Question

Two identical thin uniform rods of mass M and length L are placed in a line at separation of L. Find the gravitational force acting between them :-(seperation is measured from ends of the rod)

Answer 1

The gravitational force acting between the two rods is (GM^2)/(L^2) * ln(4/3)

Answer 2

To find the gravitational force between two identical thin uniform rods, we can use the principle of superposition by integrating the force between infinitesimal mass elements.

Let the mass of each rod be $M$ and its length be $L$.
The linear mass density of each rod is $\lambda = \frac{M}{L}$.

Let's set up a coordinate system.
Place the first rod (Rod 1) along the x-axis from $x=0$ to $x=L$.
The second rod (Rod 2) is placed in line with the first rod, with a separation of $L$ between their ends.
So, the left end of Rod 2 is at $x=L+L=2L$, and its right end is at $x=2L+L=3L$.
Thus, Rod 2 extends from $x=2L$ to $x=3L$.

Consider an infinitesimal element of mass $dm_1$ on Rod 1 at position $x$.
$dm_1 = \lambda dx = \frac{M}{L} dx$.

The gravitational force exerted by this element $dm_1$ on the entire Rod 2 can be calculated using the formula for the gravitational force between a point mass and a uniform rod.
The force between a point mass $m_p$ and a rod of mass $M_R$ and length $L_R$, where the point mass is located at a distance $a$ from the nearest end of the rod along its axis, is given by:
$F = \frac{G m_p M_R}{a(a+L_R)}$

In our case, the point mass is $dm_1$, the rod is Rod 2 (with mass $M$ and length $L$).
The distance $a$ from $dm_1$ (at position $x$) to the nearest end of Rod 2 (which is at $x=2L$) is $a = 2L - x$.
So, the infinitesimal force $dF$ exerted by $dm_1$ on Rod 2 is:
$dF = \frac{G \cdot dm_1 \cdot M}{a(a+L)} = \frac{G \left(\frac{M}{L} dx\right) M}{(2L-x)((2L-x)+L)}$
$dF = \frac{GM^2}{L} \frac{dx}{(2L-x)(3L-x)}$

To find the total gravitational force $F$, we need to integrate $dF$ over the entire length of Rod 1, i.e., from $x=0$ to $x=L$:
$F = \int_{0}^{L} \frac{GM^2}{L} \frac{dx}{(2L-x)(3L-x)}$
$F = \frac{GM^2}{L} \int_{0}^{L} \frac{dx}{(2L-x)(3L-x)}$

We can use partial fraction decomposition for the integral term:
$\frac{1}{(2L-x)(3L-x)} = \frac{A}{2L-x} + \frac{B}{3L-x}$
$1 = A(3L-x) + B(2L-x)$
Setting $x=2L$: $1 = A(3L-2L) \implies 1 = AL \implies A = \frac{1}{L}$
Setting $x=3L$: $1 = B(2L-3L) \implies 1 = B(-L) \implies B = -\frac{1}{L}$

So, the integral becomes:
$\int_{0}^{L} \left( \frac{1/L}{2L-x} - \frac{1/L}{3L-x} \right) dx = \frac{1}{L} \int_{0}^{L} \left( \frac{1}{2L-x} - \frac{1}{3L-x} \right) dx$
$= \frac{1}{L} \left[ -\ln|2L-x| - (-\ln|3L-x|) \right]_{0}^{L}$
$= \frac{1}{L} \left[ \ln|3L-x| - \ln|2L-x| \right]_{0}^{L}$
$= \frac{1}{L} \left[ \ln \left| \frac{3L-x}{2L-x} \right| \right]_{0}^{L}$

Now, evaluate the definite integral:
$= \frac{1}{L} \left[ \ln \left( \frac{3L-L}{2L-L} \right) - \ln \left( \frac{3L-0}{2L-0} \right) \right]$
$= \frac{1}{L} \left[ \ln \left( \frac{2L}{L} \right) - \ln \left( \frac{3L}{2L} \right) \right]$
$= \frac{1}{L} \left[ \ln(2) - \ln\left(\frac{3}{2}\right) \right]$
Using the logarithm property $\ln a - \ln b = \ln(a/b)$:
$= \frac{1}{L} \ln \left( \frac{2}{3/2} \right) = \frac{1}{L} \ln \left( \frac{4}{3} \right)$

Substitute this back into the expression for $F$:
$F = \frac{GM^2}{L} \cdot \frac{1}{L} \ln \left( \frac{4}{3} \right)$
$F = \frac{GM^2}{L^2} \ln \left( \frac{4}{3} \right)$