Probably an easy problem for anyone here

[Observatoren]

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 think this is a better place to ask this question, hope i haven't 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 explanation would be appreciated. And again, excuse me if I am not following the general rules for posting but Ill better myself in the future.

Thanks!

 

[Doc Al]

How do I get from 2GMr/D^3=Gm/r^2 to D=(2M/m)^1/3r

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:
$$\frac{a}{b} = \frac{c}{d}$$
Multiply both sides by b and then d.

 

[Observatoren]

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

 

[Doc Al]

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

I showed you the first step. Have you done it?

 

[Observatoren]

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

Right?

 

[Observatoren]

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

 

[Observatoren]

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...

 

[SammyS]

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

Right?

Wrong.

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

What do you get ?

Added in Edit:

OK! I see you did get the first step.

 

[SammyS]

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 want to solve D.

First solve for D³ .

 

[Doc Al]

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?

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?

 

[Observatoren]

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 expressing myself mathematicaly) take the cuberoot out of them. But i can't 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 that's to high...now I am stuck again...

 

[Doc Al]

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

Good.

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 expressing myself mathematicaly) take the cuberoot out of them.

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

But i can't take the cube root 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 that's to high...now I am stuck again...

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.)

 

[Observatoren]

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 don't 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...

 

[superdave]

Now i have: 2GMr^3=D^3Gm. Next i subtract G from both sides so i get: 2Mr^3=D^3m

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.

 

[Observatoren]

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.

Oh right of course thnx!

 

[Doc Al]

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).

Then you are making a calculation error.

And i know that the right side is supposed to look like it does above but...i don't 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...

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 = a^1/3*b^1/3

 

[Observatoren]

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. :)

 

[Observatoren]

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 :)

 

[Observatoren]

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). Cause you can't say that you multiply itself by a third of itself, so what is it that you do?

 

[superdave]

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 s^1/3, you are saying "the number you multiply by itself 3 times to get x".

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

 

[Observatoren]

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 itself. I think the notation is a bit confusing. Cant see the logic :(
A power no matter how big or small indicates that something should be multiplied to me. But i guess that powers and roots are the inverse of each other. Somewhere in the back of my head i think i know that I am ****ing this up more than i should...

Is there a way of finding roots not using a calculator btw??