How do I write this force in terms of another variable?

Question

I have the tidal force, $\vec{F}=-GMm(\frac{\hat d}{d^2}-\frac{\hat d_0}{d_0^2})$ where:

How do I express the force in terms of $\theta$? I'm given that $\vec{r}=R(\cos(\theta),\sin(\theta))$. I'm not sure how to fully eliminate $d$ and $d_0$. I could use the law of cosines, but I don't have an angle inside the triangle made by $r,d,$ and $d_0$.

Answer 1

Physically, the tidal force should take the revolution of the moon and the earth about their centre of mass into accounts.

Anyways, let's compute the expression.

\begin{align*} \mathbf{F} &= -GMm \left( \frac{\mathbf{d}}{d^3}-\frac{\mathbf{d}_0}{d_0^3} \right) \\ &= -GMm \left( \frac{\mathbf{d}_0+\mathbf{r}}{|\mathbf{d}_0+\mathbf{r}|^3}- \frac{\mathbf{d}_0}{d_0^3} \right) \\ &= -GMm \left[ \frac{\mathbf{d}_0+\mathbf{r}} {(d_0^2+r^2-2\mathbf{d}_0 \cdot\mathbf{r})^{3/2}}- \frac{\mathbf{d}_0}{d_0^3} \right] \\ & \approx -GMm \left[ (\mathbf{d}_0+\mathbf{r})\left( \frac{1}{d_0^3}+ \frac{3\mathbf{d}_0 \cdot\mathbf{r}}{d_0^5} \right)- \frac{\mathbf{d}_0}{d_0^3} \right] \\ &= -GMm \left[ \frac{3\mathbf{d}_0 \cdot\mathbf{r}}{d_0^5} \mathbf{d}_0+ \left( \frac{1}{d_0^3}+ \frac{3\mathbf{d}_0 \cdot\mathbf{r}}{d_0^5} \right)\mathbf{r} \right] \end{align*}

Answer 2

You can use

$$ \mathbf{d}_0 + \mathbf{r} = \mathbf{d} $$

To get rid of $\mathbf{d}$

$$ \mathbf{F}(\mathbf{r},\mathbf{d}_0) = - GM m \left(\frac{\mathbf{d}_0 + \mathbf{r}}{|\mathbf{d}_0 + \mathbf{r}|^3} + \frac{\mathbf{d}_0}{|\mathbf{d}_0|^3} \right) $$

However, this is as far as you can get, the problem is that the position of the moon does not depend on the position of the mass $m$, say in another words, imagine you have a satellite orbiting Earth, and you know its location. Based on this information you cannot know where the moon is, so basically you cannot express $\mathbf{d}_0$ in terms of $\mathbf{r}$