Question

Help finding gravitational attraction and speed?

Several planets (Jupiter, Saturn, Uranus) are encircled by rings, perhaps composed of material that failed to form a satellite. In addition, many galaxies contain ring-like structures. Consider a homogeneous thin ring of mass 3.0 x 10^22 kg and outer radius 6.8 x 10^8 m (the figure). (a) What gravitational attraction does it exert on a particle of mass 86 kg located on the ring's central axis a distance 4.0 x 10^8 m from the ring center? (b) Suppose that, starting at that point, the particle falls from rest as a result of the attraction of the ring of matter. What is the speed with which it passes through the center of the ring?

Answer 1

We will be using Newton's Law for Gravitation: `vec F = -(Gm_1 m_2)/|vec r_1 - vec r_2|^2 vec e_(vec r_1 - vec r_2)`

The green ring in the drawing radius `R`, mass `M` has linear density `sigma = M/(2 pi R)`.

So, for the small arc, `dl = R d theta` and `dM = sigma R d theta`.

The particle of mass `tilde m` is a distance `X` from the ring's centre sitting on its axis . That allows us to use symmetry.

The side view shows that from symmetry the distance `D` between every small arc and the particle is given by: `D^2 = R^2 + X^2`.

So from Newton's Law:

`abs(dF) = -(G \ dM \ tilde m)/D^2`

That is the magnitude of the force between the particle and the small element in the ring. Again we see symmetry demands that the particle be pulled along the axis of the ring. We need to resolve `dF` along the axis. See side view - from symmetry the angle `phi` will be the same for each small element.

`implies d vec F_X =- (G dM tilde m)/D^2 cos phi \ vece_X`, where `cos phi = X/D`

Putting it all together:

`d vec F_X =- (G dM tilde m)X/D^3 \ vece_X`

`vec F_X =- G sigma R tilde m X/(R^2 + X^2)^(3/2) int_0^(2 pi) d theta \ vece_X`

`vec F_X =- G color(red)(sigma 2 pi R) tilde m X/(R^2 + X^2)^(3/2) \ vece_X`

`vec F_X =- G M tilde m X/(R^2 + X^2)^(3/2) \ vece_X`

Part (a)

`M = 3 xx 10^22 kg`

`R = 6.8 xx 10^8 m`

`tilde m = 86kg`

`X = 4 xx 10^8 m`

`G = 6.67 xx 10^(-11) \ m^3 \ kg^(-1) \ s^(-2)`

`implies absF = 1.4 xx 10^(-4) N`

Part (b)

There are several ways to do this. We can calculate the potential energy `U` of the particle from the idea that: `vec F = - nabla U`

`implies U = G M tilde m int X/(R^2 + X^2)^(3/2) dX`

`implies Delta U= - G M tilde m [ 1/(R^2 + X^2)^(1/2) ]_(4 xx 10^8)^0`

`= -34940 \ J`

In a conservative system, that energy becomes kinetic energy for the particle:

`1/2 tilde m v^2 + Delta U = 0 implies v = 28.5 \ ms^(-1)`