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Connection Between the Angle of a Conic Section and the Energy of the Corresponding Orbit

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Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,183
If you think about a double-napped cone, and the various non-degenerate sections you can get with it:

1. Circle
2. Ellipse
3. Parabola
4. Hyperbola,

you can see that there is a progression here: increasing angle $\alpha$ that the intersecting plane makes with the horizontal. To be clear about this, $\alpha$ is not the angle that the normal to the intersecting plane makes with the horizontal, but the angle that the plane itself makes with the horizontal.

It's also true that for an inverse-square-law force, such as gravity (in the classical limit), this same progression describes increasing amounts of energy in the resulting orbit. I've long thought there must be some mathematical relationship between the two. And there is, through the eccentricity. The eccentricity of a conic section is defined as
$$e= \frac{ \sin( \alpha)}{ \sin( \beta)},$$
where $\alpha$ is the angle I've already defined, and $\beta$ is the angle that the cone makes with the horizontal.

According to Marion and Thornton's Classical Dynamics of Particles and Systems, 4th Ed., p. 304, Eq. (8.40), you also have that the eccentricity of an orbit under the gravitational force is equal to
$$e= \sqrt{1+ \frac{2E \ell^{2}}{ \mu k}},$$
where $E$ is the total energy, $\ell$ is the angular momentum, $\mu$ is the reduced mass
$$\mu= \frac{m_{1}m_{2}}{m_{1}+m_{2}},$$
and $k=Gm_{1}m_{2}$ (the numerator of the gravitational force law, although it could also be $q_{1}q_{2}/(4 \pi \epsilon_{0})$, I suppose, for a Coulomb force.)

Hence, we have that
$$ \frac{ \sin^{2}( \alpha)}{ \sin^{2}( \beta)}=1+ \frac{2E \ell^{2}}{ \mu k}.$$
So the relationship between the angle of the conic section and the energy of the orbit is that the square of the sine of the conic section angle is affinely related to the total energy.

Comments and questions should be posted here:

http://mathhelpboards.com/commentar...-conic-section-energy-corresponding-4471.html
 
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