Rigid body rotation near galactic center

[clandarkfire]

Homework Statement

Equate the expression for centripetal acceleration with the gravitational acceleration to show that the central parts of the galactic rotation curve are consistent with rigid body rotation.

The attempt at a solution
Say a star near the galactic center has mass m and the rest of the mass inside its orbit has mass M. Then:
[tex]F=ma=\frac{GMm}{r^2}[/tex]
[tex]a=\frac{v^2}{r}[/tex]
[tex]\frac{GM}{r^2}=\frac{v^2}{r}\Rightarrow{v^2} \propto{r}[/tex]
But what I want to show is that v is proportional to r, not that v^2 is proportional to r. What have I done wrong?

 

[tiny-tim]

hi clandarkfire!

M is proportional to … ?

 

[clandarkfire]

hi clandarkfire!

M is proportional to … ?

I'm actually not sure. If it's proportional to r (seems reasonable), I get [tex]G \propto {v^2}[/tex], which seems pretty nonsensical.
If it's r^2 or r^3, I still don't think I get [tex]r \propto v[/tex], which is what I'm looking for.

 

[haruspex]

I'm actually not sure. If it's proportional to r (seems reasonable), I get [tex]G \propto {v^2}[/tex], which seems pretty nonsensical.
If it's r^2 or r^3, I still don't think I get [tex]r \propto v[/tex], which is what I'm looking for.

Try r^3 again, and if it still fails please post your working.

 

[Nickie]

It appears that you have made a small mistake in your calculation. The equation for centripetal acceleration is a= \frac{v^2}{r}, not a= \frac{GM}{r^2}. Therefore, the correct equation should be:

a= \frac{v^2}{r} = \frac{GM}{r^2}

This shows that the centripetal acceleration is equal to the gravitational acceleration, which is consistent with rigid body rotation. The proportionality between v and r can be seen by rearranging the equation to v= \sqrt{\frac{GM}{r}}.

In conclusion, your approach is correct, but you have made a small error in the equation for centripetal acceleration. Keep in mind to always double check your calculations to avoid any mistakes.