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Tidal Potential & Binomial Approximation

  • Thread starter McCoy13
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Homework Statement


There is a derivation in the text that I'm having problems replicating. The text gives the formula for tidal potential as:

[tex]U_{tid}=-GM_{m}m(\frac{1}{d}-\frac{x}{d^{2}_{0}})[/tex]

Where [itex]M_{m}[/itex] is the mass of the moon, d is the distance from the CM of the moon to the point of interest on the surface of the Earth, [itex]d_{0}[/itex] is the distance between the CM of the moon and the CM of the earth, and x as the lateral distance of the point.

I'm having trouble deriving the potential for a point on the horizontal axis. That is [itex]d=d_{0}-r[/itex] where r is the radius of the earth, and [itex]x=-r[/itex]


Homework Equations


The text gives the answer as

[tex]U_{tid}=-\frac{GM_{m}m}{d_{0}}(1+\frac{r^{2}}{d^{2}_{0}})[/tex]

It suggests using the binomial approximation of

[tex](1+x)^{-1/2}=1-\frac{x}{2}[/tex]


The Attempt at a Solution



[tex]U_{tid}=-GM_{m}m(\frac{1}{\sqrt{(d_{0}-r)^{2}}}+\frac{r}{d^{2}_{0}})[/tex]

[tex]U_{tid}=-GM_{m}m(\frac{1}{d_{0}\sqrt{1-2\frac{r}{d_{0}}+\frac{r^{2}}{d^{2}_{0}}}}+\frac{r}{d^{2}_{0}})[/tex]

[tex]U_{tid}=-\frac{GM_{m}m}{d_{0}}(1+\frac{r}{d_{0}}-\frac{r^{2}}{2d^{2}_{0}}+\frac{r}{d^{2}_{0}})[/tex]

It is unclear to me where to proceed from here, as there are no like terms and putting them in terms of a common denominator does not seem to help. I thought of perhaps needing to do an additional binomial approximation, but it is unclear to me how that would help or even how I would go about doing it.
 
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Answers and Replies

  • #2
fzero
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Things are a bit confusing, since you don't explain the physical significance of the terms in

[tex]
U_{tid}=-GM_{m}m(\frac{1}{d}-\frac{x}{d^{2}_{0}})
[/tex]

From first principles, the potential due to the moon on an element on the horizontal axis is

[tex] U = =-GM_{m}m \frac{1}{d_0 -r} \sim - \frac{GM_{m}m }{d_0} \left( 1 + \frac{r}{d_0} + \frac{r^2}{d_0^2} + \cdots \right). [/tex]

The first term is constant and doesn't product a force. The second term produces a constant force, which is what keeps the Earth-Moon system bound. Since this term doesn't cause tides, your book seems to be subtracting it from the potential U to define the tidal potential

[tex]
U_{tid}= U - \left(- \frac{GM_{m}m r}{d_0^2} \right) = - GM_{m}m \left( \frac{1}{d} - \frac{r}{d_0^2} \right). ~~(*)
[/tex]

I'm not sure why you claim there is a sign difference between [tex]r[/tex] and [tex]x[/tex], but I've given a physical argument for the sign in the equation (*) above.

With the correct sign, you'll be much closer to obtaining the correct answer. However, your formula

[tex]
U_{tid}=-\frac{GM_{m}m}{d_{0}}(1+\frac{r}{d_{0}}-\frac{r^{2}}{2d^{2}_{0}}+\frac{r}{d^{2}_{0}})
[/tex]

has a couple more problems. First there is a problem with the factor of [tex]d_0[/tex] in the last term that should be straightforward to clear up. Secondly you should take your expansion

[tex]
(1+x)^{-1/2}=1-\frac{x}{2}
[/tex]

out to quadratic order since you are interested in computing terms of [tex]O(r^2)[/tex]. This will correct the coefficient of the quadratic term.
 
  • #3
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Things are a bit confusing, since you don't explain the physical significance of the terms in

[tex]
U_{tid}=-GM_{m}m(\frac{1}{d}-\frac{x}{d^{2}_{0}})
[/tex]

From first principles, the potential due to the moon on an element on the horizontal axis is

[tex] U = =-GM_{m}m \frac{1}{d_0 -r} \sim - \frac{GM_{m}m }{d_0} \left( 1 + \frac{r}{d_0} + \frac{r^2}{d_0^2} + \cdots \right). [/tex]

The first term is constant and doesn't product a force. The second term produces a constant force, which is what keeps the Earth-Moon system bound. Since this term doesn't cause tides, your book seems to be subtracting it from the potential U to define the tidal potential

[tex]
U_{tid}= U - \left(- \frac{GM_{m}m r}{d_0^2} \right) = - GM_{m}m \left( \frac{1}{d} - \frac{r}{d_0^2} \right). ~~(*)
[/tex]
The book derives the potential as an illustration as the utility of non-inertial reference frames. It says that for a mass m, the acceleration relative to the non-inertial reference frame is given by the usual sum of the forces (gravitation with both the earth and the moon) plus an inertial force (in the case the mass times the centripetal acceleration of the earth towards the moon). This leads to

[tex]m\ddot{r}=mg+F_{tid}+F_{ng}[/tex]

Where [itex]F_{ng}[/itex] are any non-gravitational forces (e.g. buoyancy) and

[tex]F_{tid}=-GM_{m}m(\frac{\hat{d}}{d^{2}}-\frac{\hat{d_{0}}}{d^{2}_{0}})[/tex]

They simply treat force as the negative gradient on d and x to get U, it seems.

I'm not sure why you claim there is a sign difference between [tex]r[/tex] and [tex]x[/tex], but I've given a physical argument for the sign in the equation (*) above.
Yeah, that was a mistake.

With the correct sign, you'll be much closer to obtaining the correct answer. However, your formula

[tex]
U_{tid}=-\frac{GM_{m}m}{d_{0}}(1+\frac{r}{d_{0}}-\frac{r^{2}}{2d^{2}_{0}}+\frac{r}{d^{2}_{0}})
[/tex]

has a couple more problems. First there is a problem with the factor of [tex]d_0[/tex] in the last term that should be straightforward to clear up. Secondly you should take your expansion

[tex]
(1+x)^{-1/2}=1-\frac{x}{2}
[/tex]

out to quadratic order since you are interested in computing terms of [tex]O(r^2)[/tex]. This will correct the coefficient of the quadratic term.
I'll take a look at this in the morning, but I think I see how it goes now. I forgot to factor out a [itex]d_{0}[/itex] of the last term when I factored it out from the square root, and so that will cancel with the other term once I fix the signs. As for fixing the coefficient, that's gonna take some algebra for me to get, and instead I'm going to sleep for now.

Thanks for your help.
 
  • #4
74
0
Taking the approximation to the second order you get a coefficient of +3/8, and the only term that survives is (2r/do)^2 giving coefficients of -1/2+3/2 = 1. Thanks again!
 

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