Your flashlight example helped me sort out all the different forces acting on the body inside the shell, and that makes it really intuitive to understand why the total force adds up to nothing. Thanks!
interesting, i was taught to integrate the dF_G that you find using the standard formula to find the total gravitational force acting on the little mass m, factoring in the thickness of the shell by substituting dM, a small piece of the large mass M, with the volume of a ring of the shell, using R, dR, an angle measure, and the circumference. :) this was great too
This is intriguing. So if I am walking on the outer surface of a hollow planet, and encounter a hole and jump in it, would that mean I would be just floating in gravity? Also if there were two spheres of same mass/material, but one is hollow (which implies bigger radius), then gravity at the surface of the hollow one would be smaller, only because the distance to the center is farther out? Am I correct?
im sorry, the video is very good, but there is a flaw in your proof. im sure you're aware of the error but let me just make things clear. when you do the aproximation to the area of a flat circle as the area illuminated by the flash light, youre making an error, because spheres are not flat and are kinda more complex, so the calculus you're using is not correct. it gets to the right result by only mere coincidence. the correct calculus is a bit more complex.
You raise a very good point. For an "illuminated circle" of infinitesimal radius, the area of the sphere's surface is the area of a flat circle. Now, I did gloss over the tricky part, which is that since the flashlight beam is not (in general) hitting the sphere perpendicular to its surface, the illuminated area will be an ellipse. The area of this ellipse will be greater than or equal to the area of a circle at that distance--my formula in light green should actually be: A_Light = pi * d^2 * tan^2(theta) / cos (alpha) where alpha is the angle of incidence of the light beam (for perpendicular incidence, alpha = 0 and you're back at the circular case). In order to make the proof work, you need to then prove that if you're floating at any point inside the sphere and pointing two flashlights in exactly opposite directions, then both flashlight beams will hit the sphere's surface at the same angle of incidence. This can be done with a little bit of geometry by building an isoceles triangle from the light beams (one line segment taken together) and the radii of the sphere to each of the illuminated points. Like I said, good point you raised. When I made the video, I don't think I even considered that, and after I read your comment it took me awhile to come up with the reasoning above. Thanks!
it looks like the grave would be stronger standing in one side,, beaus your flashlight is you shinning in all directions, so one side is stronger and the other is weaker being farther away. people sed the earth cant be round beaus we would fall off the other side,, i think you would be attracted to a 800 mile earth crust in side or out.
Your flashlight example helped me sort out all the different forces acting on the body inside the shell, and that makes it really intuitive to understand why the total force adds up to nothing. Thanks!
Talks about uniform thicknes while drawing the most crooked circle ever.
Wow. Very clear, and very intuitive explanation. Many mathematically "rigorous" explanations are simply more complicated than they need to be. Thanks!
This is an excellent explanation!
interesting, i was taught to integrate the dF_G that you find using the standard formula to find the total gravitational force acting on the little mass m, factoring in the thickness of the shell by substituting dM, a small piece of the large mass M, with the volume of a ring of the shell, using R, dR, an angle measure, and the circumference. :) this was great too
This is still awesome 12y later! Ty
This is intriguing. So if I am walking on the outer surface of a hollow planet, and encounter a hole and jump in it, would that mean I would be just floating in gravity?
Also if there were two spheres of same mass/material, but one is hollow (which implies bigger radius), then gravity at the surface of the hollow one would be smaller, only because the distance to the center is farther out? Am I correct?
To answer your first question, yes.
This video is amazing!!!
@ma00101
That would only be the case if d was the hypotenuse of the right angle triangle. It isnt, so tan (theta) = r/d, therefore r = dtan(theta).
im sorry, the video is very good, but there is a flaw in your proof. im sure you're aware of the error but let me just make things clear. when you do the aproximation to the area of a flat circle as the area illuminated by the flash light, youre making an error, because spheres are not flat and are kinda more complex, so the calculus you're using is not correct. it gets to the right result by only mere coincidence. the correct calculus is a bit more complex.
You raise a very good point.
For an "illuminated circle" of infinitesimal radius, the area of the sphere's surface is the area of a flat circle.
Now, I did gloss over the tricky part, which is that since the flashlight beam is not (in general) hitting the sphere perpendicular to its surface, the illuminated area will be an ellipse. The area of this ellipse will be greater than or equal to the area of a circle at that distance--my formula in light green should actually be:
A_Light = pi * d^2 * tan^2(theta) / cos (alpha)
where alpha is the angle of incidence of the light beam (for perpendicular incidence, alpha = 0 and you're back at the circular case).
In order to make the proof work, you need to then prove that if you're floating at any point inside the sphere and pointing two flashlights in exactly opposite directions, then both flashlight beams will hit the sphere's surface at the same angle of incidence. This can be done with a little bit of geometry by building an isoceles triangle from the light beams (one line segment taken together) and the radii of the sphere to each of the illuminated points.
Like I said, good point you raised. When I made the video, I don't think I even considered that, and after I read your comment it took me awhile to come up with the reasoning above. Thanks!
DrSmithPhysics
No thank you for taking the time to answer my question. This video was very helpfull for me. Please keep up with the good work!
Very good. Thank you.
at 7:50 why isn't it Flight = G*my*mlight/d^2?
Indeed he forgot the G (and he should put it in), but the endresult is the same.
Very interesting!
Don't give up the cleaning job
Thanks a lot
thank you very helpful!
👏👏👏
thank u
it looks like the grave would be stronger standing in one side,, beaus your flashlight is you shinning in all directions, so one side is stronger and the other is weaker being farther away. people sed the earth cant be round beaus we would fall off the other side,, i think you would be attracted to a 800 mile earth crust in side or out.
Neatness is HIGHLY overrated.
Actually. .this is just one formula that proves humans live inside earth...
It's part of Wilson's law of hollow earth, what goes down , goes out.
you are not giving the correct information about the solid sphere.
Um uhhhh so um uhhhh so so um uhhh so uhhh uhhh so so so uhhh um um
It's not live, so why be satisfied with a version where you're constantly going er...um....er..., GET ON WITH IT!
u r confushed
This is a terrible video!