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Material point describes an ellipse (with axes $a$ and $b$) under the action of force of gravity. At the end of bigger axis the velocity of point is equal to $v_0$. I need to find the force of gravity as a function of radius vector. So the force is conservative therefore the law of conservation of energy is right. Also we can describe the velocity (in cartesian coordinates, from the equation of ellipse) $v^2=\dot{x}^2+\dot{y}^2$ as a $v^2=\dot{y}^2+\frac{a^4}{4x^2} (1-\frac{2 y}{b^2}\dot{y})^2$ From which we can derive that $v_0^2=\dot{y}^2+\frac{a^2}{4}$ (simply input $x=a$ and $y=0$ and use the condition that the velocity at the end of bigger axis is equal to $v_0$. But I'm stuck, I don't know wthat to do next.

Savva Zosimov
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  • Could you double-check your equations? They're dimensionally inconsistent, e.g. $a^4/x^2$ is a squared length, not a squared speed. – J.G. Dec 15 '20 at 20:44
  • Let us consider the ellipse equation : $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ therefore $\frac{2x}{a^2}\dot{x}+\frac{2y}{b^2}\dot{y}=0$ Then we get $\dot{x}=-\frac{a^2}{2x} \frac{2y}{b^2}\dot{y}$ And $v_0^2=\dot{y}^2+\frac{a^4y^2}{x^2b^4} \dot{y}$. Yes, there was a mistake and $v_0^2=\dot{y}^2$ But still I dont know what's next – Savva Zosimov Dec 15 '20 at 20:51
  • There $\dot{y}^2$ in the expression of $v_0^2$ – Savva Zosimov Dec 15 '20 at 20:57
  • There's still a mistake; it should be$$v^2=\dot{y}^2+\left(-\frac{a^2y}{b^2x}\dot{y}\right)^2=\left(1+\frac{a^4y^2}{b^4x^2}\right)\dot{y}^2.$$ – J.G. Dec 15 '20 at 21:03

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An ellipse is often written as $r=\frac{l}{1+e\cos\theta}$, with $e$ its eccentricity and $l$ its semi-latus rectum, but the Cartesian coordinates $X,\,Y$ associated with these polar coordinates aren't the ones you're using. We have$$r=\tfrac{l}{1+eX/r}\implies r=l-eX\implies\tfrac{\left(X-\tfrac{el}{1-e^2}\right)^2}{(l/(1-e^2))^2}+Y^2=1.$$Comparing this to your coordinates gives $x=X-ea,\,y=Y,\,a=\frac{l}{1-e^2}$. With $u=\frac1r$, the Binet equation $\frac1l=u_{\theta\theta}+u=-\frac{F}{mh^2u^2}$ implies the force on a mass $m$ of specific orbital angular momentum $h$ is$$F=-\frac{mh^2}{l}u^2=-\frac{mh^2a^{-1}/(1-e^2)}{(x+ea)^2+y^2}$$(the $-$ sign denotes an attractive force). At $x=a,\,y=0$ we have $\dot{x}_0=0,\,v_0=\dot{y}_0$, so$$mh\vec{k}=ma\vec{i}\times\dot{y_0}j\implies h=av_0.$$If $a$ is the semi-major axis,$$1-e^2=\frac{b^2}{a^2}\implies F=-\frac{ma^3b^{-2}v_0^2}{(x+ea)^2+y^2}.$$

J.G.
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