But we"re just not up on our rocket science and astrophysics enough to get the joke. Any geniuses out there?

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But we"re just not up on our rocket science and astrophysics enough to get the joke. Any geniuses out there?

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Gx^2mM/R (that is x squared

The other side of the equation seems to contain the mass of 1 body and its velocity, so i would assume this equation is what u need to solve in order to put a satelite in a stable orbit around the earth, or maybe to calculate wich distances from the earth can be used as stable orbits for satelites.

or something like that :D

It`s called "phase shifting" and it`s totally physically true. You`d just have to be going extremely fast."

@nerdtchn: I know phase shifting and I get red shift/blue shift... obviously better than you. Red wouldn`t turn blue, it would shift blue. It would become a slightly oranger color of red.

delta v = SUM u*Ln(Mi/Mf) - g*t*Cos(theta)

I is smaller than 3, the function of modulus 5 x

> Rocket science isn`t difficult. No air to worry about, or kinetic friction, or rotational inertia, just objects in space

On the other hand, EVERY atom in the universe exerts a gravitational force upon every OTHER atom in the universe, which would tend to make it a little more difficult if they didn`t just cheat and discount 99.9999999% of them as being `too weak to bother with`.

Rocket science isn`t difficult. There`s minimal friction, predictable behaviors, and a lot of it can be solved through energy equations and a minimum of calculus. No air to worry about, or kinetic friction, or rotational inertia, just objects in space.

Rocket engineering on the other hand is a godawful nightmare I do not want anything to do with.

So basically, if you add positive kinetic energy (i.e. speed away from the planet) to your gravitational potential energy, it eventually becomes 0, and you escape the gravity well. The speed necessary for this is escape velocity.

So, since you`re saying:

gravitational potential energy + kinetic energy = 0

You are really saying:

-gravitational potential energy = kinetic energy

And therefore, you substitute in 1/2mv^2 for kinetic energy and GmM/R for gravitational potential energy, and end up with a wonderful equation for calculating escape velocity on any given planet.

> Hi, well the left part of the equation is the gravitational energy it takes to get from 0 to radius R with an object of mass mx away from the mass M. This is the integrated gravitational force -G_x (m_x M)/r^2 from 0 to R...

Your IQ is too high for i-am-bored. Perhaps you`d be happier over at 4chan...

[quote]Nah only kidding dude.[/quote]

Natrually enough, I asked her if the "carpe" matched the curtains, which I thought was hilarious.....she didn`t!"

Hahahaha! Although most might say that`s a fail, my inner nerd says this is a WIN.

Oh and all US government facilities buy American cars. Which includes the Pontiac GT. At LLNL and LBNL, the directorate usually drive Chrysler 300`s and Cadillac STS`s. My car was a Ford Crown Vic P71. And those are your tax dollars at work, folks!

The "x" is "times"

Also V(esc)^2 looks like their squaring the subscript and not the parameter itself.

It`s called "phase shifting" and it`s totally physically true. You`d just have to be going extremely fast.

Year 12 Astrophysics right there.

That joke never gets old.

well the left part of the equation is the gravitational energy it takes to get from 0 to radius R with an object of mass mx away from the mass M. This is the integrated gravitational force -G_x (m_x M)/r^2 from 0 to R. I guess they made a mistake here, because you normally don`t start at 0 but rather at a point with the radius of the mass m (for example the radius of the earth). Although a little extra energy won`t hurt when escaping the Earths pull, I guess that if NASA can`t do any better in math we don`t have to wonder when the space ships crash...

On the other side we have the kinetic Energy of the mass mx, thus by equating these two energies we get the escape velocity V_ESC it takes the Object of mass mx to escape the gravitational pull to mass M.

I`m an electrical engineer so this really is not my field, so you can keep any errors :-D.

It defines the relative speed of two bodies attracted toward each others by heir gravity.

So in this context the formula means:

If you can read this you`re too close you dumbass

AYE!

duffytoler had it right. Notice the "m" term vanishes, indicating escape velocity is dependent on the mass of the planet, not the mass of the spacecraft or projectile. Fun fact for the evening, I suppose.

EVERYTHING is Bush`s doing. Didn`t you know that? LOL

She was a platinum blonde and she claimed she had a tattoo which read "carpe diem" just below her bikini line.

Natrually enough, I asked her if the "carpe" matched the curtains, which I thought was hilarious.....she didn`t!

Funny bumper sticker btw!

Nice, in a kind of Big Bang Theory way.

-Sometimes the comments on IAB are just as funny if not funnier than the actual post... I love this site. "BORED MEMBERS" FTW!!!!!!

Nice, in a kind of Big Bang Theory way.

This is just the escape velocity from a 2nd order differential equation. Pretty boring, actually. Although, I would disagree on this being a common equation. You have probably seen the gravitational potential energy equation 9which is very similar) in lower division physics, but this equation doesn`t usually show up until advanced mechanics. But every prof teaches differently.

I worked at a National Lab for 5 years. And yes, most of us are def. not virgins. One of the reasons I left the lab is because everyone is sleeping with each other. Like a big giant nerdy f-fest. Pretty crazy.

You are a fool. If you knew anything about this you would know exactly what those variables are; they are very standard im afraid my uneducated friend. It is merely equating gravitational potential energy to the kinetic energy needed to overcome said energy (ie send a rocket into space)

I dont think its supposed to be a joke its just a very geeky numberplate

Yep. 1upper.

For every action there is an equal and opposite reaction.

Understand that and you understand rockets.

Where V stands for the escape velocity.

Essentially, you`re solving for V. The equation can be simplified, however, by dividing m off of both sides and multiplying both sides by 2, leaving you with

2GM/R = V^2

From there you can easily take the square root of both sides to leave yourself with V.

They made this seemingly complicated on purpose.

m = Vehicular mass (factor it out)

M = Planetary mass

R = Radius of the planet

Vesc = Escape velocity

Try v = sqrt(2GM/r) [/quote]

Thank you! Now I get it.

Meh.. I still don`t get it..

Nobody uses "x"s for multiplication, the "squared" superscript is on the "esc" instead of the whole variable Vesc, and mass is irrelevant for escape velocity.

Try v = sqrt(2GM/r)

The escape velocity of Earth comes to about 11.2 kilometers per second from the surface, assuming you launch perpendicular to the ground.

m=mass of rocket

M=mass of earth

R=distance between the two

GmM/R=gravitational force

Vesc= Escape velocity

This equation is used to calculate the escape velocity only accounting for the gravitational attraction between the two objects.

theyre finding escape velocity by substituting e=GmM/r into e= .05mv^2

Oh thank god I`m not the only one. XD

m = mass 1 (rocket)

M = mass 2 (Earth)

R = radius (distance between the center of earth and the rocket)

V(esc) = escape velocity

just sayin`

- But we`re just not up on our rocket science and astrophysics enough to get the joke. Any geniuses out there?