1. Mar 4, 2014

### QuantumX

1. The problem statement, all variables and given/known data

I had to calculate an orbital speed for an arbitrary density ρ(r)

I got:

V = r*sqrt(G*ρ(r)*4π/3)

2. Relevant equations

I am supposed to leave this in integral form, which I don't know how to do as I do not understand calculus. ρ(r) is density of r, hence the calculus required

Please help me transform this in integral form. If you could solve it for me too, that would be amazing!

Thank you so much!

2. Mar 4, 2014

### Ray Vickson

If we could solve it for you it would be more than amazing: it would be against all the PF rules. You are supposed to show your work first before we can give any hints---and only hints, nothing more.

3. Mar 4, 2014

### jackarms

Not quite sure what you mean by integral form. Do you mean that your equation gives you the speed for a certain value of r, and then you want to integrate over a range of r to get the total speed? I guess I'm wondering what V in your equation means, and also r, and also what you want to solve for.

4. Mar 4, 2014

### QuantumX

@Ray: The equation is my work. I just need to convert it to an integral.

@jackarms: Yes, the V in the equation stands for orbital speed, and r stands for the orbital radius. The density ρ(r) is the density inside the radius r, so the density is a function of the radius. I want to solve for V.

5. Mar 4, 2014

### jackarms

Is this the general equation you're working with?
$$v = \sqrt{\frac{GM}{r}}$$

6. Mar 4, 2014

### QuantumX

Yes, v = sqrt(GM/r) and M = 4π/3*r^3*ρ

Therefore, v = r*sqrt(G*ρ*4π/3)

And I need to express ρ as a function of the radius (which is why calculus is needed). So:

v = r*sqrt(G*ρ(r)*4π/3)

I apologize for not using proper notation, I'm new and still figuring things out.

7. Mar 4, 2014

### jackarms

Oh that's all right. It takes some getting used to.

Okay, I see where the calculus comes in now. The issue for this problem is that the density varies continuously in the mass, so multiplying the density by the volume won't work anymore. You can instead express a differential amount of mass using the expression:
$$dm = \rho (r) dV$$
Where V is the differential volume element. To get the total mass, you need to integrate this expression. If you haven't done much calculus before, I can walk you through it.

8. Mar 4, 2014

### QuantumX

Yes, the density varies with the radius within the mass, which is why calculus is required, otherwise if the density was uniform, it wouldn't depend on the radius... I think.

I derived the equation from a previous problem, and then this problem just said "Calculate the orbital speed for an arbitrary density ρ(r). You can leave this in the form of an integral".

Supposedly leaving it in the form of an integral means you don't have to do calculus... But I don't even know how to leave it in integral form.

9. Mar 4, 2014

### jackarms

Yes, that's right. It looks like the only thing in the problem that requires an integral is the mass, so let's just do that part.

The expression above just gives you the mass of a small bit of mass (dm) given the density at a certain radius and the volume of that small chunk (dV). Since the density varies radially outward, all points at the same radius outward have the same density, and so these points will lie along spheres. dm will be the mass between two of these spheres. If you put an integral on both sides, you get:
$$\int dm = M = \int \rho (r) dV$$
dV is the volume between two spheres -- this you can write as $4\pi r^{2}dr$ where dr is a small change in the radius. You want to sum up all of these spheres from the inside of the mass to its radius, call it R. The expression then becomes:$$M = \int^{R}_{0} \rho (r) 4\pi r^{2}dr$$
It's a lot of calculus stuff, but just trust that the expression works. Basically all you're doing is dividing the mass into shells, finding the mass of an individual shell, and then adding them all up.

10. Mar 4, 2014

### QuantumX

Thanks! I have to ask though, the question asked to calculate the velocity, not the mass...?

11. Mar 4, 2014

### jackarms

Yes, but don't you have an expression for velocity in terms of the mass?

12. Mar 5, 2014

### QuantumX

Hm, that makes sense. Thanks!