The earth's equatorial bulge

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The earth's equatorial bulge Rate Topic:

#41 29 December 2011 - 12:56 PM imatfaal

Primate

newts, on 29 December 2011 - 12:30 AM, said:

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

$Posted Image$

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 $I$ 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 $I^2$ in the denominator when there is still an $I$ outside the brackets that will multiply the interior.

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

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

$\frac{1}{2} \left( I \omega^2 - \left(\omega- \frac{\omega mr^2}{I} \right)^2 (I+mr^2) \right)$

$\frac{1}{2} \left( I \omega^2 - \left(\omega^2- \frac{2\omega^2 mr^2}{I} + \frac{\omega^2 m^2r^4}{I^2}\right) (I+mr^2) \right)$

$\frac{\omega^2}{2} \left( I - \left(1- \frac{2 mr^2}{I} + \frac{ m^2r^4}{I^2}\right) (I+mr^2) \right)$

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

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

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

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

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

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

A little learning is a dangerous thing; drink deep, or taste not the Pierian spring:
there shallow draughts intoxicate the brain, and drinking largely sobers us again.
- Alexander Pope
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#42 29 December 2011 - 05:17 PM Iggy

Meson

newts, on 29 December 2011 - 12:30 AM, said:

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

Quote

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, $I \omega = I \omega_2 + mr^2 \omega_2$ , like you did and solve it for $\omega_2$ without approximating giving,

$\omega_2 = \frac{I \omega}{I + mr^2}$

The exact equation for difference in kinetic energy is then,

$\frac{1}{2}I \omega^2 - \frac{1}{2} \left( \frac{I \omega}{I+mr^2} \right)^2 (I+mr^2)$

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

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

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

$\frac{1}{2}I \omega^2 - \frac{1}{2} \omega^2 I(1-mr^2/I)$

and there you have it,

$\frac{1}{2} \omega^2 (I-(I-mr^2)) = \frac{1}{2} \omega^2 mr^2$

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#43 29 December 2011 - 06:02 PM michel123456

Molecule

$\frac{1}{2} \omega^2 mr^2$

And the approximate factor is...

Michel

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#44 29 December 2011 - 06:49 PM Iggy

Meson

imatfaal, on 29 December 2011 - 12:56 PM, said:

+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:

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

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

michel123456, on 29 December 2011 - 06:02 PM, said:

$\frac{1}{2} \omega^2 mr^2$

And the approximate factor is...

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

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#45 29 December 2011 - 10:49 PM newts

Baryon

Iggy, on 29 December 2011 - 06:49 PM, said:

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 .

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#46 30 December 2011 - 07:58 AM michel123456

Molecule

imatfaal, on 29 December 2011 - 12:56 PM, said:

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

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

Michel

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#47 4 January 2012 - 11:40 PM Iggy

Meson

newts, on 29 December 2011 - 10:49 PM, said:

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?

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#48 28 January 2012 - 09:31 PM newts

Baryon

Iggy, on 4 January 2012 - 11:40 PM, said:

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.

michel123456, on 30 December 2011 - 07:58 AM, said:

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.

This post has been edited by newts: 28 January 2012 - 09:33 PM

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#49 29 January 2012 - 01:05 AM Iggy

Meson

newts, on 28 January 2012 - 09:31 PM, said:

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.

newts, on 28 January 2012 - 09:31 PM, said:

...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.

This post has been edited by Iggy: 29 January 2012 - 01:21 AM

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#50 29 January 2012 - 10:51 PM newts

Baryon

Iggy, on 29 January 2012 - 01:05 AM, said:

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

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.

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#51 31 January 2012 - 12:01 AM Iggy

Meson

newts, on 28 January 2012 - 09:31 PM, said:

Imatfaal's maths may well be sound, but he was under the misapprehension that I was ignoring terms which had I² on the denominator.

newts, on 29 January 2012 - 10:51 PM, said:

What I should have said is that I have ignored terms containing m²/I².

Sure.

Keep trying. I'm sure you'll figure out how this was done eventually.

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#52 31 January 2012 - 12:43 PM imatfaal

Primate

Agree with Iggy - keep soldiering on! As is probably obvious from above, I am not a great fan of losing terms too quickly unless and until the calculation becomes too difficult or they are clearly insignificant (ie if they are just cumbersome then leave and in but get a bigger sheet of paper)