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The Ask Mickeko Thread!

mickeko

Join Date 09/2008

+126

The Ask Mickeko Thread!

Ask me anything, I answer all questions (*).

You may get a stupid answer, an incorrect answer, or maybe even the RIGHT answer... Who knows...

Any answer given that is true and correct (as far as I know or in my opinion where applicable) will be marked with a .

(*) - Exceptions: Personal questions or any other questions I don't feel like answering may not be answered. Being polite might help in these cases...

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DrJBHL

Join Date 04/2002

+4054

Was he great or what?!!!

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mickeko

Join Date 09/2008

+126

Quoting CarGuy1,

reply 25

Doc...I'm thinking more along the lines of "Carnac the Magnificent"

Charlatan! May a bloated yak change the temperature of his jacuzzi!!

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DrJBHL

Join Date 04/2002

+4054

Johnny's gone, I'm afraid...but while he was here, things were merry!

Link

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mickeko

Join Date 09/2008

+126

Youtube is littered with Johnny too. As with all american talkshows, alot of the political jokes are lost on me, but there's tons of greatness to see even for me.

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HG_Eliminator

Join Date 07/2002

+354

Consider a celestial body (planet) of mass M and volume V consisting of material of uniform density M/V. If the body is spherical, then its radius is R=(3V/4{pi})^1/3, and the free-fall acceleration on its surface is g=GM/R²=GM(4{pi}/3V)^2/3, where G is the gravitational constant. Consider all possible shapes of the body. What is the largest g, that can be achieved at one point on a surface, and what is the shape of the body for which such acceleration is achieved?

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the answer is always 42

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DrJBHL

Join Date 04/2002

+4054

Quoting HG_Eliminator,

reply 5

Consider a celestial body (planet) of mass M and volume V consisting of material of uniform density M/V. If the body is spherical, then its radius is R=(3V/4{pi})1/3, and the free-fall acceleration on its surface is g=GM/R2=GM(4{pi}/3V)2/3, where G is the gravitational constant. Consider all possible shapes of the body. What is the largest g, that can be achieved at one point on a surface, and what is the shape of the body for which such acceleration is achieved?

Clearly, as R->0, g-> infinity. The maximum will be achieved as a "Black Hole" on it's event horizon, as a singularity. As R->0, M/V will also approach infinity, as V->0.

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mickeko

Join Date 09/2008

+126

Quoting HG_Eliminator,

reply 5

Why ask a question where the answer is already included? Just do the math. I'm not a calculator.

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Fuzzy Logic

Join Date 05/2002

+493

Nothing unreal exists

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superman

Join Date 03/2008

+133

Quoting HG_Eliminator,

reply 5

Max g = (4/5)(15/4)^1/3(π)^2/3M/V^2/3

Let R be the max diameter of body then its shape will be given by relating distance r from the point, where there is max g,

as a function of the angle θ from the axis of symmetry.

r² = R²cos(θ)

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HG_Eliminator

Join Date 07/2002

+354

Superman is correct.

The optimal body has axial symmetry and has the shape depicted at the picture below. The green dot denotes the point of maximal free-fall acceleration. The body can be described by relating distance r from the point at which the maximal acceleration is found as a function of the angle {theta} from the axis of symmetry (vertical black line).

r²=R²cos({theta}).

The acceleration at the green point will be g=(4/5)(15/4)^1/3{pi}^2/3M/V^2/3. This is only 2.6% larger than the gravity on the surface of a spherical planet.

The solution: Assume for a moment that we already know that the body should have axial symmetry. Consider a narrow ring of matter lying on the surface of the body. (Center of the ring coincides with the axis of symmetry and its plane is perpendicular to that axis. It is depicted on the figure below.) If the mass of the ring is dM and the distance between the points of the ring and the "green point" is r, then the contribution of the ring to free-fall acceleration will be G*{dM}cos({theta})/r².

If we remove this ring of material from the surface and place it somewhere else on the surface, it will have different r and different angle {theta}. Since the shape is optimal, it is "stationary", i.e. infinitesimal changes should not change the acceleration. This will happen only if cos({theta})/r² is constant. Consequently, r²=R²cos({theta}). The constant R denotes the maximal diameter of the body. Once the shape of the body is given, we can easily find (by an elementary integration^*) that its volume V=(4/15){pi}R³, while the acceleration at point g is given^** by the expression which has been presented above.

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Fuzzy Logic

Join Date 05/2002

+493

Aye, and if my granny had wheels she'd be a waggon...

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DrJBHL

Join Date 04/2002

+4054

In erudite terms, Baloney. That's a static solution. What's needed is a vector solution.

Ta-Da! Voila:

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Enough with the serious stuff,y'all are making my head hurt!

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superman

Join Date 03/2008

+133

You may think you want your dream girl, but really, you don't.

When you want anything, you feel a stong desire for that.

When I wanted her I felt too strong a desire for her so it was a 'want' there. True, I didn't need her as she was not really required for my subsistence as I could successfully survive without her.

It's not wrong. It may not be what you expected or wanted it to be though.

Cmon guess another. Its not me who expected anything. This quetion was asked to me by a third person. In fact I dont know what I expect but expect it to be same as what he thought.

Sometimes I think to offer him the dictionary and say that look 'ur word should be there'

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HG_Eliminator

Join Date 07/2002

+354

If π is 3.14... is it still the same with Whipped cream on top?

A bit of trivia, my Original Nick "PhenomTT" the TT was intended to emulate the Greek letter used for Pi and Phenom was short for phenomenon, together it was meant to represent an irrational number of perceived events. Ie .. me

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sydneysiders

Join Date 04/2004

+1042

does karma really ever get round to bitches? ..or we just make ourselves feel better thinking it does..

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mickeko

Join Date 09/2008

+126

Quoting sydneysiders,

reply 17

does karma really ever get round to bitches? ..or we just make ourselves feel better thinking it does..

Well, it may seem like it does at times. But it's really just random occurences.

You build up your own karma, and you are the one to eventually administer whatever is coming for you. There's nothing more to it.

(You be the judge of the accuracy of this. It's how I see things. It's a result of the insane amount of thining I've done over the years. I do not represent any theological organization, I only represent myself. )

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ArileenDesign

Join Date 08/2006

+119

Where is my damn Pizza?

ArileenDesign is looking for HG

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mickeko

Join Date 09/2008

+126

Quoting ArileenDesign,

reply 19

Where is my damn Pizza?

ArileenDesign is looking for HG

When you find the pizza, you'll also find HG. It's written on concrete.

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HG_Eliminator

Join Date 07/2002

+354

It's written on concrete.

in pizza sauce?

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DrJBHL

Join Date 04/2002

+4054

You're asking for it, HG....

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mickeko

Join Date 09/2008

+126

Quoting HG_Eliminator,

reply 21

It's written on concrete.
in pizza sauce?

I don't recommend eating concrete, pizza sauced or not.

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starkers

Join Date 05/2004

+1153

I've been thinking, if a piece of string comes up and inch short for my needs, would a rope suffice?

And if so, would starch aid in performing Indian rope tricks?

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Chasbo

Join Date 01/2005

+958

Late to the party again......

How can you be in 2 places at once when your not anywhere at all?

FIRESIGN THEATRE!!!!!!!

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