if their points were arranged like the corners of a regular tetrahedron instead of a square on a plane.
One funny thing about that 4-person arrangement is that for some values of k like k=4.2 the optimal shape _has_ holes, as a mass positioned in the center doesn't contribute as much gravity to the four points as masses positioned closer to any of the four points.
I maintain that the best thickbrick planet is the biggest hollow sphere that doesn't collapse. Assuming you value surface area. Edit: I think I mixed up one of my gravity theorems - I forgot that radius still matters for a hollow shell. I'd have to do some thinking to figure out if it's more cost effective than a solid sphere.
I'd be happy, a planet with earth like mass will become spherical by enacting gravity on itself. Soon I'lll have a regular looking planet with 7.4% lower gravity. Take that wierd agarthea rules!
Based on my attempt to cobble together a formula for converting between rectangular and polar form from high school with the inverse square law, the planet should be acorn shaped
Awesome video, one thing I don’t understand though. Why isn’t it just a cone? Surely with your squashed dew drop, you were going more out to the side than absolutely necessary at a given distance away?
The cosine is very important here, if the point directly under you is an infinitesimal away, any points on that upwards facing plane would contribute infinitely more to the horizontal component than the vertical component. If this wasn't the case for some reason I imagine you'd use a hemisphere as you'd just plot something like x² + y² = r², y
There are no constraints on the problem so I think you could have a cone, or something with the boundary of the cone. With point masses, you can arrange them in many ways. The minimization is a constraint, but you need other constraints if the little guy is not going to just live on a point.
