0
$\begingroup$

So above is a hypothetical problem I gave myself to figure out if I could do potential differences properly using Newton's Gravitation Formula and I wasn't 100% that my Math was correct (I'm basically self-taught on line integrals), so I wanted some feedback on that. The problem is that there is some mass $M$ evenly distributed in a circle with radius 1 m at some point in 2-space. What is the gravitational potential difference between the surface of this 2D "planet" and the point $(x_1, y_1)$? (Pretend there are subscripts in my diagram). For convenience purposes, I centered the "planet" at the origin.

$$V_g = \int_C \vec{F_g}(x,y) \cdot d\vec{s}$$ $$\vec{F_g}(x,y) = \frac{GM}{x_1^2 + y_1^2}\boldsymbol{\hat{\textbf{r}}}$$ $$V_g = \int_C \frac{GM}{x_1^2 + y_1^2}\boldsymbol{\hat{\textbf{r}}} \cdot d\vec{s}$$ $$ = \int_{\cos(\tan^{-1}\frac{y_1}{x_1})}^{x_1}\frac{GM}{x^2 + \sin^2(\tan^{-1}\frac{y_1}{x_1})}dx + \int_{\sin(\tan^{-1}\frac{y_1}{x_1})}^{y_1}\frac{GM}{x_1^2 + y^2}dy$$ $$=GM\bigg(\int_{\cos(\tan^{-1}\frac{y_1}{x_1})}^{x_1}\frac{1}{x^2 + \sin^2(\tan^{-1}\frac{y_1}{x_1})}dx + \int_{\sin(\tan^{-1}\frac{y_1}{x_1})}^{y_1}\frac{1}{x_1^2 + y^2}dy\bigg)$$ $$=GM\bigg(\bigg[\frac{1}{\sin(\tan^{-1}\frac{y_1}{x_1})}\tan^{-1}\bigg(\frac{x}{\sin(\tan^{-1}\frac{y_1}{x_1})}\bigg)\bigg|_{\cos(\tan^{-1}\frac{y_1}{x_1})}^{x_1} \bigg] + \bigg[ \frac{1}{x_1}\tan^{-1}\bigg(\frac{y}{x_1}\bigg)\bigg|_{\sin(\tan^{-1}\frac{y_1}{x_1})}^{y_1}\bigg]\bigg)$$

Also, yes, I realize I unnecessarily punished myself using Cartesian coordinates, so here's the whole thing in polar coordinates.

$$V_g = \int_1^{r_1} \frac{GM}{r^2}dr = \int_1^{r_1} GMr^{-2}dr = \frac{-GM}{r}\bigg|_1^{r_1} = GM - \frac{GM}{r_1}$$

$\endgroup$
0
$\begingroup$

This doesn't look like a line integral at all.

Normally you would have something more like:

$V_g = \int_C \frac{GM}{x_1^2 + y_1^2} \|d\mathbf r\|\\ \mathbf r(t) = R(\cos t, \sin t)\\ d\mathbf r = R(-\sin t, \cos t)\ dt\\ \|d\mathbf r\| = R \ dt\\ V_g = \int_0^{2\pi} \frac{GMR}{x_1^2 + y_1^2} \ dt$

But I am not sure about our equation. You say, $x_1,y_1$ is a fixed point.

You need the distance from some $x_1,y_1$ in the interior to each $x,y$ on the rim.

$V_g = \int_C \frac{GM}{(x-x_1)^2 + (y-y_1)^2} \|d\mathbf r\|\\ \int_0^{2\pi} \frac{GM}{R^2 - 2x_1R\cos t - 2y_1 R\sin t + x_1^2 + y_1^2} R \ dt$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.