I don't want to take on the second and third laws at this time. They are interesting in their own right, but otherwise this would turn from a gem to a tome. It is sufficient for my purpose to state that the second law proved that gravity was a centrally-directed force, and the third law proved that it required a force proportional to the inverse square of radius and not at all dependent on azimuth.

The first law is where things get interesting. For all its flaws, I still recommend The Mechanical Universe, in particular episode 22. This derivation follows that presented starting in at about 18 minutes.

Now having selected a central, inverse square field, what is the shape of the orbit produced? That is the derivation above, and in a sense proves that the first law is a consequence of the second and third laws.

## Problem statement

**Given:**Newton's laws of motion and the law of universal gravitation, in particular that the force between two bodies is attractive, directed along the line between them, proportional to both their masses, and inversely proportional to the distance between them, applied to two bodies of constant mass which interact only through this gravitation.**Show that:**The path of one of the bodies relative to the other is a conic section with one focus at the other body.

## Solution

The middle column is the cheesy "poetry" that goes with this derivation in The Mechanical Universe. It really does add to the effect, though, especially towards the end. The right column is a description of the derivation.\[\vec{F}=-G\frac{MM_0}{r^2}\hat{r}\] | The universal law of gravitation | Specifically between two bodies, one of which has mass \(M_0\) and is at the coordinate center. We will call this one the center, and the other one the planet. The vector \(\vec{r}\) points from the center to the planet, so the equation in this form describes the force on the planet. |

\[\vec{F}=\frac{-D}{r^2}\hat{r}\] | Combine constants \(D=GMM_0\). | |

\[M\vec{a}=\frac{-D}{r^2}\hat{r}\] | The second law of motion | Specifically, applied to the planet \(\vec{F}=M\vec{a}\). |

\[\vec{a}=\frac{-D}{Mr^2}\hat{r}\] | Divide both sides by mass of planet. | |

\[\vec{L}=Mr^2\frac{d\theta}{dt}\hat{k}\] | The conservation of angular momentum | Specifically of the planet, and not particularly conservation, just angular momentum of planet. Variable \(\theta\) is the true anomaly of the planet in its orbit, and vector \(\hat{k}\) is perpendicular to the plane of the oribt. Since gravity is directed purely at the center, there is no torque, so angular momentum \(\vec{L}\) is constant. |

\[\vec{a}\times\vec{L}=\frac{-D}{Mr^2}\hat{r} \times Mr^2\frac{d\theta}{dt}\hat{k}\] | Vectors cross together, like Shakespeare's star-crossed lovers. | Left side is cross product of two vectors, right side is cross product of their definitions. |

\[\vec{a}\times\vec{L}=\frac{-D}{Mr^2}Mr^2\frac{d\theta}{dt}\hat{r} \times \hat{k}\] | Associativity of scalars across a cross product | |

\[\vec{a}\times\vec{L}=-D\frac{d\theta}{dt}\hat{r} \times \hat{k}\] | In a combination as potent as Michelangelo's Hand of God... | Cancellation of terms |

\[\vec{a}\times\vec{L}=D\frac{d\theta}{dt}\hat{k} \times \hat{r}\] | Anti-commutativity of cross product | |

\[\hat{k} \times\hat{r}=\frac{d\hat{r}}{d\theta}\] | Setting off a ballet of vectors as spirited as a Beethoven symphony. | Substitution. This one has a complicated geometric justification. The vector \(\hat{r}\) is already presumed to lie in a plane perpendicular to \(\hat{k}\). The length is fixed to 1, since it is a unit vector, and the plane is fixed, so the only degree of freedom left is to spin around in a circle. When it does so, the derivative of the unit vector with respect to the angle around the circle is perpendicular to \(\hat{r}\). The length of the derivative vector is 1 since the radius is 1 and the angle is given in radians. So, the derivative vector is completely constrained: It is perpendicular to \(\hat{r}\), perpendicular to \(\hat{k}\) since it is in the same plane as \(\hat{r}\), and has length 1. This happens to be the vector described by \(\hat{k} \times \hat{r}\). |

\[\vec{a}\times\vec{L}=D\frac{d\theta}{dt}\frac{d\hat{r}}{d\theta}\] | Plug in the above substitution. | |

\[\vec{a}\times\vec{L}=D\frac{d\hat{r}}{dt}\] | Chain rule | |

\[\frac{d\vec{v}}{dt}\times\vec{L}=D\frac{d\hat{r}}{dt}\] | The equations of the heavens find their simplest expression in the elegant language of differential calculus. | Definition of acceleration |

\[\frac{d}{dt}\left(\vec{v}\right)\times\vec{L}=D\frac{d\hat{r}}{dt}\] | ||

\[\frac{d}{dt}\left(\vec{v}\times\vec{L}\right)=D\frac{d\hat{r}}{dt}\] | Permitted only because \(\vec{L}\) is a constant vector | |

\[\vec{v}\times\vec{L}=D\left(\hat{r}+\vec{e}\right)\] | Integral to the grand design, and constant in addition. The ellipses of the heavens are revealed as an expression of the perfect circle of underlying geometry. | Integrate both sides. Introduce vector \(\vec{e}\) as constant of integration, it will happen to be the eccentricity vector, which points from the center to periapse, and has length equal to the eccentricity of the orbit. In a sense, it shows that \(\vec{v}\times\vec{L}\) follows an epicycle. |

\[\vec{r}\cdot\vec{v}\times\vec{L}=D\vec{r}\cdot\left(\hat{r}+\vec{e}\right)\] | The product of these events... | Dot multiply both sides by \(\vec{r}\) |

\[\vec{r}\times\vec{v}\cdot\vec{L}=D\vec{r}\cdot\left(\hat{r}+\vec{e}\right)\] | ...once order is exchanged... | Left side is scalar triple product, so the operations may be switched around |

\[\frac{M\vec{r}\times\vec{v}\cdot\vec{L}}{M}=D\vec{r}\cdot\left(\hat{r}+\vec{e}\right)\] | ...and mass allowed its play... | Reintroduce mass of planet |

\[\frac{\vec{L}\cdot\vec{L}}{M}=D\vec{r}\cdot\left(\hat{r}+\vec{e}\right)\] | Definition of angular momentum of planet | |

\[\frac{L^2}{M}=D\vec{r}\cdot\left(\hat{r}+\vec{e}\right)\] | Dot product of vector with itself is square of vector length | |

\[\frac{L^2}{DM}=\vec{r}\cdot\left(\hat{r}+\vec{e}\right)\] | ||

\[\frac{L^2}{DM}=\left(\vec{r}\cdot\hat{r}+\vec{r}\cdot\vec{e}\right)\] | Distribute dot product | |

\[\frac{L^2}{DM}=\left(r+\vec{r}\cdot\vec{e}\right)\] | Dot product of a vector with its hat vector is length of vector | |

\[\frac{L^2}{DM}=\left(r+re \cos \theta\right)\] | Definition of dot product | |

\[\frac{L^2}{DM}=r\left(1+e \cos \theta\right)\] | ||

\[\frac{L^2/DM}{1+e \cos \theta}=r\] | ...comes to resemble an old friend -- a curve of familiar aspect. Ellipse, parabola, and hyperbola, all find their arcs described on the celestial tapestry. | Polar form of conic section equation, with center at one focus of the conic |

I think that there is an implicit assumption that the mass of the center is so much greater than the mass of the planet that the center doesn't move at all. Except now it's not implicit.

## No comments:

## Post a Comment