Let's discuss Mathematics

[Fifty]

Tell us what's wrong with them first, that's the easier procedure. My first thought is that you read/the book printed some quantification wrong.

NEVERMIND it was a misread by me in the statement of the empty set axiom (I thought it said that the empty set was an ELEMENT of every set, rather than a SUBSET of every set).

My issue was that if the empty set was an element of every set, then the axiom of pairing can't be true. Let a = {1} and b = {2}, then by the axiom of pairing there is a set C with only {1} and {2} as elements. But if the empty set is an element in every set, then C also has the empty set as an element, so it can't just have {1} and {2} as elements.

 

[pboily]

Wow! There's a prestigious mathematician in the thread now!

Its so nice being noticed!

[wiki=http://en.wikipedia.org/wiki/Johann_Peter_Gustav_Lejeune_Dirichlet]Johan Peter Gustav Lejeune Fifty[/wiki]

 

[PlutonianEmpire]

A Boeing 747-400 is at the Earth's rotational Equator, flying in a straight line heading due east following the equator, at typical cruising speed, Mach 0.84.

The plane has just crossed the terminator into night, with the sun having just set, and it is spring equinox. With the plane's speed and the planet's rotation in mind, how long is it before the plane crosses the terminator again and the sun rises, assuming the plane doesn't deviate from its course following the equator and has enough fuel. How many hours?

 

[sanabas]

Plane's flying at ~1029 km/h.

Terminator's moving at ~532 km/h.

So plane is moving at ~497 km/h relative to the terminator, and needs to cover ~6378 km. So ~ 12 hours, 50 minutes. Plus another 2 minutes because flying 10 km up makes the diameter 10 pi km longer.

 

[PlutonianEmpire]

Umm, are you sure? The Earth's rotation is 24 hours, so at the equator, nights are only 12 hours, and since the plane is at the equator, and is going WITH the Earth's rotation, not against it, I'd think it would cross the terminator again in less than 12, 13 hours.

 

[Olleus]

Plane's flying at 1029 km/h

Terminator's moving at (earth circumference)/(1 day) = 1663 km/h

Change frame of reference so terminator is moving at 0, and plane at 2692 km/h.

Plane needs to travel 1/2 of earths circumference which is 19958 km.

Therefore, it takes 7.414h = 7h 25min.

As a check, if the plane is moving at 0 at the start of the question, there are 12h between dusk and dawn, which must be right.

 

[sanabas]

Umm, are you sure? The Earth's rotation is 24 hours, so at the equator, nights are only 12 hours, and since the plane is at the equator, and is going WITH the Earth's rotation, not against it, I'd think it would cross the terminator again in less than 12, 13 hours.

It'd work better if I used the earth's circumference, not diameter, and had the terminator moving the right way.

 

[PlutonianEmpire]

Lolllllllllllll.....

I had a really hard time figuring out the equations, so because of that, all my planes were going way too fast, but I knew I was missing something.

I googled how to get the circumference of a world, and it turned out I was neglecting to multiply by 2. Then your posts made sense.

Thanks guys.

 

[Ajidica]

This is pretty easy, but I can't remember how to do it.
How do I remove the denominator from -5/(7x^2)?
I know if the 7 wasn't there I could shift the x^2 up to the numerator and have the answer be -5x^-2 but I can't remember how to handle the seven.

(If it is any help I'm doing derivatives and I started with f(x)=5/7x.)

 

[sanabas]

This is pretty easy, but I can't remember how to do it.
How do I remove the denominator from -5/(7x^2)?
I know if the 7 wasn't there I could shift the x^2 up to the numerator and have the answer be -5x^-2 but I can't remember how to handle the seven.

(If it is any help I'm doing derivatives and I started with f(x)=5/7x.)

You've already got it. You can't change the -5/7 part, you can put the x^2 on top or bottom by changing the sign. The answer's (-5/7)x^-2, or -5x^-2/7, or -5/(7x^2), depending how you want to write it.

 

[Ajidica]

Thanks. I was always terrible at exponents.

 

[PlutonianEmpire]

How do I calculate the oblateness of a particular gas giant, real or fictional?

An explanation of the formula/equation would be welcome.

 

[dutchfire]

This http://en.wikipedia.org/wiki/Oblate_spheroid ?

 

[ParadigmShifter]

I think he wants to be able to calculate how oblate a gas giant would be given its properties.

I'm guessing things that would influence it would be rotational speed, wind speed, density, drag, gravity in the region, etc.

I don't know much about astrophysics though.

Truronian might know something about it he did fluid mechanics IIRC.

 

[Truronian]

Unfortunately I didn't do anything like that. I wouldn't have thought it would be too hard to find an equation for it though, it should only depend on the mass and rotational velocity. I think.

A quick google turned up this thread from another forum...

EDIT: Or if you fancy something slightly heavier... http://www.barnesos.net/publications/papers/Oblateness.pdf

 

[PlutonianEmpire]

Unfortunately I didn't do anything like that. I wouldn't have thought it would be too hard to find an equation for it though, it should only depend on the mass and rotational velocity. I think.

A quick google turned up this thread from another forum...

EDIT: Or if you fancy something slightly heavier... http://www.barnesos.net/publications/papers/Oblateness.pdf

I visit that other forum on a regular basis. It's the support forum for Celestia, the astronomy program I use for my worldbuilding. A quick search should reveal my presence there.

Thank you for the help.

 

[Gigaz]

I had a course lately which approached the problem for liquid drops, it is usually too difficult to calculate it by hand. For a gas giant it is a different problem because:
1. gravity is a long range interaction unlike the dipole or Van-der-Waals forces between molecules.
2. A big planet can not be modelled as composed of incompressible units. Compression energy becomes important.
You'd have to find the minimum of the sum of E(compression), E(rotation) and E(gravitation) depending on the eccentricity of the ellipsoid to determine the equilibrium state.

 

[PlutonianEmpire]

Ok, here's a new one.

Lilith and Hades orbit each other every 10.02 Earth days.

Belle Hades' sidereal rotation period is 26.4483 Earth hours. Its year is 7196 Earth hours, or 299.83 Earth days.

Based on just these numbers, how many Belle Hadean SOLAR (not sidereal, based on when the Lilith/Hades barycenter reaches zenith, and Belle Hades orbits the barycenter too, since it is circumbinary) days does it take for Lilith and Hades to complete their revolution around each other, as observed by a Belle Hadean?

And how did you calculate that number?

 

[Gigaz]

I think this problem is rather ill-posed. But I'll assume that the three bodies are in a plane.
The problem is composed of two seperate problems:

First one: How much time passes for a Belle Hadean until the Lilith/Hades Barycenter is at the same position again? This is the Belle Hadean Solar.
Second one: How much time passes for a Belle Hadean between two Hades-Lilith-Belle Hades linear events? This is the revolution as observed by the Belle Hadean.

Observed revolution expressed in Belle Hadean Solar is the final solution to the problem.

 

[dutchfire]

But I'll assume that the three bodies are in a plane.

Any three bodies lie in a plane.