Wiploc

Rather like a man on stilts using the stairs, awkwardly.


You can do math! The cavalry is here!


Okay.


Now this part I understand. Thank you, thank you.


This is intuitively obvious to me, but I was unable, as far as I know, to make the case for those for whom it was not intuitively obvious.

Now let me ask another question. This discussion has gotten me curious about what shape the earth would have to be to maximize a person's weight. A cone segment, right (Those darn segments again!) but of what angle, how wide? No, it would need a rounded bottom, like an upside-down ice cream cone that someone has licked nearly flat --- only wider, more squat, at least ninety degrees. Does that seem right?

And, more topically, if the earth weren't turning, would people weigh more at the poles or on the equatorial bulge?
crc

SULPHUR

Read the post above you and explain why gravity on mt everest could be the the same as at the dead sea

Calling for another geophysicist

Lobstrosity

Maximization problems are fun, but you need to set up proper constraints before you can do meaningful calculations. One constraint, I assume, would be that we're dealing with a fixed total mass that we are going to sculpt? This isn't sufficient to make the problem non-trivial, however, for the shape that would maximize your gravity on the surface would just be an infinitely-dense, infinitely small sphere. The constraint that would make this problem non-trivial would involve allowed mass density. How how about this for a maximization problem:

Let's say the earth has total mass M and total volume V. Also, we assume the Earth has uniform mass density everywhere given by rho = M/V. Now, you are allowed to re-scuplt the Earth any way you like so long as the total volume is still V and density everywhere remains uniform, given by M/V (which means your total mass remains M). If you wish to maximize gravity at a point, what shape should you sculpt and where would it be relative to the point of maximum gravity?

I would actually have to think a bit about this question. It could prove rather difficult as the integrals required can get a bit hairy if you don't have enough symmetry. I intuitively feel like the solution to this problem should have a high degree of symmetry, however, so there's a chance I might be able to do it.

Well, first off, if the Earth weren't turning, there would be no equatorial bulge The bulge destroys the spherical symmetry of the problem and makes Gauss' law unusable. To know the answer for sure, you'd need to integrate over all the mass. It seems like this would be somewhat challenging, so I have turned to google. This web page implies that gravitional acceleration is larger at the poles (I assume that means they're not accounting for centrifugal acceleration in the numbers). I wanted to see why this was, so I looked farther and found this (the info you want is in Appendix A, on pages 8-10), though these equations seem to take as parameters the gravity at the poles and the equator, so I'm not sure how much help this is. Basically, as you can see, it's a non-trivial problem.

SULPHUR

It is common fact that force of attraction is greater at the poles than the equator there the weights vary. The force can vary locally due to massive mineral deposits and so is is used in exploration. Overall the force is directed at the centre of the earth. that is straight down. The equations for F above are valid and I cannot see why they are being discounted. Could you explain wiploc or lobstrsity.

Lobstrosity

Your equation for F is not generally applicable, Sulphur. Your equation for F is only valid for point masses, but it also happens to hold true if you are outside of a perfectly isotropic sphere due to the excessive symmetry of the problem. Furthermore, if you are very far from an object such that it appears as a point mass for all intents and purposes, then you can use the approximation F = G m1 m2 / d². When dealing with gravity on the surface of an oblate Earth, you do not have spherical symmetry (well, the Earth is still exceedingly spherical, but you cannot treat it as such if you're looking for perturbations due to its oblateness) and you are not sufficiently far away that you can treat it as a point mass. Thus you cannot use your equation. For example, if were dealing with an extremely oblate Earth that was approaching infinite girth and zero height, use of F = Gmm/d² would give you a weight that was approaching infinity at the poles (since the numerator is remaining finite and non-zero while d is approaching zero). This is not the correct limit for such a scenario (the correct limit would be for the force to approach a constant).

There are two ways to calculate forces explicity for objects of spatial extent:


Eq (1) is Gauss's law. It is always true but only useful if the problem exhibits extraordinary symmetry (such as planar, spherical, or cylindrical). Eq 2 is the general formulation of Newton's law of universal gravitation as applied to bodies with spatial extention. We make the assumption that gravity is a linear field and then add up the contributions the infinitesimal mass located at each point within our object. You can also integrate the gravitational potential over all space and then compute the force as the negative gradient of the gravitational potential (at least that way you're integrating scalars and not vectors).

Does this make sense?

As an example to illustrate, a question for you: Let's say we have a spherical planet of uniform density, total mass M, and radius R. Now this particular planet happens to have a spherical void where its core should be (i.e. there is zero mass between r = 0 and R/2). What is the gravity at points within this cavity (i.e. g(r) = ? for 0 < r < R/2)? How would you answer this with your technique?

SULPHUR

You must forget about the surface and concentrate one mass at the centre of the earth. the other mass is at tne pole is x.At the equater the distance is y If y is a greater distance then force is less. It has been proved experimently.
Why do you think gravity measement is used in geological exploration WE are looking for gravitational abnormalities.They are small but they are there even in mN

Lobstrosity

I'm not saying that you're wrong about gravity's being greater at the pole. I'm just saying that your understanding of the real theory behind this result is lacking. You're reducing the problem to simplistic approximations that happen to work adequately only because the Earth only just slightly oblate. And you still didn't answer my question about what gravity would be inside a spherical cavity centered within a spherical planet. Do you think you could solve such a problem with g = GM/d²?

Basically you asked what Wiploc and I were talking about and I told you. You might not have to concern yourself with integrals because the Earth is spherical to a reasonable approximation, but in general, for problems of computing gravity due to arbitrary mass distributions, F = GMm/d² just doesn't cut it--you need to integrate.

SULPHUR

I am glad that I am right with my simple solution. I am used to dealing with practical problems. I admit I didn't see you problem.Heres one from my old uni days.How many degrees of freedom in a four phase diagram. I can stil remember the answer If you want to check.

SULPHUR

A good start is at,

http://www.earth.man.ac.uk/CAL/products/phase.html

Yggdrasill

I've been busy during the weekend, so I haven't had a oppurtunity to post. So, to tie up some loose ends:

I used "paint", and to attach it, you just save your illustration in one of these formats: gif jpg png txt bmp or jpeg, and then choose the file with the "browse" function. I was merely pointing out one reason why I can't agree with the situation posted by wiploc. (Where digging down is the same as taking mass from the opposite side and placing it on top of yourself.) And I can honestly say that I have never seen Gauss' law before in my entire life. Ok then, perhaps I overextended myself a bit when I tried to look at wiploc's assertions. Perhaps I should have mentioned that I wasn't quite sure, and that I was posting a hypothesis that I formed during a physics class, while I was annoyed by the way the physics I am taught use point masses. (As you know, if you use this system you would weigh the most in the center.)

The one thing I dislike the most about physics is it's use of approximations, because it really goes a long way to prove that all known physics are just primitive mental constructs, and that we don't know anything. We don't even know the true value of pi.