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Probably an easy problem for anyone here

  1. Nov 29, 2012 #1
    Ill start by appologizing for not using LaTex, but i havent figured out how it works yet and im a little bit impatient.
    The equation is probably a simple one. Im not very trained in math but happen to have some interest in physics, especially astronomy, and this is used in astronomy but i think this is a better place to ask this question, hope i havent posted this thread all in the wrong place.

    To the question. How do I get from 2GMr/D^3=Gm/r^2 to D=(2M/m)^1/3r
    A step by step explaination would be appreciated. And again, excuse me if im not following the general rules for posting but Ill better myself in the future.

    Thanks!!
     
  2. jcsd
  3. Nov 29, 2012 #2

    Doc Al

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    Since you have a fraction equal to a fraction, why not start by multiplying both sides by the denominators. Then you can isolate D^3.

    Example:
    [tex]\frac{a}{b} = \frac{c}{d}[/tex]
    Multiply both sides by b and then d.
     
  4. Nov 29, 2012 #3
    I appreciate your pedagogic approach but would it be rude of me to ask for a step by step explaination?? :s
     
  5. Nov 29, 2012 #4

    Doc Al

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    I showed you the first step. Have you done it?
     
  6. Nov 29, 2012 #5
    So what I get would be something like this; (2GMr)2/D^6=(Gm)3/r^6

    Right?
     
  7. Nov 29, 2012 #6
    You really have to approach me like a child. I have no Uni ed. Just curiosity!
     
  8. Nov 29, 2012 #7
    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...
     
  9. Nov 29, 2012 #8

    SammyS

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    Wrong.

    Take the equation, [itex]\displaystyle \ \frac{2GMr}{D^3}=\frac{Gm}{r^2}\ \ [/itex] and multiply both sides by D3 and r2.

    What do you get ?

    Added in Edit:

    OK! I see you did get the first step.
     
  10. Nov 29, 2012 #9

    SammyS

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    You want to solve D.

    First solve for D3 .
     
  11. Nov 30, 2012 #10

    Doc Al

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    Exactly!

    Now, as SammyS said, since you eventually want to solve for D, first get D^3 by itself. What would you divide both sides by to get D^3 by itself on the right hand side of that equation?
     
  12. Nov 30, 2012 #11
    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, im not really sure...but to get rid of the cube of D and r i have to (hope i express this right now, im from sweden so i have some problem expressing myself mathematicaly) take the cuberoot out of them. But i cant take the cuberoot out of the whole right side. But if i change the value of r by taking the cuberoot out of it the rest of the variables are going to get values thats to high...now im stuck again...
     
  13. Nov 30, 2012 #12

    Doc Al

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    Good.

    Right. You need to take the cuberoot of both sides.

    You won't be changing the value of r. The cube root of r^3 is just r. What's the cube root of the rest? (Look at what you were trying to show from your first post. You are almost done.)
     
  14. Nov 30, 2012 #13
    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 right side is supposed to look like it does above but...i dont see how to get there. I mean, what it says above is that i take the cuberoot out of r but raises the rest to one third power. Why? To my knowledge a big cuberoot sign over the whole right side as it is when 2Mr^3/m would be enough...
     
  15. Nov 30, 2012 #14
    You got the right result, but just be careful. You divide G from both sides, not subtract. G is being multiplied by something, so you can't just subtract it out. But like i said, you got the right result anyway in this case, just be careful in the future.
     
  16. Nov 30, 2012 #15
    Oh right of course thnx!!
     
  17. Nov 30, 2012 #16

    Doc Al

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    Then you are making a calculation error.

    In case you weren't aware: Raising something to the 1/3 power is how you express taking a cube root.

    (a3)1/3 = a

    Also: (a*b)1/3 = a1/3*b1/3
     
  18. Nov 30, 2012 #17
    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. :)
     
  19. Nov 30, 2012 #18
    It could be that im 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 :)
     
  20. Nov 30, 2012 #19
    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 doin when you heighted something to a fraction (or rational number). Cause you cant say that you multiply itself by a third of itself, so what is it that you do???
     
  21. Nov 30, 2012 #20
    So when you say x3 you are saying "the number you get when x is multiplied by itself 3 times". You got that. So when you say s1/3, you are saying "the number you multiply by itself 3 times to get x".

    Think about 27. 273 = 19683. 271/3 = 3. There are 9 "1/3's" in 3, right? 1/3 * 9 = 9/3 = 3. So 39 = 279/3 = 273 = 19683
     
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