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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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