Second Kepler law for hyperbolas

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SUMMARY

Kepler's Second Law, also known as the area law, applies to all types of orbits, including hyperbolas and parabolas, in the context of central force motion. The discussion provides a mathematical proof demonstrating that the rate at which area is swept out by a particle under a central force remains constant, regardless of the orbit type. The proof utilizes polar coordinates and the relationship between position, force, and angular momentum to establish that the area swept per unit time is a constant value. It is crucial to distinguish between hyperbolas and hyperboles, as the latter refers to a different concept entirely.

PREREQUISITES
  • Understanding of central force motion
  • Familiarity with polar coordinates
  • Basic knowledge of angular momentum
  • Concept of area swept out in orbital mechanics
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  • Study the implications of Kepler's Laws on different orbital shapes
  • Explore the mathematical derivation of angular momentum in polar coordinates
  • Investigate the differences between hyperbolas and hyperboles in mathematics
  • Learn about the applications of central force motion in astrophysics
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Students of physics, astronomers, and anyone interested in orbital mechanics and the mathematical foundations of celestial motion.

jaumzaum
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A body can describe 3 types of orbits around another (considering only the gravitational force), a elipse, a hyperbole or a parabola.
Does the second kepler law (area law) work for hyperboles or parabolas too?
 
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Yes. Kepler's 2nd law holds for any central force motion. Here is one proof:

Consider a particle moving under a central force ##F = f(r)\hat{r}##. We see immediately that ##\frac{\mathrm{d} L}{\mathrm{d} t} = r\times F = 0## hence ##L = r\times p## is a constant of motion. Now consider the position of the particle at an instant ##t## and another instant ##t + \Delta t## so that the position of the particle is given in polar coordinates by ##(r,\theta)## and ##(r + \Delta r, \theta + \Delta \theta)## respectively. For small ##\Delta \theta##, the area swept out during this interval will be given by the triangle swept out of side length ##r + \Delta r## and height ##r\Delta \theta## so ##\Delta A = \frac{1}{2}r^{2}\Delta \theta + \frac{1}{2}r\Delta r \Delta \theta ## hence ##\frac{\Delta A}{\Delta t} = \frac{1}{2}r^{2}\frac{\Delta \theta}{\Delta t} + \frac{1}{2}r\frac{\Delta r \Delta \theta }{\Delta t}## so ##\frac{dA}{dt} = \lim_{\Delta t\rightarrow 0 }\frac{\Delta A}{\Delta t} = \frac{1}{2}r^{2}\frac{d\theta}{d t} = \text{const.} ## as desired.
 
Spelling error. You are talking about hyperbolas. Hyperbole is a completely different thing.
 

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