# Finding the gravitational field of a rod

1. Apr 18, 2012

### jjr

1. The problem statement, all variables and given/known data
A homogeneous rod with even thickness has mass M and length L.
a) Find the gravitational field g(x) at a distance x from the rod on an axis normal to the midpoint of the rod.
b) Find the gravitational field g(y) at a distance y from the rod's midpoint on an axis through the rod in the direction of its length.

2. Relevant equations

Gravitational potential V(r) = -GM/r (1)
Gravitational field g(r) = -∇V(r) (2)

Where G is the gravitational constant, M is the mass of the relevant object and r is the distance from the object.

3. The attempt at a solution

a) My plan was to first work out the gravitational potential through integrating the contributions from infinitesimal lengths of the rod (from -L/2 to L/2) and then use equation (2) to find the field. The mass per length of the road is μ = M/L, so an infinitesimal piece of the rod dm = μ dl, where dl is an infinitesimal length. I got dV(r) = -$\frac{G dm}{r}$ = -$\frac{G μ dl}{r}$. The r is (pythagoras) √(x2 + l2), where l is the distance from the midtpoint of the rod. The problem arises when I try to integrate this function dV(r) = -$\frac{G μ dl}{√(x^2 + l^2)}$ from -L/2 to L/2. Here's when I get in trouble. This integral is a bit hairier than expected, and I find it hard to get a reasonable answer. Could someone point me in the right direction here?
I'll post my attempts at the solution for b) if I can't figure it out after getting help with a).

Thanks,
J

2. Apr 18, 2012

### Curious3141

I didn't really check the rest of your work, but if you want to integrate $\frac{G μ dl}{√(x^2 + l^2)}$,

try expressing that as $\frac{\frac{G μ}{l} dl}{√(1 + \frac{x^2}{l^2})}$ then making the substitution $\frac{x}{l} = y$ to get it into a simpler form, then substitute $y = \sinh{u}$.

3. Apr 19, 2012

### jjr

Tried solving the integral and evaluating it from -L/2 to L/2, but it doesn't look legit. I think my reasoning has failed me at some point while setting up the expression for dV(r). The answer is suppose to be g(x) = -2 $\frac{G M}{L x}$ * $\frac{1}{√(1+(2x/L)^2)}$ i. Suggestions?

J

4. Apr 19, 2012

### ehild

I think the integral is easier if you determine the field intensity directly.

ehild

Last edited: Apr 19, 2012
5. Apr 19, 2012

### jjr

Right you are, I was perhaps a bit vague at this point; if I can find the scalar gravitational potential field the grav. vector field is given to be its gradient.

6. Apr 19, 2012

### ehild

Yes, I noticed it at the end. Try to find the field directly. It leads to an easier integral.

ehild

7. Apr 19, 2012

### ehild

And calculate with theta, using y=xtan(θ), r=x/cos(θ).

ehild

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