Homework Help: Working out the units, need help please

CavortingMonk · Aug 24, 2012

Please bare with me, i'm not sure how to put in the maths on a forum like this.

Ok so i was asked to combine two equations to get a third, this I have done. The equation is M = v squared x r and then the producet of those divided by G.

M is the mass of a black hole
v squared is the velocity of an object orbitting the black hole
r is the radius of the object from the centre of the black hole
G is newtons universal constant of gravitation given as 6.7 x 10 to the minus 11.

3. The attempt at a solution

The maths I don't need help with, i'm sure I have that right, it's the units i'm struggling to understand.

v is given as Km s-1
r is given simply as Km
G is given as 1 m3 kg-1 s-2

This would give units of
(Km s-1)2 x Km

m3 kg-1 s-2

I'm sure Kg is the actual unit I should end up with but i don't see how i get there with thee units. Again I apologise for not being able to enter the maths a little better, hopefully i've explained it well enough.

mbrmbrg · Aug 24, 2012

Welcome aboard!

Your units do indeed work out.
I suggest taking your equation
[tex]M=\frac{v^2×r}{G}[/tex]
and replacing your variables with the appropriate units in the form
[tex][unit_1]=\frac{[unit_2]^2×[unit_3]}{[unit_4]}[/tex]
then simplify and you should get kg=kg.

Please post details of your work if you're still having problems. You can click "QUOTE" in the bottom right-hand corner of my post to see the code I used to get clean-looking equations.

mbrmbrg · Aug 24, 2012

Oh. On re-reading this, I now understand what you mean by:

CavortingMonk said: ↑

This would give units of
(Km s-1)2 x Km

m3 kg-1 s-2

So far, that is fine. Here it is looking prettier:

[tex]\frac{[\frac{km}{s}]^2 × [km]}{[m^3]×[\frac{1}{kg}]×[\frac{1}{s^2}]}[/tex]

Simplify that. You will need to convert m to km (or km to m) to make them cancel with each other.

CavortingMonk · Aug 24, 2012

I see what you're saying, thank you also for putting it into a much neater form :)

I did think about converting the meters to kilometers but where I struggled to understand it is I thought m3 was meters cubed. How do i convert that to kilometers? I'm sure I was told how but it seems to have completely left my brain. Obviously I can convert a meter to a kilometer, but m3?

Thanks for the help, this is the only part tripping me up and i'm sure it's the simplest bit lol.

Ok I got som coffee and took another go at it. After simplufying I was left with the units Km3, m3, s2, s2, and Kg. I understand the s2, cancels out. Which leaves me with Km3, m3 and Kg.

So I can simply convert the m3 unit to Km3 right? Then I'm left with Kg and I'm done.

mbrmbrg · Aug 24, 2012

Ok I got som coffee and took another go at it. After simplufying I was left with the units Km3, m3, s2, s2, and Kg. I understand the s2, cancels out. Which leaves me with Km3, m3 and Kg.

So I can simply convert the m3 unit to Km3 right? Then I'm left with Kg and I'm done.

Exactly!

In the spirit of this forum, instead of showing you how to convert m³ to km³, I'll show you how to convert cm² to m², which is the same concept, and then you can do the actual conversion for your homework yourself.

If I wanted to convert 15cm into m, I would simply multiply by the conversion factor, like so:
[tex]\frac{15 cm}{1}×\frac{1 m}{100 cm}=0.15 m[/tex]
Multiplying by the conversion factor doesn't change the value of anything, because the conversion factor is just a convenient form of 1, and anything multiplied by 1 retains its value.

Our case is slightly trickier, because there are exponents involved. We will do the same thing, but raise our conversion factor to whatever power makes everything cancel. (This is perfectly legal, because 1^anything still equals 1). So to convert 43cm² to m² we multiply by the conversion factor squared:
[tex]\frac{43 cm^2}{1}×\left(\frac{1 m}{100 cm}\right)^2\\
=\frac{43 cm^2}{1}×\left(\frac{1^2 m^2}{100^2 cm^2}\right)\\
=\frac{43 cm^2}{1}×\left(\frac{1 m^2}{10000 cm^2}\right)\\
=0.0043 m^2[/tex]

Make sense?

CavortingMonk · Aug 24, 2012

Yep that makes sense. I had a go at it before you posted your reply and it seems to match up well with what you did, so I'm pretty happy :)

Thanks for your help.