Solving the Tidal Force Problem

[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 can't 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 can't get it to match the force they said it should, I am pretty sure that my problem is coming from not really knowing how to handle the d,d_0 unit vectors.

thanks for the help

 

[Andrew Mason]

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.

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

 

[wais]

Hi there,

I understand that you are having some difficulty solving the tidal force problem. I will try my best to explain the solution in a step-by-step manner.

Firstly, let's rewrite the expression for d as d = d_0(1-R/d_0). This is just a rearrangement of the given expression.

Next, let's expand the binomial term (1-R/d_0)^2 using the binomial expansion formula. This gives us 1 - 2R/d_0 + (R/d_0)^2.

Now, let's plug this expanded expression into our original expression for tidal force:

F_tid = -G*M*m[(d_unit vector/d^2) - (d_0 unit vector/d_0^2)]
= -G*M*m[(d_unit vector/(d_0(1-R/d_0))^2) - (d_0_unit vector/(d_0^2))]
= -G*M*m[d_unit vector/(d_0^2(1 - 2R/d_0 + (R/d_0)^2)) - (d_0_unit vector/(d_0^2))]
= -G*M*m[d_unit vector/(d_0^2 - 2Rd_0 + R^2) - (d_0_unit vector/(d_0^2))]
= -G*M*m[d_unit vector/(d_0^2) - 2Rd_0_unit vector/(d_0^2) + R^2_unit vector/(d_0^2) - d_0_unit vector/(d_0^2)]
= -G*M*m[(d_unit vector - 2Rd_0_unit vector + R^2_unit vector - d_0_unit vector)/(d_0^2)]

Now, we can simplify the expression in the brackets by grouping similar unit vectors together:

(d_unit vector - 2Rd_0_unit vector + R^2_unit vector - d_0_unit vector) = (d_unit vector - d_0_unit vector) - 2R(d_0_unit vector) + R^2_unit vector
= (d_unit vector - d_0_unit vector) - (2Rd_0_unit vector - R^2_unit vector)
= (d_unit vector - d_0_unit vector) - R(2d_0_unit vector - R_unit vector)

Note that