Gravitationally Equivalent Point Particle

Classical Mechanics Level 4

There are two objects with the same mass $M$ , situated on the $x$ -axis.

One is a thin rod with its mass uniformly distributed between $x = 0$ and $x = 1$ .

The other is a point particle at $x = 2$ .

If we wanted to replace the rod with a point particle of mass $M$ which would exert the same gravitational force on the point particle at $x = 2$ , what should its $x$ -coordinate be (to 3 decimal places)?

The answer is 0.586.

1 solution

Guilherme Niedu
Nov 30, 2016

Let us call the rod mass linear density as $\rho$ , which will be equal to $\frac{M}{L}$ , where $L$ is the rod lenght, equal to $1$

The gravitational pull felt by the point particle is:

$g = \frac{GM(r) }{r^2}$

Being $r$ the distance between the rod and the point particle

$\large g = \int_{r_{min}}^{r_{max}} \frac{G\rho(r)dr}{r^2}$

$\large g = \int_{2}^{1} \frac{GMdr}{r^2}$

$\large g = \frac{GM}{r} \Big|_2^1$

$\large g = \frac{GM}{2}$

This should be the same as a pull from a distance $r^*$ . So:

$\large \frac{GM}{2} = \frac{GM}{r^{*^2}}\$

$r^* = \sqrt{2}$

And the x coordinate is:

$\color{#3D99F6} x = 2 - r^* = \fbox{0.586...}$

Gravitationally Equivalent Point Particle

The answer is 0.586.

1 solution

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