ThePoshFart 19 Posted July 18, 2015 A little bit of backstory first this is something GodOfInternetLag, RyanBlack and I were talking about after RyanBlack said something about teleporting to Australia. After that I asked the following question. If you were to be teleported to the opposite side of a sphere with gravity(the earth, from the USA to Australia), would you fall head first towards the ground when you got there or would you land on your feet? Basically would you wind up being the top or bottom image. RyanBlack did all this math and said you'd land on your feet. <02:59:02> "Tate Hollows [Warden]": F = -\frac{GMm}{r^2} <02:59:15> "Tate Hollows [Warden]": M = \rho \frac{4}{3}\pi r^3 <02:59:34> "Tate Hollows [Warden]": F = -\left(\frac{Gm}{r^2}\right)\left(\rho \frac{4}{3}\pi r^3\right) = -\left(\frac{4}{3}G\rho \pi\right) mr <02:59:53> "Tate Hollows [Warden]": r(t) = R\cos{\left(t\sqrt{\frac{4}{3}G\rho \pi} \right)} 0 Share this post Link to post Share on other sites
ThePoshFart 19 Posted July 18, 2015 Agreed. With? 0 Share this post Link to post Share on other sites
Hollows 450 Posted July 18, 2015 According to the calculations if you were to tunnel through the Earth, as following what he quoted me saying: <02:59:02> "Tate Hollows [Warden]": F = -\frac{GMm}{r^2} <02:59:15> "Tate Hollows [Warden]": M = \rho \frac{4}{3}\pi r^3 <02:59:34> "Tate Hollows [Warden]": F = -\left(\frac{Gm}{r^2}\right)\left(\rho \frac{4}{3}\pi r^3\right) = -\left(\frac{4}{3}G\rho \pi\right) mr <02:59:53> "Tate Hollows [Warden]": r(t) = R\cos{\left(t\sqrt{\frac{4}{3}G\rho \pi} \right)} Then... You would keep going all the way to the other side. In general, you can tell if there’s friction around because it slows everything down, but it doesn’t reverse the direction of things. (Yes, I used possible theoretical physics to answer the question about teleportation. You'd land on your feet, in essence, theoretically, according to the laws of science.) 0 Share this post Link to post Share on other sites
Preacher 4 Posted July 18, 2015 According to the calculations if you were to tunnel through the Earth, as following what he quoted me saying: <02:59:02> "Tate Hollows [Warden]": F = -\frac{GMm}{r^2} <02:59:15> "Tate Hollows [Warden]": M = \rho \frac{4}{3}\pi r^3 <02:59:34> "Tate Hollows [Warden]": F = -\left(\frac{Gm}{r^2}\right)\left(\rho \frac{4}{3}\pi r^3\right) = -\left(\frac{4}{3}G\rho \pi\right) mr <02:59:53> "Tate Hollows [Warden]": r(t) = R\cos{\left(t\sqrt{\frac{4}{3}G\rho \pi} \right)} Then... You would keep going all the way to the other side. In general, you can tell if there’s friction around because it slows everything down, but it doesn’t reverse the direction of things. (Yes, I used possible theoretical physics to answer the question about teleportation. You'd land on your feet, in essence, theoretically, according to the laws of science.) Yes, yes I see, I see that I do not understand any of this. 0 Share this post Link to post Share on other sites
Hollows 450 Posted July 18, 2015 F=GmM/r2 = Earth's Mass. M = \rho \frac{4}{3}\pi r^3 = The Density (Of Earth). F = -\left(\frac{Gm}{r^2}\right)\left(\rho \frac{4}{3}\pi r^3\right) = -\left(\frac{4}{3}G\rho \pi\right) mr. (Which is what you are 'falling' at, and or would be the speed of your teleportation, in theory.) Then we use Hooke's Law - F = -kX. Hooke's law is a principle of physics that states that the force F needed to extend or compress a spring by some distance X is proportional to that distance. (Cited from Wikipedia to put it in a summary.) The answer is... a = r(t) = R\cos{\left(t\sqrt{\frac{4}{3}G\rho \pi} \right)}. Where R is the radius of the Earth, and t is how long you’ve been falling and or teleporting. Cosine, it’s worth pointing out, is sinusoidal. (This thanks to The Physicist.) 0 Share this post Link to post Share on other sites