If you dug a 2000 mi hole in the earth what would the gravitational

>>8695464

These instructions are using generalities:

For the tower 2000 miles tall, consider the Earth as a point. That point will be located at the Earth's center. The displacement from the top of the tower to the point will be 2000 miles + the radius of the Earth (~4000 miles) = 6000 miles. Next, find the mass of the Earth. Use the Universal Gravity equation to find the force of gravity acting on whatever body is placed at the top of the tower.

In considering the gravity's pull inside of a 2000 mile hole, set up an equation that condenses Earth's mass to a single point at the center of the Earth (call it G[1]) then subtract out the gravity that is above you using the mass fraction of the Earth in a conic shape above you(call it G[2]). The force of gravity in your hole will be G[1] - G[2].

>>8695560
Dumb this down for me

>>8695464
>Is the mass above you pulling you upward negating some of that?
Nope. The mass on the other side cancels it out.

>The same as a planet with a radius of the distance from yours in the hole to the center of the earth?
Yes. Specifically the gravity due to all the mass in the sphere below you, concentrated at the center. Gauss' law in terms of gravity, basically.

To get the gravity on the tower just take into account its position from the center of the earth while using Newton's law of gravitation.

>>8695813
Sure thing! Although, I don't know how much physics and mathematics you've had....

First: to simply calculations, you can reduce any object to single point that has the same mass. It will be located at that objects center of gravity. For the Earth, it is assumed to be the exact center of the Earth; for a person, somewhere near the navel.

Second: Newton's Law of Gravitation says that every particle in the universe is exerting gravity on every other particle in the universe. (This one is quite mind-boggling to think that the most distant piece of matter is pulling you, here on Earth, towards it light-eons away and you are pulling at it with the same exact force.) The equation is F = Gm[1]m[2]/(r^2), where F is the force of gravity, G is the gravitational constant = 6.674x10^-11 N (m/kg)^2, m[1] is the mass of one of the objects (the Earth), m[2] is the mass of the second object (assumed to be you), and r is the distance between the two objects.

My first equation setup says to consider the entire mass of the Earth as a single point at the center of the Earth. The distance from the top of your 2000 mile tower to the center of the Earth is 6000 miles. You now know that r = 6000 miles (convert it to metric then convert to meteres because the gravitational constant is given in metric.). Multiply the gravitational constant by the mass of the Earth in kilograms, and multiply it again by the mass of you in kilograms, then divide it by the square of the distance. Punch all those into the calculator and you'll find the Force being exerted by your mass on the Earth and by the Earth on your mass.

The second scenario considers the force of gravity if you're in the 2000 mile hole. Every particle that makes up the Earth is pulling at you. Some of the force that is pulling you towards the center of the Earth is being counteracted by the gravity of the particles above you. So if you were to step on a scale, you'd weigh less.

>>8695464
>Into the earth
Force decreases linearly as you move towards the center of the earth precisely for the reason you state (there's mass above you).

>Above the earth
Goes with 1/r^2

>>8696007
It's not linear. Linear calculations are for pussies who don't know calculus and assumes that the pull of the material overhead is equal and opposite of the pull of corresponding material on the other side of the sphere. NEWS FLASH: the material overhead is closer to you and has stronger pull than it's counterparts from the other side.

>>8696018
>It's not linear.

Shell theorem mah nigga. But you'll be right, it won't be precisely linear, but it'll be a decent approximation.

>>8696018
>the material overhead is closer to you and has stronger pull than it's counterparts from the other side
yeah but there is more mass on the other side of the planet than the conic above you so the two factors cancel

>>8696018
It's not linear but at the scale we're talking about it is effectively linear

>>8696021
>Shell theorem
Shell theorem is a linear approximation of the dynamic change as you go deeper into a sphere.
>Decent approximation
Sure, if you're not interested in the dynamics.

>>8696037
Consider if you're between two equally massive particles and that you're closer to one than the other, the closer is going to exert more force on you than the more distant. They do not cancel out. Yes, there is still more mass below you in the hole than above...that is why you're still being pulled down. However, "the two factors cancel" is false because the mass above you is closer than the corresponding mass on the other end of the sphere. Therefore, it has a stronger pull on you and you would weigh less than the simple linear approximation from the shell theorem.

>>8696072
>Sure, if you're not interested in the dynamics.

You'd still be able to get the dynamics of a particle moving down the tunnel though. They just wouldn't be as accurate as a more realistic model (say assuming a linear change in density over a constant density).

>>8696072
Nah they cancel exactly. If you're inside a hollow sphere, there is no gravity at all, even if you're close to the edge. A solid sphere can be decomposed into a bunch of nested hollow spheres. Thus any mass in the hollow sphere above your radial distance from the earth's center exerts no force on you.

>>8696072
https://en.wikipedia.org/wiki/Shell_theorem#Inside_a_shell

>>8695464
>gravitational pull be at the bottom?
about 10 m/s^2, slightly more than at the surface.

>>8695464
>2000mi tower on the earth?
Acceleration at the top would be between 4.33 m/s^2 (pole) and 4.28 m/s^2 (equator).