Question #f66fb

The most easy way to show its effects is considering 4 equal masses (m each of them) placed in every corner of a square of side `2a` and calculate their interaction with a distant mass (M). The distance from the center of square to this mass is `r`. It is very important that `r` is huge in comparison with `a`.

By using the Newton's gravitational law

`F=G*(M*m)/R^2`

`r`, the distance between the mass M and the center of square, can be written in its components `x` and `y`

`r^2 = x^2+y^2`

the same for the distance from mass M to every corner of square

`r_1^2=(x-a)^2+y^2`
`r_2^2=x^2+(y-a)^2`
`r_3^2=(x+a)^2+y^2`
`r_4^2=x^2+(y+a)^2`

the total gravitational force in the system is

`F=G*M*m*[1/r_1^2+1/r_2^2+1/r_3^2+1/r_4^2]`

Taken in account the approximation formula

`1/(x+epsilon) ~~ 1/x *(1-epsilon/x)`

`F~~G*(M*4m)/r^2*[1-a^2/r^2] =`

`= G*(M*4m)/r^2 -G*(M*4m*a^2)/r^4`

In this expression, we have 2 elements. The first one

`G*(M*4m)/r^2` is the interaction of 4 masses (m) as if they were

concentrated in a single point, just in the center of the square. The

second one `Q =−G⋅(M⋅4m⋅a^2)/r^4` is the gravitational

quadropole for this example. See that this element is negative and decrease faster with distances.

This concept can be generalised to any geometrical configuration with 4 masses and also to larger number of masses (octopole, hexapole, etc.). When we have a extensive mass, its gravitational effects can be approximated in this way (point mass plus gravitational dipole plus gravitational quadrupole plus ...).