Newtons Dynamic Laws

I. DYNAMIC LAWS DERIVATION

This note summarizes Newton's derivation for the planetary dynamics. I have followed [5] very closely and stripped it of all historical notes for convenience. Newton's goal, in particular, was to find out the nature of the centripetal force that causes the planet to revolve around the Sun in an elliptical orbit. He argued that, without any external influence, the planet will move in straight line at constant velocity. It is this force that imparts acceleration to the body to make it adhere to orbit.

The following figure depicts the needed graphics for analysis. It looks very crowded but just in one figure it depicts all the essential parameters and the variables of our problem.

Figure 1: Based on Newton's diagram for problem 3. A planet $P$ moves in an elliptic orbit APQB about a center of force $S$ located at a focus of the ellipse.

It shows the Sun, S, resides at one focus of the ellipse. The planet P moves along the elliptical curve APQ, the tangent at P is ZPR, and the centripetal force is along PS. Newton explains that at point P, if the force vanishes, the planet will move along the tangent to R. It is the centripetal force that will move R back to Q on the ellipse.

From Galileo's experiments, a body under the influence of central force will travel a distance proportional to the acceleration times the square of the traveled time. Conversely in time, $\delta t$, this acceleration, is proportional to $QR$, divided by $(\delta t{)}^2$.

Meanwhile, for any centripetal force, Newton proved that the line of force sweeps equal areas in equal times. This allowed him to describe the time geometrically by an area proportional to the time. That is $\delta t\propto SP\cdot QT$. Therefore, the acceleration is given by

1. Newton's lemma:

Proof is given in Appendix A.

1. $\mathrm{\Delta }PXV$ and $\mathrm{\Delta }PEC$ are similar, hence $PX/PE=PV/PC$

1. Apollonius: Prop 15, Book1 $PV\cdot GV/Q{V}^2=P{C}^2/C{D}^2$

Aside from the proofs given in [7], Appendix B provides a little simple proof.

4. The shape QRPX is quadrilateral,

Thus, the last four equations yield,

5. Apollonius: Prop 31, Book7

Proof is given in Appendix C.

1. $\mathrm{\Delta }QTX$ and $\mathrm{\Delta }PFE$ are similar, hence

Substituting from Eqs. (1.2) and (1.7) into (1.8) gives

Now, Eqs. (1.6) and (1.9) yield

In the limit Q approaches P. Consequently V and X will approach P. That is

Using the above equation in Eq. (1.10) gives

Here, $L=C{B}^2/(2\mathbb{A}AC)$ is the latus rectum of the ellipse. Finally, Eq. (1.1) results in,

In words, the centripetal force is inversely proportional to the Sun/Planet radius squared.

Kepler's Third Law

Newton then gave a proof to Kepler's third law – the orbit's period is proportional to one and half the power of the ellipse major axis diameter. Herein, we provide a proof that is a little variant from Newton's.

Newton has shown that for any centripetal force, a planet will sweep an area proportional to time. Thus from the diagram, the area swept for the infinitesimal time, $\delta t$, is

\begin{array}{l} \text {Swept area in \delta t = SP\cdot QT\Rightarrow} \\\text {rate of Swept area = SP\cdot QT / \delta t} \end{array}\tag{1.14}

Since ellipse area = π · CB · AC, then the time period T is

Squaring each side of Eq. (1.15) yields,

Substituting for QT from Eq.(1.12) into the above gives

Since $L=C{B}^2/(2\mathbb{A}AC)$, the above can be simplified to

Equation (1.13) implies $QR\cdot S{P}^2\propto (\delta t{)}^2$ and therefore, makes (1.18)

That is the orbit's period squared is proportional to the cube of the ellipse semi axis diameter.

Appendix A

Newton's Lemma

Proof: $H$ is the $2^{\text{nd}}$ foci of the ellipse from which we draw $HI$ parallel to $RPZ$. Ellipse reflection property, [8], states $\mathrm{\angle }HZP=\mathrm{\angle }SPR$. Therefore, $\mathrm{\angle }PHI=\mathrm{\angle }HZP=\mathrm{\angle }SPR=\mathrm{\angle }HIP$. Hence $\mathrm{\Delta }HIP$ is equal sided triangle and $PH=PI$.

Moreover:

Since HI parallel to RPZ and thus to DK and since SC=CH, then SE=EI. Therefore

or,

Chord Bisector

Apollonius Proposition 15

The eccentric circle is that one that shares the ellipse's major axis. Herein, we project the ellipse points $P$, $Q$, $D$, $G$ and $K$ onto $P^{\prime }$, $Q^{\prime }$, $D^{\prime }$, $G^{\prime }$ and $K^{\prime }$ on the eccentric circle. Consequently, the lines $P^{\prime }C{G}^{\prime }$ and $D^{\prime }C{K}^{\prime }$ become two perpendicular diameters in the eccentric circle. The point $V$ on the $CP$ is mapped onto $V^{\prime }$ on the line $P^{\prime }{Q}^{\prime }$ so that $Q^{\prime }{V}^{\prime }$ is perpendicular to the diameter $P^{\prime }C{G}^{\prime }$, hence

Notice that $\mathrm{\Delta }CV{V}^{\prime }$ is similar to $\mathrm{\Delta }CP{P}^{\prime }$, thus

Likewise $\mathrm{\Delta }CG{G}^{\prime }$ is similar to $\mathrm{\Delta }CP{P}^{\prime }$, thus

(B.2)

The above two equations give

Substituting from (B.1) yields

Figure 2b: projecting the ellipse points P, Q, D, G and K onto $P^{\prime }$, $Q^{\prime }$, $D^{\prime }$, $G^{\prime }$ and $K^{\prime }$ on the eccentric circle.

What remains is to relate QV to $Q^{\prime }{V}^{\prime }$. We know that QV is parallel to DC, so from Q we draw a parallel to PC until it meets DC in q, therefore Cq=QV. Project q onto point $q^{\prime }$ on $D^{\prime }C$, and it is straight forward to prove that $C{q}^{\prime }=Q^{\prime }{V}^{\prime }$. Now $\mathrm{\Delta }CD{D}^{\prime }$ is similar to $\mathrm{\Delta }Cq{q}^{\prime }$, Thus

Substituting in (B.5) results in

But $C{P}^{\prime }=C{D}^{\prime }=a$ implies that

Parallelogram area Equivalence Apollonius Proposition 31 Book 7 With reference to Fig. 3 below, we introduce two proofs.

Method 1: geometric algebra

Let $P^{\prime }$ and $D^{\prime }$ be the points on the eccentric circle from which their projection $P$ and $D$ on the ellipse are obtained. The coordinates of $\bar{P}$ and $\bar{D}$ are then given by

and,

Since $x_d=y_p$ and $y_d=x_p$ then $C{P}^{\prime }$ and $C{D}^{\prime }$ are perpendicular and thus the area enclosed by $C{P}^{\prime }$ and $C{D}^{\prime }$ $=C{K}^2=a^2$. Therefore the area enclosed by CP and $CD=\frac{b}{a}C{K}^2=ab$ as desired.

Method 2: vector analysis: In Fig. 3b, let P and D denote the points $P_1(x_1,y_1)$ and $P_2(x_2,y_2)$ respectively. The area PCD is then given by the cross product of vectors CP and CD. Thus

Figure 3b: Projecting the ellipse points P, D, G and K onto $P^{\prime }{D}^{\prime }$, $G^{\prime }$ and $K^{\prime }$ on the eccentric circle.