[Q] A question about gravity.

[Glenn Lyvers]

(Correct me if I'm wrong please.) I understand that gravity is one of the four identified forces (electromagnetism, gravity, strong and weak), and that the effect of gravity is relative to the mass of an object. Unlike density, which can be measured by mass per volume, the gravity itself is a pure number, and it's effect is measured as being numerically equivalent to the acceleration of objects being influenced by its' effect. In our planet, it is relative to the diameter, and therefore, if you were to dig a hole half way between the crust and the core, and stop, the gravity at that point would be much different (lower), because the diameter of the earth at that location would be much less.

 

I know it might sound intuitive, and maybe I just over thought it, but I have a very simple question. On Earth, if you stand on a mountain top, is the gravitational effect at that location different than if you were standing at the foot of the mountain?:)

[Boof-head]

Actually if you dug a hole halfway to the center, all the mass around you would pull on your mass.

 

Mass and gravity are like a variable and a function; with Newton's G, you have either G, or G(m).

There is no "m" unless G(m), IOW.

 

This is fundamentally different in the quantum domain where mass is a function, and G disappears, so that m,Q -> Q(m) and gravity is missing since it is so weak the other forces are dominant; you are in the realm of "frequency" - of mass, charge and spin; they all "vibrate" against each other: result 'one' particle, mass m.

[Glenn Lyvers]

Yes I'm aware of Newton's G, and you are obviously correct that the pulling force halfway to the core would be, not only down, but also surrounding you. Neat to think about! But would the standard average (value of gravitational effect at the earths surface) equation g = 9.8 m/s2 = 32.2 ft/s2 demonstrate a difference in effect under the conditions I suggest?

 

Because standing on the top of the mountain does not change the diameter of the earth, does the distance from the top of the mountain, and the foot of the mountain change the gravitational effect? It seems like it would but I'm not clear in this point because it's not my specialty.

 

Thanks for your reply.

[Boof-head]

Standing on a mountain changes the gravitational 'diameter' or field around you. Gravity depends on distance between 2 centers of mass G(m,m'); the second term is "you", so if you know your own mass (because you determined it previously) you can calculate G(m) - G(m').

 

Newton's formula; the field ("gravitational field") is different below the surface because of the effect described from all surrounding G(m) mass:= mass 'in' G.

[freeztar]

Yes, the difference in elevation causes time to speed up or slow down and causes mass to likewise be distorted.

 

For evidence of this, please see time dilation and general relativity.

[Pyrotex]

Hey Glenn,

that's a good question.

Yes, if you were at the top of, say, Mauna Kea in Hawaii, you should be able to measure that the gravity there is different than at the bottom of the mountain. Measurable, but a very small difference.

 

Gravity at the surface of the Earth, and above, varies as the factor 1/r squared.

Let's keep this all simple by using approximations.

If you were to climb a mountain that was 400 miles high, you would have increased your distance from the center of the Earth by 10%. (The Earth's radius is ~4,000 miles.)

So, your gravity went from 1/® squared to 1/(1.1*r) squared.

Or, from 1/® squared to 1/1.21 *® squared.

If you take 1 and divide it by 1.21, you get about 0.83 or 83%

You will feel only 83% of "surface" gravity at the top of a 400 mile mountain.

 

Now, going down into the Earth is different. One of the most fascinating uses of integral calculus is proving that if you are inside a spherical shell of matter, then the total gravitational pull from all parts of the shell cancel out. The gravity inside a shell is zero. Doesn't matter how thick the shell is, doesn't matter where you are inside the shell. Zero. It's a lovely proof.

 

So, if you go down into a hole in the Earth, say 10 miles down, you are now inside a "shell" of the Earth exactly 10 miles thick. The gravitational pull from all that mass cancels out to zero. You will feel only the gravitational pull from the remainder of the Earth: a sphere just 3,990 miles in radius.

 

If you go down 1,000 miles, you will feel the pull only from that part of the Earth that is a sphere just 3,000 miles in radius. As you get nearer and nearer to the center of the Earth, all gravitational force quickly cancels out to zero.

 

Does that help?

[Pyrotex]

...Mass and gravity are like a variable and a function; with Newton's G, you have either G, or G(m). There is no "m" unless G(m), IOW.....

Boof, I've been studying physics for over 40 years, and I'm sorry to say, but your post is totally incoherent. It just doesn't relate to anything in a real physics book. Would you like to try again?

[Boof-head]

Define mass: m

Define gravitational force: G

 

How would a physics book describe these things? One is "real mass", one is "a real constant".

 

As I stated (incomprehensibly) IF G, then G(m); let's see someone do this the other way around and get "m" all by itself, and G all by itself. You can do the latter, but you can't (by propositional calculus) do the former - you have to use the constant, so "there is no mass unless G", just like someone said.

 

Any clearer

(p.s. I'm leaning towards conclusions that "science forums have people in them, who are allowed to insult other people's intelligence". I don't feel all that concerned, but someone here as accused me of being incomprehensible, I find this is incomprehensible to me. I have absolutely no need to post, or reply or even be a contributor; I am posting these words to try to get someone to wake up and read it again, I don't "do" physics, I used to. I don't do particle physics, I never have. Someone here informed me that a coordinate system oriented along the x axis is meaningless, and I find this response meaningless since a coordinate system is always oriented (in physics), so, please do not insist that you are all people who have 'studied' physics, I have, and such responses lead me to conclude you haven't done any after all)

[Hasanuddin]

Hi Glenn,

 

Its funny how a simple question can uncover a can of worms. Your original guess was correct, gravity is decreased at the top of a mountain. The fact can be proven by a simple pendulum. Because of the diminished gravity the time it takes the bob to complete one full cycle (period) will increase. Using the well-known period formula the exact value for period can be calculated.

 

An even bigger question (that is tied to your original inquiry) is what is gravity fundamentally and how does gravity relate between fundamental particles, i.e., between matter and antimatter? That question is a truly giant one. Currently that question is being considered at http://hypography.com/forums/alternative-theories/18910-the-dominium-model-by-hasanuddin.html

 

Come over and join the fun. That thread will either answer or generate questions concerning gravity for you

[Glenn Lyvers]

Thanks all. I am happy to report my question was answered and I understand now. Thank you Pyrotex, Hasanuddin, Freezetar and Boof-Head.

 

Very kind of you all to reply. I thank you for your help in my understanding of this issue.

 

Peace,

Glenn