The earth's equatorial bulge

[newts]

I think I made myself a bit unpopular on the speculations forum, by criticising physicists' religious beliefs, and making fun of their imaginary beings (dark matter, quarks, gluons, and Higgs).

 

So to show that I am truly penitent, I have been trying to act like a proper physicist, by doing some pointless calculations about the size of the earth's equatorial bulge. The article is too long to post here, but can be seen at http://squishtheory....uatorial-bulge/.

 

The highlights are, that by assuming the mass of the earth to be uniformly distributed, I actually got the same value as Newton calculated in the Principia, though by a very different method. To get the measured value of 21.36 km, I had to assume that the core of the earth is actually much denser than the crust.

Edited November 27, 2011 by newts

[mathematic]

I think I made myself a bit unpopular on the speculations forum, by criticising physicists' religious beliefs, and making fun of their imaginary beings (dark matter, quarks, gluons, and Higgs).

 

So to show that I am truly penitent, I have been trying to act like a proper physicist, by doing some pointless calculations about the size of the earth's equatorial bulge. The article is too long to post here, but can be seen at http://squishtheory....uatorial-bulge/.

 

The highlights are, that by assuming the mass of the earth to be uniformly distributed, I actually got the same value as Newton calculated in the Principia, though by a very different method. To get the measured value of 21.36 km, I had to assume that the core of the earth is actually much denser than the crust.

Your assumption about density is valid in fact.

[newts]

Your assumption about density is valid in fact.

I know that, there is an graph of the earth's density here http://en.wikipedia.org/wiki/File:RadialDensityPREM.jpg

 

On other forums there are discussions about how to calculate the size of the bulge, and nobody seems to have a proper answer. I thought since I had already been posting on speculations, I would experiment to see if anybody here was interested in the bulge. The answer is apparently no, since nobody has clicked the link; maybe I should have been a bit more tactful in the way I started the thread.

[Enthalpy]

It would be interesting if you could determine through its bulge if Jupiter has a big dense kernel.

Its bulge is huge and well documented. No liquid nor solid surface has been observed but the upper atmosphere's properties are rather well known.

 

As you developed at the linked site, mass distributed at the outer shells (upper atmosphere there) creates a higher bulge than mass concentrated in a dense kernel. Jupiter has a low average density, so a rock or metal kernel would be small as compared with the atmosphere's radius.

 

What is not known (or wasn't):

- If such a kernel exists (astronomers bet on "yes" presently), if it's metallic hydrogen, rock, metal...

- What the atmospheric density and temperature profile is (the big red spot is colder)

- What kind of gas composes the deeper atmosphere (molar mass depends on altitude on Earth)

- Where the very strong magnetic field (1T) originates. People desire a metal kernel for that.

- Where the excess heat originates. Jupiter radiates seriously more infrared than it gets Sunlight.

 

Well, I can't tell you much more... Maybe someone has already investigated that. I'm a dilettante for astronomy as well, and astronomy improves quickly.

Good luck!

Edited December 6, 2011 by Enthalpy

[michel123456]

(...)So to show that I am truly penitent, (...)

 

You ment "pertinent", surely.

[StringJunky]

You ment "pertinent", surely.

 

pen·i·tent/ˈpenitnt/

Adjective:

Feeling or showing sorrow and regret for having done wrong; repentant.

Noun:

A person who repents their sins or wrongdoings and (in the Christian Church) seeks forgiveness from God.

Synonyms:

adjective. repentant - contrite - remorseful - regretful

noun. repentant sinner

 

per·ti·nent/ˈpərtn-ənt/

Adjective:

Relevant or applicable to a particular matter; apposite.

Synonyms:

relevant - apposite - appropriate - suitable - proper

 

He used the the correct word for what he wanted to convey.

[michel123456]

pen·i·tent/ˈpenitnt/

Adjective:

Feeling or showing sorrow and regret for having done wrong; repentant.

Noun:

A person who repents their sins or wrongdoings and (in the Christian Church) seeks forgiveness from God.

Synonyms:

adjective. repentant - contrite - remorseful - regretful

noun. repentant sinner

 

per·ti·nent/ˈpərtn-ənt/

Adjective:

Relevant or applicable to a particular matter; apposite.

Synonyms:

relevant - apposite - appropriate - suitable - proper

 

He used the the correct word for what he wanted to convey.

 

I ment that he is not penitent, he is pertinent. Penitence is his excuse. A truly penitent would have say sorry I was wrong I will never do that again. Newts choosed to show he knows something in some random subject to attract attention on his favorite subject where he thinks he is right.

[StringJunky]

I ment that he is not penitent, he is pertinent. Penitence is his excuse. A truly penitent would have say sorry I was wrong I will never do that again. Newts choosed to show he knows something in some random subject to attract attention on his favorite subject where he thinks he is right.

 

Ok Michel. My apologies. I know English is not your native language and thought you didn't know the difference.

[Iggy]

I think I made myself a bit unpopular on the speculations forum, by criticising physicists' religious beliefs, and making fun of their imaginary beings (dark matter, quarks, gluons, and Higgs).

 

So to show that I am truly penitent, I have been trying to act like a proper physicist, by doing some pointless calculations about the size of the earth's equatorial bulge. The article is too long to post here, but can be seen at http://squishtheory....uatorial-bulge/.

 

The highlights are, that by assuming the mass of the earth to be uniformly distributed, I actually got the same value as Newton calculated in the Principia, though by a very different method. To get the measured value of 21.36 km, I had to assume that the core of the earth is actually much denser than the crust.

Could you summarize your approach so that one isn't forced to wade through all the finer points to determine where a potential error might lie?

 

It's common practice to start with an abstract, then an introduction which is usually filled with statements like "section 2 will calculate "______" using "_____" and show why that implies "______" before getting into all the gory details.

 

If your link had more of that type of format it would be much more approachable.

Edited December 7, 2011 by Iggy

[michel123456]

Ok Michel. My apologies. I know English is not your native language and thought you didn't know the difference.

No problem, I appreciate your concern.

 

I ment that he is not penitent, he is pertinent. Penitence is his excuse. A truly penitent would have say sorry I was wrong I will never do that again. Newts choosed to show he knows something in some random subject to attract attention on his favorite subject where he thinks he is right.

which is not a bad thing.

Re-reading my own post I found it could be misunderstood.

[newts]

I ment that he is not penitent, he is pertinent. Penitence is his excuse. A truly penitent would have say sorry I was wrong I will never do that again.

My claim to be penitent was meant ironically, certainly I am impenitent and impertinent too.

 

Newts choosed to show he knows something in some random subject to attract attention on his favorite subject where he thinks he is right.

Your assessment is very near the mark, except that I am not displaying knowledge, rather I worked out a solution myself from very basic physics principles.

 

In my book I wrote a chapter pointing out that Newton's achievements far outstripped Galileo's; one example being that Galileo thought, in so far as he was capable of thinking, that the rotation of the earth caused the tides; whilst Newton correctly figured out both problems. Then I started wondering why Newton predicted a bulge of 17 miles instead of the correct value of 13 miles or 21 km. I then did the calculation myself, and came up with 11 km based on centrifugal forces alone, so I searched the web to find my error, but mostly came up with people suffering from the same problem.

 

Eventually I got to Lubos Motl's site where he pointed out that the gravity of the bulge increases its size; but he used a computer program to calculate, and reached no positive conclusions about the mass distribution of the earth from his calculations. My calculation based on a uniform distribution of matter, does actually agree with an expression that has subsequently appeared on Wikipedia, so is almost certainly correct. Wikipedia gives a link to a site which derives it much more succinctly than I do, but it probably only makes sense to people with a high degree of mathematical knowledge.

 

My calculation of how the mass would have to be distributed in order to give the correct value, is probably not completely right, but nobody else on the web has tried to calculate it at all, so at least I made an attempt. On Speculations I was recently accused of being incapable of passing physics 101, and of being no match for a physics PhD, so drawing attention to the fact that I solved a problem that other physicists could not, is part of the propaganda war.

[newts]

Could you summarize your approach so that one isn't forced to wade through all the finer points to determine where a potential error might lie?

Thanks for advice, I have put in an index with page jumps. Errors probably lie in the section titled Calculation of the core bulge.

You are welcome to look for them, but you will probably need to read some of the rest to understand my method.

 

It would be interesting if you could determine through its bulge if Jupiter has a big dense kernel.

Its bulge is huge and well documented. No liquid nor solid surface has been observed but the upper atmosphere's properties are rather well known.

I had been thinking that I should have given Jupiter a mention. Here is what I added to my blog. Hopefully it answers some of your questions:

 

Jupiter's equatorial bulge

 

Since Jupiter has irregular features on its surface, the fact that is spins, was presumably noticed soon after the invention of the telescope in 1608. With an equatorial diameter greater by about 1/15 than the polar diameter, its elliptical shape should also have been soon apparent; certainly in the Principia Newton refers to measurements of this. It was of course Newton who had the imagination to correlate the two things, and realise that the elliptical profile resulted from the planet's rotation.

 

The size of an equatorial bulge depends on the inverse of a planet's density, and the square of its rotation speed. The average density of the earth is 4.15 times as great as that of Jupiter. Also Jupiter rotates once in 9.9 hours. (24/9.9)² = 5.87, multiply that by 4.15 = 24.4. Therefore Jupiter should have a bulge 24.4 times that of the earth.

 

The earth's bulge based on centrifugal force alone is 1/578. So if all the mass of Jupiter were situated at the centre, and the surface had negligible mass, then Jupiter's bulge would be 24.4/578 = 1/24 of its radius.

 

The earth's bulge if its mass was uniformly distributed would be 1/230; so if Jupiter's mass was uniformly distributed then its bulge would be 24.4/230 = 1/9.4. The earth's actual bulge is close to 1/300, so if Jupiter had a similar mass distribution to the earth, it would have a bulge of 24.4/300 = 1/12.3.

 

Since Jupiter's actual bulge is 1/15, we must conclude that on average much more of Jupiter's mass is situated close to the centre than is the case for the earth. Using the equation: bulge due to centrifugal forces alone + surface density (3/5 bulge) = bulge, we get 1/24 + surface density (3/5 times 1/15) = 1/15, which solves to give a surface density of 45/72 that of the average density. Since Jupiter's average density is 1.33, the surface density would be about 0.83 that of water; assuming that all the upper crust of Jupiter is of uniform density, and that the remaining dense core of the planet is situated very close to the centre, things which are probably not even approximately true.

[michel123456]

Thanks for advice, I have put in an index with page jumps. Errors probably lie in the section titled Calculation of the core bulge.

You are welcome to look for them, but you will probably need to read some of the rest to understand my method.

 

 

I had been thinking that I should have given Jupiter a mention. Here is what I added to my blog. Hopefully it answers some of your questions:

 

Jupiter's equatorial bulge

 

Since Jupiter has irregular features on its surface, the fact that is spins, was presumably noticed soon after the invention of the telescope in 1608. With an equatorial diameter greater by about 1/15 than the polar diameter, its elliptical shape should also have been soon apparent; certainly in the Principia Newton refers to measurements of this. It was of course Newton who had the imagination to correlate the two things, and realise that the elliptical profile resulted from the planet's rotation.

 

The size of an equatorial bulge depends on the inverse of a planet's density, and the square of its rotation speed. The average density of the earth is 4.15 times as great as that of Jupiter. Also Jupiter rotates once in 9.9 hours. (24/9.9)² = 5.87, multiply that by 4.15 = 24.4. Therefore Jupiter should have a bulge 24.4 times that of the earth.

 

The earth's bulge based on centrifugal force alone is 1/578. So if all the mass of Jupiter were situated at the centre, and the surface had negligible mass, then Jupiter's bulge would be 24.4/578 = 1/24 of its radius.

 

The earth's bulge if its mass was uniformly distributed would be 1/230; so if Jupiter's mass was uniformly distributed then its bulge would be 24.4/230 = 1/9.4. The earth's actual bulge is close to 1/300, so if Jupiter had a similar mass distribution to the earth, it would have a bulge of 24.4/300 = 1/12.3.

 

Since Jupiter's actual bulge is 1/15, we must conclude that on average much more of Jupiter's mass is situated close to the centre than is the case for the earth. Using the equation: bulge due to centrifugal forces alone + surface density (3/5 bulge) = bulge, we get 1/24 + surface density (3/5 times 1/15) = 1/15, which solves to give a surface density of 45/72 that of the average density. Since Jupiter's average density is 1.33, the surface density would be about 0.83 that of water; assuming that all the upper crust of Jupiter is of uniform density, and that the remaining dense core of the planet is situated very close to the centre, things which are probably not even approximately true.

You should put all that into a graph.

[Iggy]

Thanks for advice, I have put in an index with page jumps. Errors probably lie in the section titled Calculation of the core bulge.

Good deal. I was looking more for a summary of your method, but I suppose we can start at the beginning and take things piecemeal. I have a question/problem with the first thing:

 

A simple way to calculate the size of the bulge, is to use the equation v² /r to work out the effective reduction in the mass of the water in the equatorial tunnel, resulting from the centrifugal force. The speed of a particle on the equator is: 2p ´ 6,378 km (the earth’s radius) per day (actually 23hrs 56 minutes). In metres per second it works out as 465 m/s.

 

If we square that and divide it by the earth’s radius we get a centrifugal force of .0339, which is 1/289 of the force of gravity at the earth’s surface.

 

At the centre of the earth, there is no centrifugal force; and the equation v² /r means that a particle a third of the earth’s radius from the centre, would have a centrifugal force of a third, because it is going round at a third of the speed. So the average centrifugal force over the whole tunnel, is half of 1/289, that is 1/578. Therefore for the pressure in the two tunnels to balance, the column of water in the equatorial tunnel must be longer by 1/578 of the earth’s radius. This gives us an equatorial bulge, due to centrifugal forces of about 11.035 km.

 

Both the centrifugal and gravitational force of a rigid body of uniform density changes linearly with r, so I don't think you're making good sense. To put that another way.... you say "at the center of the earth, there is no centrifugal force". That is true enough, but there is also no gravitational force at that spot either. I don't believe the ratio of centrifugal force divided by gravitational force changes with r in the way you say... it should stay constant since they are both linear.

 

For example, at r=3189050m (half of the equatorial radius) the gravitational force [you can calculate with gm/r²] is 4.90 m/s². The centrifugal force is 0.01696 m/s². The ratio there is the same as at the surface: 0.003.

 

At r=637810m (10% of the equatorial radius) you get 0.00339 m/s² for the centrifugal force and 0.98 for the gravitational field. Again the ratio is 0.003.

 

The average of the ratio is not 1/578 because the ratio doesn't change from 0 to 1/289 with r. The ratio stays a constant 1/289 with r (from the center to the surface), and that's how much lighter the equatorial column of water would be (1/289) if the earth were spherical and of constant density.

 

Do you agree, and how would this affect your reasoning?

[newts]

so I don't think you're making good sense.

What I wrote does not make sense, but the value of 11,035 km is correct because I calculate it later using a valid method.

 

Everything you wrote does make sense and is correct, but I do not think it leads directly to a method of calculating the bulge. I need to rethink and rewrite that bit.

[newts]

I don't believe the ratio of centrifugal force divided by gravitational force changes with r in the way you say... it should stay constant since they are both linear.

I have added an extra sentence to try to explain what I am calculating a bit better. You are absolutely right that the ratio does not change; but I am not using ratios, all I am doing is summing up the total centrifugal force, and then balancing that with the extra height of the column.

[Iggy]

What I wrote does not make sense, but the value of 11,035 km is correct because I calculate it later using a valid method.

Ok. I'll be interested to see the valid method, but it'll have to wait till tomorrow.

 

To be clear, the correct answer to that first thing would be 27.9 km, not 11.035 km. With uniform density, following your thought experiment, the equatorial radius is longer by 1/229, not 1/578 as you suspect.

 

Newton correctly calculated this.

 

The equation that compares the ellipticity, [latex]\delta[/latex] (the factor by which the equator is longer) to the ratio between centrifugal and gravitational force, [latex]\phi[/latex] (1/289 -- as we found) is:

 

[latex]\frac {\delta}{1/100} = \frac{\phi}{4/505}[/latex]

 

Newton gave this equation in the Principia book 3 and it was later proved by Clairaut and other mathematicians correct for slowly rotating bodies of uniform density.

 

As you can see, the equatorial column isn't longer in the same amount (i.e. [latex]\delta[/latex] isn't equal to) the ratio of centrifugal and gravitational force ([latex]\phi[/latex]). Nor is it equal to half phi. It is 1.26 times [latex]\phi[/latex].

 

Where [latex]\phi = \frac{1}{289}[/latex], [latex]\delta = \frac{1}{229}[/latex].

 

The equator is longer by 1/229 where the ratio of centrifugal force to gravitational force at the equator is 1/289. Roughly 28 km.

[michel123456]

The equator is longer by 1/229 where the ratio of centrifugal force to gravitational force at the equator is 1/289. Roughly 28 km.

But Newts speaks about radius, not about perimeter.

[Iggy]

But Newts speaks about radius, not about perimeter.

Right. Sorry about the imprecise language. I also meant the equatorial radius. The equatorial column of water in the thought experiment would be 1/229 longer where the ratio of centrifugal force to gravitational force at the equator is 1/289. That was the result Newton got.

 

--------------

 

I'm reading onward in the article. Sorry I'm so slow. I don't have much time.

 

To work out the mass of the bulge, we need to calculate its volume. To do this we actually need to know how high it is, so to make things simple we will start with the measured value of roughly 21 km, which is about 1/300 of the earth’s radius. If we were to add a thickness of 1/300 over the whole surface of the earth, that would increase its volume by about 3/300. However we are adding nothing at the poles; so taking account the elliptical shape of the bulge, we will assume that it would have a mass 2/300 of the rest of the earth.

-equatorial bulge

I believe it would be 1/299 precisely....

 

...yes, because the volume of a sphere is [latex]4/3 \pi R^3[/latex] and a spheroid is [latex]4/3 \pi {R_1}^2 R_2[/latex], if [latex]R_2[/latex] is larger than [latex]R_1[/latex] by 1/300, the volume of the spheroid would be 300/299 times the volume of the sphere. It would increase the volume by 1/299.

 

 

If the particle drops down the polar tunnel, then when it reaches the centre it will be a distance R from every part of the ring. So the energy gained from the ring will be (1- 1/Ö2)R, or about 0.29R, times the mass of the ring, which is 2/300 the mass of the earth.

 

Ok, the gravitational potential should be,

 

[latex]\phi = \frac{GM}{(R^2 + X^2)^{1/2}}=-\frac{GM}{Y}[/latex]

 

where X is the distance from the center to the point under consideration along the polar axis and Y is the distance along the hypotenuse that you've labeled [latex]\sqrt{2}R[/latex].

 

At X=R (i.e. at the north pole) it has,

 

[latex]\phi = -\frac{GM}{\sqrt{2}R} = -\frac{(6.67 \times 10^{-11})(2 \times 10^{22})}{\sqrt{2}(6378100)} = -1.478 \times 10^5 J/kg[/latex] due to the ring.

 

At X=0, Y=R, (at earth's center) it has,

 

[latex]\phi = -\frac{GM}{R} = -\frac{(6.67 \times 10^{-11})(2 \times 10^{22})}{6378100} = -2.0900 \times 10^5 J/kg[/latex] due to the ring.

 

Mass gains 61,367 Joules per kilogram by moving from the north pole to the center due solely to a ring at the equator having a mass 1/299 of the earth.

 

Therefore the energy gained from the trip to the centre, is sufficient to increase the height of the water at the equator by 0.58/300 times the earth’s radius, which is about 11.6 km.

 

The gravitational potential of a sphere is [latex]\phi = -\frac{GM}{R}[/latex], so we have:

 

[latex]-\frac{GM}{R_2} - -\frac{GM}{R_1} = 61367[/latex], solving for R₂ gives,

 

[latex]R_2 = - \frac{R_1GM}{61367 R_1 - GM} = 6384370 \ m[/latex]

 

so that,

 

[latex]R_2 - R_1 = 6384370 - 6378100 = 6270 \ m[/latex]

 

So, the energy gained from dropping down a polar shaft from a ring of mass at the equator alone would be enough to lift the same quantity of mass roughly 6.3 km off the surface of the equator.

 

Why do you suppose we differ here?

 

If we add this to the 11.035 km bulge caused by the centrifugal force, we are already above our required value of 22 km.

 

I'm still not seeing where you got that 11.035. It is clearly half of the centrifugal force at the equator divided by the gravitational force at the equator times the polar radius... but why does half have any meaning?

 

EDIT-->

 

To be precise and consistent, we might establish:

 

R_polar = 6356.7 km

R_equator = 6356.7 + 21 = 6377.7 km

 

Since you used 21 km and earth's actual polar radius is 6356.7 this would make sense.

 

This makes the equatorial radius longer by 1/304. The volume of the spheroid would then be larger by 1/303. In other words, a spheroid with r1 = 6356.7, r2 = 6377.7 would be 304/303 times the volume of a sphere with r = 6356.7. The volume would be 1/303 larger.

 

If we both reasoned with those numbers I think we'd come out alright. If I recalculate with those numbers I get a slightly different 6151 m, or roughly 6.2 km. Might using those same numbers give you the same?

Edited December 15, 2011 by Iggy

[newts]

You should put all that into a graph.

I am not sure how I could put that on a graph. I did research Jupiter to see if it would be possible to do a useful calculation, but it appears that Jupiter is mostly made from metallic hydrogen, which is liquid in nature. Since the density of metallic hydrogen is not really known, there is no way to calculate whether Jupiter has a rocky core.

 

It would seem likely that somebody who had studied maths and the earth's equatorial bulge, would have come up with a better way to calculate than I did; but it was only after I published my page, that Wikipedia even included a formula for calculating the bulge. So maybe experts are not that keen to share their expertise.

 

 

The equation that compares the ellipticity, [latex]\delta[/latex] (the factor by which the equator is longer) to the ratio between centrifugal and gravitational force, [latex]\phi[/latex] (1/289 -- as we found) is:

 

[latex]\frac {\delta}{1/100} = \frac{\phi}{4/505}[/latex]

 

Newton gave this equation in the Principia book 3 and it was later proved by Clairaut and other mathematicians correct for slowly rotating bodies of uniform density.

Because I have not studied the subject, what I have done is figure out a solution from first principles. You are quoting a formula which I do not know, but I would assume that it takes into account both the centrifugal effect, and the fact that once a bulge has been created by centrifugal forces its size is increased by its own gravity. Newton's value is the same as my final value for an earth of uniform density. Because my method is quite basic, it can be adjusted to take account of the fact that the centre of the earth is denser than the crust; I do not know if that is the case with Newton's method.

 

Following somebody else's maths is never easy, that is why I tried to describe in careful detail what I doing; but even so anybody who really wanted to understand my method would probably need to actually do the calculations themselves. If you read to the end, you will see that the answers I come up with are not far off the measured value.

Edited December 16, 2011 by newts

[Iggy]

You are quoting a formula which I do not know, but I would assume that it takes into account both the centrifugal effect, and the fact that once a bulge has been created by centrifugal forces its size is increased by its own gravity.

It takes into account Newton's universal law of gravity and the centrifugal force. It assumes uniform density and slow rotation. That is all.

 

Newton's value is the same as my final value for an earth of uniform density.

Then let's move on.

 

...I tried to describe in careful detail what I doing; but even so anybody who really wanted to understand my method would probably need to actually do the calculations themselves.

Would you agree that this calculation,

 

Therefore the energy gained from the trip to the centre, is sufficient to increase the height of the water at the equator by 0.58/300 times the earth’s radius, which is about 11.6 km.

should be 6.2 km instead, and would that difference affect your results?

[newts]

Would you agree that this calculation, should be 6.2 km instead, and would that difference affect your results?

I certainly got that calculation wrong, I have no idea how I got 11.6. The correct calculation for the numbers i used is 12.3. you seem to have calculated half that value. thanks for finding the error, but 12.3 is certainly right for what i was calculating, which is probably why i did not recheck the value of 11.6.

 

It would not really affect my results anyway, because that bit is just a rough calculation to try to make it easier for people with little knowledge of maths to follow my method.

[Iggy]

...12.3 is certainly right...

you might find a problem here,

 

To work out the mass of the bulge, we need to calculate its volume. To do this we actually need to know how high it is, so to make things simple we will start with the measured value of roughly 21 km, which is about 1/300 of the earth’s radius. If we were to add a thickness of 1/300 over the whole surface of the earth, that would increase its volume by about 3/300. However we are adding nothing at the poles; so taking account the elliptical shape of the bulge, we will assume that it would have a mass 2/300 of the rest of the earth.

 

Adding 1/300 to earth's radius adds what to its volume? You guessed "2/300". Perhaps that is the error...? How much bigger is the right side of the equation:

 

[latex]\frac{4}{3} \pi R^3 < \frac{4}{3} \pi R^2 \left( \frac{301R}{300} \right)[/latex]

 

The left side is the volume of a sphere of radius R and the right side is the volume of the sphere with a 1/300 bulge [volume of a spheroid]. Can you agree the volume isn't 2/300, but 1/300 more?

 

So the energy gained from the ring will be (1- 1/√2)R, or about 0.29R, times the mass of the ring, which is 2/300 the mass of the earth. Therefore the energy gained from the trip to the centre, is sufficient to increase the height of the water at the equator by 0.58/300 times the earth’s radius, which is about 12.3 km.

.29 * 6371 * 1/300 = 6.16

 

Would you agree?

 

It would not really affect my results anyway, because that bit is just a rough calculation to try to make it easier for people with little knowledge of maths to follow my method.

I was, in fact, going to suggest some editing. We're 1500 words in, and so far nothing has appeared relevant to your conclusions.

 

By the way, when you said that your equation was on wikipedia, can you point me to that, because I'm having a hard time reading the way the equations are rendered on the site you gave. Thank you.

Edited December 17, 2011 by Iggy

[michel123456]

you might find a problem here,

 

 

 

Adding 1/300 to earth's radius adds what to its volume? You guessed "2/300". Perhaps that is the error...? How much bigger is the right side of the equation:

 

[latex]\frac{4}{3} \pi R^3 < \frac{4}{3} \pi R^2 \left( \frac{301R}{300} \right)[/latex]

 

From the wiki page, I understand that the equatorial radius is squared. In your formula you have squared the polar radius.

Edited December 18, 2011 by michel123456

[Iggy]

From the wiki page, I understand that the equatorial radius is squared. In your formula you have squared the polar radius.

Oh my! Could that be right...

 

How would that work out, Newts?