Gravitation between particle and rod. Simple integration. Problem with answer.

[elvinc]

Hi,

I have attempted a basic gravitation question from “Classical Mechanics” R. Douglas Gregory (1st ed). I get the answer provided at the back of the book, but it doesn't make sense to me. Can someone help me interpret the answer?

Question: Pg 71, Q3.3

A uniform rod of mass M and length 2a lies along the interval [-a, +a] of the x-axis and a particle of mass m is situated at the point x = x' (*). Find the gravitational force exerted by the rod on the particle.
<< end of first part of question >>

(*) Although not explicitly stated this means the particle is lying on the rod on the x-axis.

Approach:

Using the method from a previous example, split the rod into short segments of length \delta x then applying the inverse square law for universal gravitation the force of attraction between the particle and that small section of the rod is:

\frac{Gm}{(x^{\prime} - x)^2}.\frac{M\delta x}{2a}

Rearranging slightly and summing all these – this becomes an integral

\frac{GmM}{2a} \int_{-a}^{+a} \! \frac{dx}{(x^ \prime - x)^2}

That boils down to

force of attraction = \frac{GmM} {(x^ \prime)^2 - a^2}

which is the answer in the back of the book. Good. Well maybe not.

If I substitute in a couple of values

x^\prime = 0 (the particle is placed at the centre of the rod)

then the expression for the force becomes

-\frac{GmM} {a^2}

a negative force. I don't understand that.


x^\prime = +a (the particle is placed at one end of the rod)

then the expression for the force becomes

\frac{GmM} {a^2 - a^2}

an “infinite” force. I don't understand that.

What is wrong here?


Thanks,

Clive

 

[DrGreg]

If you've quoted the book correctly, the book is wrong. Clearly the force ought to be zero when x' is zero.

The problem is the integral doesn't take account of the direction of the force. You need to split it into two integrals from -a to x' (positive force) and from x' to a (negative force).

Are you sure you haven't misread the question?

 

[rcgldr]

If the particle is inside the rod at the center of the rod, the force is zero. If the particle is on the surface of the rod, then as the diameter of the rod -> 0, then the force -> ∞.

Also

\int \frac{dx}{(x^ \prime - x)^2} = \frac{1}{x^\prime - x}

which aproaches ∞ as x approaches x^'.

To get around this problem assume that the forces equal distant from x^' cancel. If x^' = 0, then the force is zero. If x^' < 0, then the net force is the force from (2 x^' + a) to (+a). If x^' > 0, then the net force is the force from (-a) to (2 x^' - a). If x = -a or +a, then there there are no forces to cancel and the force is ∞. If x < -a or x > +a, then the net force is the force from -a to +a.

If the rod had finite diameter, then the rod could be considered as a series of thin disks, and the math for intensity is covered here:

http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elelin.html#c3

For gravitational intensity, replace k with G, and σ becomes mass per unit area of the disk. Force equals intensity times mass of particle. In this case, as the distance from a disk of finite radius approaches zero, the intensity approaches a constant (2 π G σ ) instead of ∞, so the problem is solvable.

 

[elvinc]

I'm going to approach this problem by cancelling ``balanced'' forces then integrating what is left.

If the particle is at position x^\prime then its distance from the right-hand end (+a) is a - x^\prime. I'm assuming that x^\prime is in the positive part of the x-axis but I'm pretty confident the argument works wherever it is located along the rod.

So the forces from that part of the rod to the right of the particle at position x^\prime are equal and opposite to the forces from same length of rod to the left of the particle. The left-hand point on the rod giving equal length to the portion of the rod to the right of the particle ends at position x^\prime - (a - x^\prime) = 2x^\prime - a. Therefore we can ``cancel" the right-directed and left-directed forces acting on particle between 2x^\prime - a and +a on the rod. So the integral for the ``remaining force'' at the far left of the rod is


\frac{GMm}{2a}\int_{-a}^{2x^\prime - a} \! \frac{dx}{(x - x^\prime)^2} \, dx


I have laboriously written out the algebra, in small steps, so it may be easier to spot an error in my algebra.


=\frac{GMm}{2a}\left[-(x - x^\prime)^{-1} \right]_{-a}^{2x^\prime - a}

=\frac{GMm}{2a}\left[-(2x^\prime -a - x^\prime)^{-1} + (-a -x^\prime)^{-1} \right]

=\frac{GMm}{2a}\left[+\frac{1}{(a - x^\prime)} - \frac{1}{(a + x^\prime)} \right]

=\frac{GMm}{2a}\left[\frac{(a + x^\prime) - (a - x^\prime)}{(a^2 - (x^\prime)^2)} \right]

=\frac{GMm}{a}\left(\frac{x^\prime}{a^2 - (x^\prime)^2} \right)

=\frac{x^\prime}{a}\left(\frac{GMm}{a^2 - (x^\prime)^2} \right)

This is subtly different from the expression in my first post because of this \frac{x^\prime}{a} term multiplying the whole expression for the force.


Now I will substitute some particular values.

If {x^\prime} = 0, this is equivalent to the particle being positioned at the middle of the rod. The \frac{x^\prime}{a}\left(\frac{GMm}{a^2 - (x^\prime)^2} \right) equals zero. This makes sense because the forces pulling the particle to the left and right are equal and opposite.

If {x^\prime} = +a, this is equivalent to the particle being positioned at the right-hand end of the rod. The \frac{x^\prime}{a}\left(\frac{GMm}{a^2 - (x^\prime)^2} \right) now has a zero denominator and we are back into the ``infinity'' situation.

Can this question be solved by this method or is there no solution in this ``zero thickness rod'' situation?

Regards,

Clive Long

 

[Sourabh N]

The infinity solution is bound to come, due to the test particle being infinitesimally close (infact sitting on) to the rod. For example, assume that in place of rod, there are two particles, one at x = -a, one at x = +a. You can find finite expression for force at all places except x = -a or +a, because force is infinite there.

 

[elvinc]

<< snip >>
You can find finite expression for force at all places except x = -a or +a, because force is infinite there.

What is special about the end positions of the rod? Surely, the point is exactly on the rod at any point between -a and +a and the infinities would arise at any position?

Clive

 

[Sourabh N]

They are special because they are the end points (lol)

Suppose the test particle is at (a - \delta(a very small value)). You agree the force won't be infinite as long as \delta is finitely small?

For any other point inside the rod, you assumed that the -x' to +x' part cancels out, right? In doing this, you really jumped over the point where your test particle was sitting. (sorry if the language is not perfect, I'm bad at this ) For the end point, you can't do this because there nothing on the other side.

I hope you get the point I'm trying to make here.