New answers tagged

3

How long time does it take before three planets achieve the same relative position? The answer is never, except for the case when their orbital periods can be expressed with low integers, like the 4:2:1 resonance of Io, Europa and Ganymede However, what you are asking about is when they are going to be in almost the same position again, a quazi-period. To ...


2

The gravitational force you would feel inside the hollow planet is zero. Let's prove it. Let's call the interior of the planet $P$ (it is an open ball in $\mathbb{R}^3$. 1) Gravitational force is conservative, which means that $F = - \nabla \Phi$, where $F$ is the gravitational force, and $\Phi$ is a smooth function (the potential). 2) By Gauss' Theorem ...


1

C. Newton's Shell theorem states that if you're inside a hollow spherical shell that has matter even distributed in the shell, then the net gravitational pull you'd feel would be 0. This is because the attraction to the small near part of the shell is equal to the attraction to the large far part of the shell.


2

Have them pull out their smart phone and open up Google maps and touch the icon to zoom to their current location. Remind them that what they just did is only made possible by satellites in orbit using clocks specially tuned according to general relativity to maintain accurate synchronization with clocks on the surface (i.e., in the smart phone). Welcome ...


20

Related to the Andrea di Biagio answer. Here is a typical flight path for the Beunos Aires-Auckland route. The distance is approximately 10,300 km by the shortest route along a sphere. Direct flights are offered by Air New Zealand and take 11h40m - an average speed of 882 km/h. Looking at the map you present, the distance from Auckland to Buenos Aires looks ...


9

You need three flat Earth theorists on different continents (or otherwise well separated) all in on a (video) conference call, and have them measure the azimuth and elevation angles of the sun at the same time. The set of angles that they report will be inconsistent with their locations being on a flat disk laid out as in the map in the question. This is a ...


3

If the sun is a disk (or sphere) that is a certain distance above the earth, close enough that you get it at different angles above the horizon depending on where you stand, then it has to look like a different size to different people. If it looks the same size, then you cannot explain that it's at a different height above the horizon. See this diagram: ...


11

Foucault Pendulum is a great example. The original purpose of this experiment was to prove that earth rotates relative to the stars and not the other way around, yet nevertheless it proves that the earth rotates in a way which contradicts the "flat earth" theorem. This experiment can be easily recreated at home, if you don't live close enough to equator.


5

If you believe that anyone lives on antarctica at all, or really anywhere south of the antarctic circle, then you can use the "midnight sun" effect, which should not happen in the flat earth theory (because the sun moves around the "antarctica ring" during northern winter). Surveillance cameras will also work for this, if eyewitness accounts are not ...


2

Try https://www.insecam.org/ .Take a flat map of earth and make them go to the cities and watch the light and the online clocks. Nothing but a sphere fits the data. If the sun is going around a disk there would be night almost simultaneously over the flat disk, whereas the cameras will show progressive changes and night for half of their flat earth.


3

Take an arrow which you put tangent to the earth (assume it to be a perfect sphere). Now, travel around in such a way that 1) you enclose some finite area, 2) do not rotate the arrow locally. Coming back to your starting point you will notice the following: Although you did not rotate your arrow locally, you will end up with an arrow which is rotated ...


5

Can you convince your friend that time zones are for real? If he will believe that it is nighttime in China when it is daytime in the US, then his disk theory can't be correct.


6

Airplanes. Note that dozens of flights cross every ocean and the equator in all directions every day. If the Earth were a disk, it would actually be impossible to do that. For example, if the border of the disk were more or less where the borders of a planisphere are, it would be impossible to cross the Pacific. How would your friend explain how flights ...


27

I live close to Lake Erie and often see scenes like in this picture. Note that the bottom of the cargo ship cannot be seen due to the curvature of the Earth.


1

The number of tides stays just about the same, as the earth turns under the tidal bulges. On a water earth we would have two tides per day. Local landforms impact that greatly, varying from place to place. The tidal force is the difference of the moon's gravity on the near and far sides of the earth. It falls off as $\frac 1{r^3}$, so the if the moon ...


0

The tides will get smaller as the moon moves away having less gravitational effect. As long as there are tides there will be two per day because this is based on the Earths rotation not the moons distance. As the Earth spins the tides rise on the side facing the moon and the opposite side because of gravity. Its these tidal forces along with the Earths spin ...


1

Just imagine there would be no moon at all. Obviously there would be no tides due to the moon. As Dr. Chuck pointed out in the comments, there would still be tides due to the sun, but the tidal forces of the Sun are only about 46% of the moons tidal forces (taken from german Wikipedia about Tides: https://de.wikipedia.org/wiki/Gezeiten). So the farer the ...


-3

So this is really really easy and I don't know why the top answer doesn't explain what's going on. I have a lamp on my desk, which is opposite the headboard of my bed. Sometimes I lie there reading with the lamp on to illuminate the room, but with my lowered vantage point, oh no, a first world problem hits - there's a bright light in my field of vision ...


0

According to wikipedia the orbital period of Planet 9 is 10,000–20,000 years. This means it would take a minimum of 5,000 years to switch between perihelion and aphelion. So - even our best historical observation data of kuiper belt objects is just a tiny fraction of the time it takes for Planet 9 to move in it's orbit. There's not really sufficient ...


1

It's a combination of a few things. Firstly when we are looking for exo-planets we know we are not going to observe them based on their luminoscity, therefore we use different techniques based on how the exo planet will effect the light we observe from their sun. This method works brilliantly if the star and exoplanet are in relatively close proximity but ...


5

Pretty simple reason really. We only see exoplanets under extremely lucky circumstances. So we are only seeing a tiny tiny fraction of all exoplanets. If for example we are only seeing 0.1% of all exoplanets in each star system we look at, that is a HECK of a lot worse than the 8 out of 9 in our own star system.


23

The reason why we can see exoplanets 13,000 light years away but not a planet 200 AU away (about 30 light-hours) is because these planets are found using different techniques. The planet discussed in the article I linked was discovered using a technique known as "microlensing," which requires a star to pass behind another star with a planet around it. The ...


67

The problem with finding a new planet in our solar system is not that it is too faint, but knowing where to look in a big, big sky. This putative planet 9 is likely to be in the range 20-28th magnitude. This is faint (especially at the faint end), but certainly not out of reach of today's big telescopes. I understand that various parts of the sky are ...


6

We haven't detected planets millions of light years away. Right now the most distant is less than 20,000 light years away. Even for the planets we have detected, they are for the most part not "seen" or imaged directly. Instead they are found by the effect they have on the parent star (usually gravitational wobble or transit detection). In both cases, ...


2

"Space is big. Really big. You just won't believe how vastly, hugely, mind-bogglingly big it is." It seems like the night sky isn't so very big, so it should be easy to observe objects, right? Wrong. Once you use telescopes, the night sky becomes huge and if you don't know what you are searching for, you'll only find out about it by coincidence. The ...


3

All we can tell (assuming of course that the conclusions of the CalTech team are correct) is that there is a large mass in a distant orbit around the Sun. The mass could in principle be anything, but some things are more likely than others. It seems very plausible that the mass could be a planet that got ejected from an orbit nearer the Sun because: we ...


1

In principle, any body with the mass of the proposed planet would have the same gravitational effect as the planet. Therefore, it would explain the orbit of those other bodies equally well. We know of a lot of planets in orbit around stars (and we have theories about how they form). However, I don't think we've ever seen or theorized black holes in orbit ...


3

If at any time the speed of the planet in the reference frame of the star exceeds the escape velocity $\sqrt{2GM_\star/r}$, where $M_\star$ is the mass of the star and $r$ is the distance from the star to the planet, it will escape in a hyperbolic trajectory (or straight line if $M_\star\rightarrow0$). As noted in the other answers, the result of the ...


4

The scenario you suggest is of course hypothetical, but in all cases you must conserve angular momentum and mass/energy. So for example: If you have a way of removing mass from a star in such a way that the mass disappears outside the orbit of the planets (in astrophysics this is accomplished simply by mass loss - either the star has a wind that expels mass ...


1

Your final question very much correlates with a famous thought experiment.If the Sun was suddenly removed the planet s will still continue to stay in orbit. For 8 minutes and 20 seconds. This is because the speed of the space time fabric or simply putting gravity travels at the speed of light. That is, the earth will be devoid of sunlight and will move ...


-2

A coarse correction would have to be made because of the difference in the curvature of space in that section of space when a planet is deleted. The sun holds the planets in orbit and they would fly away as if the string to a tether ball was cut to all the planets.


0

My intuitive answer is that such a torus would be unstable, because an increase in density at one point on the surface would accumulate ever more mass, thinning out the opposite site, by conservation of momentum moving towards the centre of mass until the planet would be a very eccentric (because of the angular momentum) oblate spheroid.


1

I've got to chime in on this one for reasons of discussion only, because I can think of two reasons why this would not happen. Collisions are usually depicted as a center mass to center mass action. when we think of the earth and moon, it's thought to be slightly off center. yet, there can be a 'collision' when two bodies enter each others gravitational ...


-4

I studied the subject for 5 years and came to the conclusion that in average the toppling movement of the earth occurs each 12,000 years in average. The mass of the meteorite hitting the earth must be equalor more then 10 12th kg See for details in www.couldthesunriseinthewest.com/- Johan Leupen



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