# Tidal force

1. Jan 28, 2005

### matpo39

I am having a little trouble with this problem it is as follows:

consider the tidal force #(F_tid=-G*M*m[(d_unit vector/d^2)-(d_0 unit vector/d_0^2)]) on a mass m at the position P. write d as (d_0-R(radius of earth))=d_0*(1-R/d_0) and use binomial approximation to show that F_tid= -(2*G*M*m*R/d_0^3)x_unit vector.

sorry i cant get the picture up but all it is is the earth with center at (0,0) and point P is located all the way to the left edge of the earth on the x axis and and 0 on the y axis. the moon is to the left of the earth and is also on the x axis. which is why it is easy to see that the force will be in the -x direction.

first i used the binomial expansion and got d_0(1+2*R/d_0) and i replaced d in equation # with that value anf got this

-G*M*m[d_unit vector/(d_0(1+2R/d_0))^2 - d_0_unit vector/(d_0^2)]
I have been fiddleing with it all day and cant get it to match the force they said it should, im pretty sure that my problem is coming from not really knowing how to handle the d,d_0 unit vectors.

thanks for the help

2. Jan 29, 2005

### Andrew Mason

I gather that the question is asking you to determine:

$$F_{tide}=-GMm(\frac{\hat d}{d^2} - \frac{\hat d_0}{d_0^2})$$

which can be rewritten:

$$F_{tide}=-GMm(\hat d\frac{d}{d^3} - \hat d_0\frac{|\vec d - \vec R|}{|\vec d - \vec R|^3})$$

I can see why you are having problems. That denominator $|\vec d-\vec R|^3$ is messy and requires a complicated binomial expansion to solve. I had to look up the derivation in a mechanics text - Barger and Olsson, Classical Mechanics (First ed.) at pages 268-270 - it is not trivial to find the general solution. For the situation where d and d0 are colinear, it reduces to:

$$F_{tide}=-GMm\hat d(\frac{d}{d^3} - \frac{d - R}{(d - R)^3})$$

But its still a lot of work.But you can see that the numerator is the order of R and the denominator in the order of d^3.

AM

Last edited: Jan 29, 2005