The earth's equatorial bulge

[newts]

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?

This is what i calculated:

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

but because the new shape is an ellipse not a sphere i only added 2/3 of that.

 

 

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.

maybe i have not explained what i have done very well, but i do not see how less explanation would make it easier.

 

The equation is the last one under the heading: Accurate calculation of bulge, assuming the mass of the earth to be evenly distributed

[Iggy]

This is what i calculated:

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

but because the new shape is an ellipse not a sphere i only added 2/3 of that.

Interesting.

 

 

I think Michel hit the nail right on the head. The correct relationship would be:

 

[latex]\frac{4}{3} \pi R^3 < \frac{4}{3} \pi R_P (R_E)^2[/latex]

 

Where [latex]R_P[/latex] is the polar radius and [latex]R_E[/latex] is the equatorial radius. Since the equatorial radius is 301/300 times the polar radius all we have to do is solve for C:

 

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

 

which gives,

 

[latex]C = \left( \frac{301}{300} \right)^2[/latex]

 

Your approximation is quite good.

 

 

maybe i have not explained what i have done very well, but i do not see how less explanation would make it easier.

No, I agree. More explanation is better.

 

The equation is the last one under the heading: Accurate calculation of bulge, assuming the mass of the earth to be evenly distributed

I'd still love to see the wiki page

 

 

Let read the next bit...

 

To calculate how much slower, we need to equate the total angular momentum when a drop of water is on the North Pole, with the total angular momentum when it has moved to the equator:

 

Iw = Iw (2) + mR² w (2)

Does the "I" on the left side of the equation have the same value as the "I" on the right?

[Enthalpy]

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.

Very interesting!

 

Because Earth consists of little compressible solids or liquids, allegedly with a uniform Ni-Fe composition over much of its radius.

In contrast, Jupiter is commonly considered as a ball of gas, more compressible than rock.

A perfect gas would differentiate more the density, making Jupiter more spherical than the extrapolation from Earth.

 

A first answer is that gas near to liquid density isn't very compressible - but still much more than a liquid.

(In contrast, a plasma is nearly a perfect gas)

A second reason would be a high temperature gradient, consistent with the excessive heat radiated by Jupiter.

But my gut feeling is that 1/15 versus 1/12.3 doesn't allow a heavy kernel of metal or ceramic.

 

If you could put some numerical constraint on the mass of a solid kernel, it would be worth a paper on arXiv.

With some nearly-reasonable assumptions, like an adiabatic convective "equilibrium" over altitude, as in Earth's atmosphere.

Alas, both Van der Waals' and virial's equations are very inaccurate near liquid density. Bad joke. Better equations must exist.

 

Segregation by gravitation over the molar mass makes things more complicated.

The upper atmosphere does it on Earth, despite the gravity well isn't as deep.

Maybe Jupiter's flattening can put a numerical constraint on molar mass segregation, telling that convection is efficient - the planet's aspect speaks in favour.

 

-----

 

Could an analytic nearly-solution be just little more complicated than the model with an equatorial ring?

Like: the altitude of equal potential differs from a sphere by thatmuch*cosine(2*latitude), and if really needed by thatlittle*cosine(6*latitude)?

I may have read somewhere that the field is then a Bessel function. For sure, it has already been done. Less usable for a simpler book.

[newts]

Your approximation is quite good.

When I first read Michel's comment, I thought he was saying the error came from using 6378 km, instead of 6357 km. It is so easy to misinterpret what people are saying, especially when one does not read things carefully.

 

What I did originally was just guess the value of 2/3, though it can easily be obtained by integration. But following your line of thinking, I could amend the article to point out that increasing the polar radius would involve increasing the radius in only 1 dimension, so would only increase it by 1/300; whilst increasing the equatorial radius involves increasing it in 2 dimensions, so is 2/300.

 

Does the "I" on the left side of the equation have the same value as the "I" on the right?

What I am doing is working out the energy of the final drop of water making the journey from the pole, so yes.

 

Because Earth consists of little compressible solids or liquids, allegedly with a uniform Ni-Fe composition over much of its radius.

In contrast, Jupiter is commonly considered as a ball of gas, more compressible than rock.

A perfect gas would differentiate more the density, making Jupiter more spherical than the extrapolation from Earth.

 

A first answer is that gas near to liquid density isn't very compressible - but still much more than a liquid.

(In contrast, a plasma is nearly a perfect gas)

A second reason would be a high temperature gradient, consistent with the excessive heat radiated by Jupiter.

But my gut feeling is that 1/15 versus 1/12.3 doesn't allow a heavy kernel of metal or ceramic.

What my calculations show, is that if for instance the top ¼ of the radius was very light, then there would be no need for a heavy core. I would guess that what is considered to be the surface of Jupiter would need to be liquid, otherwise would it not be transparent like the earth’s atmosphere?

 

According to Wikipedia, the variation of the density of the earth is known fairly accurately, but my calculations do not seem to agree with their assumptions. If my calculations are correct, it looks like more of the earth’s density must be nearer the centre than is commonly thought.

 

Actually I am pretty ignorant about physics, other than those things I have figured out for myself. I only did the calculations in the first place, because I wanted find out why Newton got the wrong answer in the Principia. The main reason for publishing the page, was to try to lure people to my blog, to introduce them to my theory of particle physics, but as expected it has not worked.

 

 

Edited December 21, 2011 by newts

[Iggy]

What I am doing is working out the energy of the final drop of water making the journey from the pole, so yes.

Would you say I includes the drop?

[newts]

Would you say I includes the drop?

The idea is that the drop has negligible moment of inertia when at the pole, because the radius there is virtually 0.

[Iggy]

The idea is that the drop has negligible moment of inertia when at the pole, because the radius there is virtually 0.

that makes very good sense

 

The initial kinetic energy was ½Iw ² , and the final kinetic energy is:

 

½[w (1-mR² /I)]² times I+mR² .

 

This gives us a surplus energy of ½ mR² w ²

did you use some force of algebra to get you there?

[Enthalpy]

Figures about Jupiter's depths there

http://www.futura-sciences.com/fr/news/t/astronomie/d/le-coeur-de-jupiter-pourrait-etre-en-train-de-disparaitre_35400/

like 16,000K and 4TPa. Nobody went there to measure, but at least it can be present-day consensus supposition.

 

The paper at Futura-Sciences tells "solid" and "liquid" but this must be a simplification for "dense gas and plasma".

Notice taken, though, that said liquids aren't the ordinary form - hydrogen being a metal there instead.

 

And in 2016, Juno shall orbit Jupiter to make a precise gravity map and give clues to the planet's depth.

[newts]

did you use some force of algebra to get you there?

To try to keep the story flowing, i have ommitted most of the algebra; on the basis that anybody sufficiently interested could do it themselves.

 

And in 2016, Juno shall orbit Jupiter to make a precise gravity map and give clues to the planet's depth.

I calculated for Saturn, with a rotation period of 10.57 hours and a mean density of .687. It should have a bulge 41.3 times that of the earth.

 

If all saturn's mass was at the centre, its bulge would be 1/14, if its mass was uniformly distributed its bulge would be 1/5.6, in fact its bulge is 1/10.2; this suggests that even more of Saturn's mass is situated near the centre than is the case for Jupiter. So perhaps Saturn has more of a solid core than Jupiter.

[Iggy]

To try to keep the story flowing, i have ommitted most of the algebra; on the basis that anybody sufficiently interested could do it themselves.

but, you do agree that reducing,

 

[latex]\frac{1}{2}I \omega^2 - \frac{1}{2} \left(\omega(1-mr^2/I) \right)^2 (I+mr^2) [/latex]

 

to

 

[latex]\frac{1}{2}mr^2 \omega^2[/latex]

 

via algebra is what you did, and the correct way to go?

 

//edited latex

Edited December 22, 2011 by Iggy

[newts]

but, you do agree that reducing,

 

[latex]\frac{1}{2}I \omega^2 - \frac{1}{2} \left(\omega(1-mr^2/I) \right)^2 (I+mr^2) [/latex]

 

to

 

[latex]\frac{1}{2}mr^2 \omega^2[/latex]

 

via algebra is what you did, and the correct way to go?

 

//edited latex

Yes, it was a few lines of algebra, but I did not include it. If I get round to rewriting the page perhaps I should do?

 

If Saturn had a comparable mass distribution to Jupiter, it would have a bulge of 1/8.8 rather than 1/10.2, so Saturn must on average have a lot more of its mass nearer the centre. It also is puzzling that Saturn and Jupiter should have such different average densities, when they are both thought to be made mostly of hydrogen.

[Iggy]

Yes, it was a few lines of algebra, but I did not include it.

Could you please demonstrate. I can't seem to get that.

[newts]

but, you do agree that reducing,

 

[latex]\frac{1}{2}I \omega^2 - \frac{1}{2} \left(\omega(1-mr^2/I) \right)^2 (I+mr^2) [/latex]

 

to

 

[latex]\frac{1}{2}mr^2 \omega^2[/latex]

 

via algebra is what you did, and the correct way to go?

 

//edited latex

Sorry, I did not check last time. It actually reduces to minus [latex]\frac{1}{2}mr^2 \omega^2[/latex], as it is a loss of energy.

 

Is that what you got? I approximated by ignoring all terms including an m^2

[Iggy]

Sorry, I did not check last time. It actually reduces to minus [latex]\frac{1}{2}mr^2 \omega^2[/latex], as it is a loss of energy.

 

Is that what you got? I approximated by ignoring all terms including an m^2

ouch

 

Yeah, I'm sure I could simplify that to [latex]-\frac{1}{2}mr^2 \omega^2[/latex] if I approximated [latex]\omega(1-mr^2/I) \approx \omega[/latex]. But, that can't be right. That would be assuming earth doesn't change speed, and it would make all of this meaningless:

 

But what we have not considered is where the energy comes from, and the answer is of course that all the energy that the water gains, must have been lost by the fact that the whole earth ends up spinning slower. To calculate how much slower, we need to equate the total angular momentum when a drop of water is on the North Pole, with the total angular momentum when it has moved to the equator:

 

Iw = Iw (2) + mR² w (2)

 

which can be simplified and approximated to give:

 

w(2) = w (1 - mR² /I)

 

 

I trust you know what your math means. It's like you're trying to figure it out. I don't get it.

 

It actually reduces to minus [latex]\frac{1}{2}mr^2 \omega^2[/latex], as it is a loss of energy.

because we're subtracting the final state from the original state negative is a gain of energy. If you start with 10 apples and end up with 11, you could subtract 10-11 = -1... negative in that case means you've gained an apple.

 

This should make sense because, if you move mass from the pole to the equator without slowing the earth, the earth will have gained kinetic energy at the end.

[newts]

I trust you know what your math means. It's like you're trying to figure it out. I don't get it.

I think understanding maths is hard work, I find following other people's maths almost impossible, even when I have calculated the same answer myself.

 

Here is the algebra relating to

 

 

 

 

½[w (1-mR² /I)]² multiplies out to give ½w ² [(1-2mR² /I)+(mR² /I)² ]. The final term in the brackets is insignificant.

 

½w ² (1-2mR² /I) times I+mR² , multiplies out to give ½w ² [(I+mR² ) - 2mR² - (2m² R^4/I²)]. Again the final term in the brackets is insignificant. The rest is just addition/subtraction.

[imatfaal]

I think understanding maths is hard work, I find following other people's maths almost impossible, even when I have calculated the same answer myself.

 

Here is the algebra relating to

 

 

 

 

½[w (1-mR² /I)]² multiplies out to give ½w ² [(1-2mR² /I)+(mR² /I)² ]. The final term in the brackets is insignificant.

 

½w ² (1-2mR² /I) times I+mR² , multiplies out to give ½w ² [(I+mR² ) - 2mR² - (2m² R^4/I²)]. Again the final term in the brackets is insignificant. The rest is just addition/subtraction.

Newts

 

I can only get that algebra to work out if I ignore any term with [imath]I[/imath] in the denominator - and that cannot be correct. Perhaps you can go long hand and explain - personally I think it is a little rash to ignore terms with [imath]I^2[/imath] in the denominator when there is still an [imath]I[/imath] outside the brackets that will multiply the interior.

 

[math] \frac{1}{2}I \omega^2 - \frac{1}{2} \left(\omega(1-mr^2/I) \right)^2 (I+mr^2) [/math]

 

[math] \frac{1}{2} \left( I \omega^2 - \left(\omega(1-mr^2/I) \right)^2 (I+mr^2) \right)[/math]

 

[math] \frac{1}{2} \left( I \omega^2 - \left(\omega- \frac{\omega mr^2}{I} \right)^2 (I+mr^2) \right)[/math]

 

[math] \frac{1}{2} \left( I \omega^2 - \left(\omega^2- \frac{2\omega^2 mr^2}{I} + \frac{\omega^2 m^2r^4}{I^2}\right) (I+mr^2) \right)[/math]

 

[math] \frac{\omega^2}{2} \left( I - \left(1- \frac{2 mr^2}{I} + \frac{ m^2r^4}{I^2}\right) (I+mr^2) \right)[/math]

 

[math] \frac{\omega^2}{2} \left(I- \left( I -\frac{2Imr^2}{I}+\frac{Im^2r^4}{I^2}+mr^2-\frac{2m^2r^4}{I}+\frac{m^3r^6}{I^2} \right) \right)[/math]

 

[math] \frac{\omega^2}{2} \left(I- I +\frac{2Imr^2}{I}-\frac{Im^2r^4}{I^2}-mr^2+\frac{2m^2r^4}{I}-\frac{m^3r^6}{I^2} \right)[/math]

 

[math] \frac{\omega^2}{2} \left( +\frac{2Imr^2}{I}-\frac{Im^2r^4}{I^2}-mr^2+\frac{2m^2r^4}{I}-\frac{m^3r^6}{I^2} \right)[/math]

 

[math] \frac{\omega^2}{2} \left( +\frac{2mr^2}{1}-\frac{m^2r^4}{I}-mr^2+\frac{2m^2r^4}{I}-\frac{m^3r^6}{I^2} \right)[/math]

 

[math] \frac{\omega^2}{2} \left( +2mr^2 -mr^2-\frac{m^2r^4}{I}+\frac{2m^2r^4}{I}-\frac{m^3r^6}{I^2} \right)[/math]

 

[math] \frac{\omega^2}{2} \left( +mr^2+\frac{m^2r^4}{I}-\frac{m^3r^6}{I^2} \right)[/math]

[Iggy]

I think understanding maths is hard work, I find following other people's maths almost impossible, even when I have calculated the same answer myself.

 

Here is the algebra relating to

 

 

 

 

½[w (1-mR² /I)]² multiplies out to give ½w ² [(1-2mR² /I)+(mR² /I)² ]. The final term in the brackets is insignificant.

 

½w ² (1-2mR² /I) times I+mR² , multiplies out to give ½w ² [(I+mR² ) - 2mR² - (2m² R^4/I²)]. Again the final term in the brackets is insignificant. The rest is just addition/subtraction.

There is a very commonly used trick for accurate approximations that you used here:

 

Iw = Iw (2) + mR² w (2)

 

which can be simplified and approximated to give:

 

w(2) = w (1 - mR² /I)

 

It's quite hard to understand why you didn't use the same trick again.

 

There really is, however, only one approximation you have to make. If I were doing this I would start with, [latex]I \omega = I \omega_2 + mr^2 \omega_2[/latex], like you did and solve it for [latex]\omega_2[/latex] without approximating giving,

 

[latex]\omega_2 = \frac{I \omega}{I + mr^2}[/latex]

 

The exact equation for difference in kinetic energy is then,

 

[math]\frac{1}{2}I \omega^2 - \frac{1}{2} \left( \frac{I \omega}{I+mr^2} \right)^2 (I+mr^2) [/math]

 

cancel terms, pull an I out of the denominator, and cancel again:

 

[math]\frac{1}{2}I \omega^2 - \frac{1}{2} \left( \frac{I \omega^2}{1+mr^2/I} \right)[/math]

 

then make your approximation... [math]\left( \frac{I}{1+mr^2/I} \right) \approx I(1-mr^2/I)[/math] giving,

 

[math]\frac{1}{2}I \omega^2 - \frac{1}{2} \omega^2 I(1-mr^2/I)[/math]

 

and there you have it,

 

[math]\frac{1}{2} \omega^2 (I-(I-mr^2)) = \frac{1}{2} \omega^2 mr^2[/math]

[michel123456]

[math]\frac{1}{2} \omega^2 mr^2 [/math]

 

And the approximate factor is...

[Iggy]

Newts

 

I can only get that algebra to work out if I ignore any term with [imath]I[/imath] in the denominator - and that cannot be correct. Perhaps you can go long hand and explain - personally I think it is a little rash to ignore terms with [imath]I^2[/imath] in the denominator when there is still an [imath]I[/imath] outside the brackets that will multiply the interior.

+1

 

Newts, I think the term you ignored, 2m² R⁴/I² should have been 2m² R⁴/I. That term is larger than mr²/I, so you could, by your reasoning, just eliminate mr²/I from:

 

[math] \frac{1}{2}I \omega^2 - \frac{1}{2} \left(\omega(1-mr^2/I) \right)^2 (I+mr^2) [/math]

 

as insignificant. But, that should I think lead to the same bad result we had a couple posts ago.

 

[math]\frac{1}{2} \omega^2 mr^2 [/math]

 

And the approximate factor is...

if you multiplied that by I/I+mr^2 you should get the exact answer I'd expect

[newts]

Newts, I think the term you ignored, 2m² R⁴/I² should have been 2m² R⁴/I. That term is larger than mr²/I

Yes, I can never do maths without including errors, I do not know why I squared the I.

m is the mass of the drop of water, so all I have ignored is two terms which include m² .

[michel123456]

[math] \frac{\omega^2}{2} \left( +mr^2+\frac{m^2r^4}{I}-\frac{m^3r^6}{I^2} \right)[/math]

imatfaal has the most healthy approach IMHO. I wonder why he stopped where he stopped.

[Iggy]

Yes, I can never do maths without including errors, I do not know why I squared the I.

m is the mass of the drop of water, so all I have ignored is two terms which include m² .

what?

[newts]

what?

I was trying to convey the fact that you did identify an error in the maths I posted above. Also that the terms I ignored were ones that contain m² , which are insignificant because the mass of the water drop (m) is such a tiny fraction of the mass of the earth.

 

Thanks for your criticism. I originally thought that nobody would be much interested the details of the maths, but perhaps that is wrong, so if I revise the page I may include some more algebra.

 

 

imatfaal has the most healthy approach IMHO. I wonder why he stopped where he stopped.

Imatfaal's maths may well be sound, but he was under the misapprehension that I was ignoring terms which had I² on the denominator. Probably my fault for not explaining things properly, but it is not a particularly easy calculation, which is probably why not many other webpages deal with it.

Edited January 28, 2012 by newts

[Iggy]

Also that the terms I ignored were ones that contain m² , which are insignificant because the mass of the water drop (m) is such a tiny fraction of the mass of the earth.

I'm sorry... you're not making any sense. This is what you said before:

 

½[w (1-mR² /I)]² multiplies out to give ½w ² [(1-2mR² /I)+(mR² /I)² ]. The final term in the brackets is insignificant.

 

½w ² (1-2mR² /I) times I+mR² , multiplies out to give ½w ² [(I+mR² ) - 2mR² - (2m² R^4/I²)]. Again the final term in the brackets is insignificant. The rest is just addition/subtraction.

 

All of the terms have m in them. The answer has an m in it. m² is more significant as a term than m. Ignoring terms with m, or m², is not what happened, nor could happen.

 

It's like you're trying to figure out how this math was done and having quite a lot of trouble figuring it out.

 

...but it is not a particularly easy calculation, which is probably why not many other webpages deal with it.

 

Calculating the change in speed is an extremely easy calculation. The method of approximating it is given in post 42.

Edited January 29, 2012 by Iggy

[newts]

I'm sorry... you're not making any sense. This is what you said before:

 

½[w (1-mR² /I)]² multiplies out to give ½w ² [(1-2mR² /I)+(mR² /I)² ]. The final term in the brackets is insignificant.

 

½w ² (1-2mR² /I) times I+mR² , multiplies out to give ½w ² [(I+mR² ) - 2mR² - (2m² R^4/I²)]. Again the final term in the brackets is insignificant. The rest is just addition/subtraction.

 

All of the terms have m in them. The answer has an m in it. m² is more significant as a term than m. Ignoring terms with m, or m², is not what happened, nor could happen.

 

It's like you're trying to figure out how this math was done and having quite a lot of trouble figuring it out.

It is a lot easier to do maths than to explain to somebody else what one is doing. What I should have said is that I have ignored terms containing m²/I². If m/I was a millionth, then m²/I² would be a trillionth, the moment of inertia of the drop of water is very small compared to the moment of inertia of the whole earth.