Center of the Earth

Doc Al

I think I'm missing your point with your off-center sphere, so I can't comment. Is it a hollow sphere? An imaginary sphere? Beats me. Please note that all of these arguments about zero gravity inside a hollow cavity within the earth depend on a spherically symmetric distribution of mass.

I'll repeat: Assume a spherically symmetric distribution of mass, of radius R. Now imagine an imaginary sphere of radius r (r < R) co-centered with the big sphere of radius R. The mass outside of the small sphere will exert no gravitational field within the small sphere. So any gravity experienced at any point within the small sphere is entirely due to the mass within that small sphere. (We are of course ignoring any other bodies in the universe.)

Cyrus

Hey Doc,

Did you click on the picture i attached that displays what I mean. For simplicitys sake lets just think of them as circles in a common plane. At that every point inside the big circle has a point mass that is equal.

According to what you said, my red dot, is a point mass that is not located in the center of the big circle. The black dot IS the center of the big circle. According to you, the gravitational force will equal to the sum of all the point masses inside the circle that I can draw around the red dot, with a center of the black dot. (this corresponds to the blue circle in the picture.) I hope this part is clear.

Now, what I am saying is that, what if i IGNORE the black dot, and consider my RED dot as the center of some arbitrary circle. Then the maxiumim circle I can draw with the center at the RED dot without going outside of my big circle will equal the radius of the big circle, minus the distance from the red dot to the black dot.

Now If I draw this circle, (based on my picture), it has a radius of 22miles. And what I am saying is that at the red dot inside this circle of 22 miles, there should be no gravitaional forces present because for every point inside the circle of 22 miles, there is the same point opposite it to cancel out any net force due to gravity.

So clearly, it can be seen that the only things that cant cancel out, will be the points that lie within the green area. And what I am wondering is that if I sum the distances squared times the GMm for each particle of the green area, and sum it for all of the points in the green space, should this not produce a gravitational force that is equal to the way we do it your way, by including the effects of all the points in the blue circle and adding them up.

It says 0 views on my picture so I think you did not see it in my previous post. Please click on it as it will make things alot clearer than my confusing text ( sorry about that )

Doc Al

I get you now.

Yes, I'd say you are correct, assuming that the large sphere has uniform density. If you want to find the gravitational field at point x (red dot) then it is equal to the field from a sphere of mass of radius 3 miles (in your example). But you can certainly draw a sphere around the red dot and say that all mass inside the sphere will contribute no field at the red dot. So the sum of the field from the remaining mass in the big sphere must add up to the same. Very clever!

Cyrus

YAY, I was about to try and derive some form of a proof of this, but I fast realized that it would be a MAJOR pain in the butt to do LOL! Ill just take your expert word for it and save myself the trouble Another side note, I am just reading on coulombs law right now. And wow is it EXactly the same as newtons law of gravity, well except for the sign convention. How is it that these two separate things are so AMAZINGLY simliar? I assume newton came up with his formula first? Becuase coulomb had the added advantage of charging two spheres and seeing how the affect of the force is 1/r^2. but how on earth could newton determine its a 1/r^2 force? The forces of gravity are either so minute you cant detect them, or there so large you cant use them experimentally. Newton couldent see the effects of two plannets by moving them close or far appart :-P. I just dont see where newton got the idea for this formula through practical logic, I do see how coulmb could though. He had an experiment he could play around with.

nrqed

It's actually quite simple (although quite hard to explain without a picture, so I'll just mention the basic idea....if you want more details I'll try to be more specific but I don't have a scanner to show you a hand drawing unfortunately). The basic idea is that Newton understood that an object in orbit around the Earth (or the Sun or whatever) is actually constantly falling. The Moon, for example, is in constant free fall. Knowing the distance to the Moon (The Greeks already knew that!) and that the sidereal month (I won't get into this for now) is about 27.3 days, it's easy to calculate that in one second, the Moon falls about 1.35 millimeter! (what I mean by it falls 1.35 mm is this: Imagine that the force of gravity would cease to exist. The Moon would then keep going along a straight line. But instead it follows a circle (approximately). Now follow the Moon for one second and compare the position it would have if it was flying off along a straight line and the position it actually has. It's easy to use Pythagora's theorem to find the 1.35 mm (it's crucial that it be a small time interval hence a small distance. With large values, say one day, the geometry would be more complicated).


What does this have to do with the 1/r^2 law? well, on the surface of the Earth, an object released from rest (an apple let's say!) falls 4.90 m (1/2 a t^2).

So the apple has an acceleration (acceleration is proportional to distance when released from rest) which is about 4.9/0.00135 = 3630. or about 3600 if I round off a bit. On the other hand, the Moon is 384 000/63 80 = 60 times farther from the center of the Earth than the apple.

But 3600 is 60^2! So being 60 times farther reduces the acceleration (so the force) by a factor of 60^2. Therefore, it's an inverse square law.


Actually, it's also easy to get the same result by using Kepler's third law and assuming circular orbits (a very good approximation for most planets). One simply has to use that the centripetal acceleration is v^2/r and one finds an inverse square law again.

Regards

Pat

Cyrus

Very Interesting! Thanks Pat. I find it more meaninful to learn ABOUT formulas than to learn formulas. Any trained monkey can use a formula. I find the beauity in how did someone manage to think up such a correct idea mathematically with such hard things to deal with. Those greeks were pretty smart. I think carl sagan was right in saying that if they had had more time, they would have made up process scientifically 1000 years. Its too bad that never happened, maybe warp 9 --> engage, would not be fantasty today :-).

Gara

I think what Doc Al is saying is the gravity below you would not "only" be pulling on your legs, while the gravity above you would not "only" be pulling your head.

Cyrus

Pressure is defined as force per unit area, but as we approach the infitesmal point mass with point area at the center of the earth we would have pressure =force/zero,

but i guess the force would be infinite and the area would be zero, and we would have an indeterminate, of the form infinity over zero, so in oder to find the pressure we would have to use l'hospital ( thats EL hospital's :-) )rule, and that anwser would give us a definite anwser as we did the limit and what not, because obviously there is finite mass and so there will have to be finite pressure.

chroot

Cyrus:

You're making it way too complicated. At the center of the earth, there is no gravitational force. There is, however, enormous pressure.

- Warren

marlon

OOOhhhh, wow,
this is real science...
cyrusabdollahi,
listen to the answer of chroot, though, he is right...

Remember that physics is constructed in order to make the world more easy, understand it...

regards
marlon

Cyrus

Sorry for asking , I will not post in this topic anymore what do you mean by real science? I think the question I posed in terms of an indeterminate is a valid, provided I were to have a function of the area and the force and knew the limits of my integral.

stevo72

Sorry but this is not entirely true.

Gravity is "perceived" to be generated by the attraction of one mass against another. A body at the centre of the Earth would be attarcted by, and to, every particle that makes up the Earth around it. Thus gravity is in effect in every direction. However the NET gravitational effect would be ZERO.

Now pressure is created by the acceleration of a body against another body. The NET acceleration created by gravity at the centre of the Earth is zero, for the reason given above. So if only gravity was the cause of pressure, the pressure would be zero. However, their is both heat and motion in the Earth's core caused by friction, chemical reaction and gravity in the layers of the Earth beyond the centre. These energy sources generate acceleration of Earth particles which would apply pressure unevenly to any object at the centre of the Earth. As a result the contents of the Earth core are in constant motion. If these other sources of energy where not in place then the natural shape a planet such as the Earth would take is a hollow shell since a gravity inflexion point would be reached at a certain distance below the surface where there would not be sufficient gravity to hold the particles in place.

lalbatros

Assuming a liquid planet, the pressure at the center is equal to the gravitational potential at the surface of the planet.

D H

Two back-to-back nonsense posts to a five year old thread!

The first of the two is utter nonsense. You are ignoring the pressure from above, stevo72. The second doesn't make sense at all. Energy and pressure are different things with different units.

lalbatros

Pressure is a density of energy.
Indeed: Pa/m² = Pa.m/m³=J/m³.
I was simply suggesting to use the Bernouilli equation for this question.

Cyrus

I already got an answer to this question, years ago. I don't know why you're using Bernoulli for my question, as I didn't ask anything with regards to Bernoulli...

lalbatros

I was answering to stevo72, if you permit.