Recent content by Observatoren

Isnt the solarsystem the remnants of an accretion-disc from the sun during its birth? Maybe someone has already mentioned this but I mention it anyways :) And what I want to point out by this, if this is the case, is that then it seems logic that all the planets orbit around the sun in much the...

Thank you Naty1, very guiding! :)

I thought I might post a question on the subject. When I read about this experiment I often see that we can deduce from it that observation makes reality collapse into what we observe. That the very action of observation does this. But by this they mean observation as in measuring right? With...

When i look at x^3 i know that means x*x*x But i can't know that, for example 27^1/3 means 3*3*3, without knowing that 3 is the cuberoot out of 27. See what i mean? There is now obvious way to do this calculation in your head like with x^3. I want to think of x^1/3 like x times one third of...

Ive thought a little today about (a^3)^1/3=a When you square something you take a number and multiply itself by itself. When you make a number cubed you multiply the number 3 times...and so on. But what are you actually doing when you heighted something to a fraction (or rational number)...

It could be that I am not using my scientific calculator. But ill think about this for a bit and see if i get it. Your help is very much appreciated my friend, very much indeed! Couldnt feel more welcome :)

It could be that i am not using my scientific calculator. Ill think about this a bit and see if i get it. Your help is very much appreciated my friend! very much indeed. :)

Oh right of course thnx!

Normally i would have done this without question. But when i put some hypothetical values for the variables and calculate it like: (2M/m)^1/3r, i get another value than what i get when i just take the cuberoot out of the whole right side as it is (with the same hyp. values). And i know that the...

Now i have: 2GMr^3=D^3Gm. Next i subtract G from both sides so i get: 2Mr^3=D^3m Now i divide both sides with m, which gives: D^3=2Mr^3/m Now, I am not really sure...but to get rid of the cube of D and r i have to (hope i express this right now, I am from sweden so i have some problem...

Ooh no I get what you mean now its too late to do this now sorry! You meant multiplying so that I would get rid of the fractions of course! Sorry... well: 2GMr^3=D^3Gm Right? Sorry for spamming...

You really have to approach me like a child. I have no Uni ed. Just curiosity!

So what I get would be something like this; (2GMr)2/D^6=(Gm)3/r^6 Right?

I appreciate your pedagogic approach but would it be rude of me to ask for a step by step explanation?? :s

Ill start by appologizing for not using LaTex, but i haven't figured out how it works yet and I am a little bit impatient. The equation is probably a simple one. I am not very trained in math but happen to have some interest in physics, especially astronomy, and this is used in astronomy but i...