Elementary Newtonian Mechanics

Abstract

This chapter deals with the kinematics and the dynamics of a finite number of mass points that are subject to internal, and possibly external, forces, but whose motions are not further constrained by additional conditions on the coordinates.

Appendix: Practical Examples

1. Kepler Ellipses. Study numerical examples for finite motion of two celestial bodies in their center-of-mass frame (Sect. 1.7.2).

Solution. The relevant equations are found at the end of Sect. 1.7.2. It is convenient to express $m_{1}$ and $m_{2}$ in terms of the total mass $M=m_{1} +m_{2}$ and to set $M=1.$ The reduced mass is then $\mu =m_{1} m_{2} .$ For given masses the form of the orbits is determined by the parameters

$$\begin{aligned} p={l^{2}\over A\mu } \quad \mathrm{and} \quad \varepsilon = \sqrt{1+{2El^{2}\over \mu A^{2}}} \;, \end{aligned}$$

(A.1)

which in turn are determined by the energy E and the angular momentum. It is easy to calculate and to draw the orbits on a PC. Figure 1.6a shows the example $m_{1} =m_{2}$ with $\varepsilon =0.5,$ $p=1,$ while Fig. 1.6b shows the case $m_{1} =m_{2} /9$ with $\varepsilon =0.5,$ $p=0.66.$ As the origin is the center of mass, the two stars are at opposite positions at any time.

2. Motion of a Double Star. Calculate the two orbital ellipses of the stars of the preceding example pointwise, as a function of time, for a given time interval $\varDelta t.$

Solution. In Example 1 the figures show $r(\varphi )$ as a function of $\varphi .$ They do not indicate how the stars move on their orbits as a function of time. In order to obtain r(t),  one returns to (1.19) and inserts the relative coordinate $r(\varphi ).$ Separation of variables yields

$$\begin{aligned} t_{n+1} -t_{n} ={\mu p^{2}\over l} \int ^{\phi _{n+1}} _{\phi _{n}} {\mathrm{d}\varphi \over (1+\varepsilon \cos \varphi )^{2}} \end{aligned}$$

(A.2)

for the orbital points n and $n+1.$ (The pericenter has $\phi _{\mathrm{P}} =0.)$ The quantity $\mu p^{2} /l$ has the dimension of time. Introduce the period from (1.23) and use this as the unit of time,

$$ T=2\pi {\mu ^{1/2}a^{3/2}\over A^{1/2}} =\pi {A\mu ^{1/2}\over 2^{1/2}(-E)^{3/2}} \;. $$

Then

$$ {\mu p^{2}\over l} =\bigl (1-\varepsilon ^{2} \bigr )^{3/2} {T\over 2\pi }\;. $$

The integral in (A.2) can be done analytically. Substituting

$$ x\,{\buildrel {\scriptstyle \mathrm{def}}\over =}\, \sqrt{{1-\varepsilon \over 1+\varepsilon }} \mathrm{tan} {\varphi \over 2} $$

one has

$$\begin{aligned} I\equiv & {} \int {\mathrm{d}\varphi \over (1+\varepsilon \cos \varphi )^{2}} ={2\over \sqrt{1-\varepsilon ^{2}}} \int \mathrm{d} x {1+[(1+\varepsilon )/(1-\varepsilon )]x^{2}\over (1+x^{2})^{2}} \\= & {} {2\over \sqrt{1-\varepsilon ^{2}}} \left\{ \int {\mathrm{d}x\over 1+x^{2}} + {2\varepsilon \over 1-\varepsilon } \int {x^{2}\mathrm{d}x\over (1+x^{2})^{2}} \right\} \;, \end{aligned}$$

whose second term can be integrated by parts. The result is

$$\begin{aligned} I= & {} {2\over (1-\varepsilon ^{2})^{3/2}} \mathrm{arctan} \left( \sqrt{{1-\varepsilon \over 1+\varepsilon }} \mathrm{tan} {\varphi \over 2} \right) \nonumber \\&-{\varepsilon \over 1-\varepsilon ^{2}} \, {\sin \varphi \over 1+\varepsilon \cos \varphi } +C\;, \end{aligned}$$

(A.3)

so that

$$\begin{aligned} {t_{n+1}-t_{n} \over T}= & {} {1\over \pi } \Biggl [\mathrm{arctan} \left( \sqrt{{1-\varepsilon \over 1+\varepsilon }} \mathrm{tan} {\varphi \over 2} \right) \nonumber \\&-{1\over 2} \varepsilon \sqrt{1-\varepsilon ^{2}} {\sin \varphi \over 1+\varepsilon \cos \varphi } \Biggr ]^{\phi _{n+1}} _{\phi _{n}} \;. \end{aligned}$$

(A.4)

One can compute the function $\varDelta t(\varDelta \phi ,\phi ),$ for a fixed increment $\varDelta \phi$ and mark the corresponding positions on the orbit. Alternatively, one may give a fixed time interval $\varDelta t/T$ and determine succeeding orbital positions by solving the implicit equation (A.4) in terms of $\varphi .$

3. Precession of Perihelion. (a) For the case of bound orbits in the Kepler problem show that the differential equation for $\varphi =\varphi (r)$ takes the form

$$\begin{aligned} {\mathrm{d}\varphi \over \mathrm{d}r} ={1\over r} \sqrt{{r_{\mathrm{P}}r_{\mathrm{A}}\over (r-r_{\mathrm{P}})(r_{\mathrm{A}}-r)}} \;, \end{aligned}$$

(A.5)

where $r_{\mathrm{P}}$ and $r_{\mathrm{A}}$ denote pericenter and apocenter, respectively. Integrate this equation with the boundary condition $\varphi (r=r_{\mathrm{P}})=0.$

(b) The potential is now modified into $U(r)=-A/r+B/r^{2} .$ Determine the solution $\varphi =\varphi (r)$ and discuss the precession of the pericenter after one turn, in comparison with the Kepler case, as a function of $B\lessgtr 0$ where $\vert B\vert \ll l^{2} /2\mu .$

Solutions. (a) For elliptical orbits, $E<0,$ and one has

$$ {\mathrm{d}\varphi \over \mathrm{d}r} ={1\over \sqrt{2\mu (-E)}} \, {1\over r} \,{1\over \sqrt{-r^{2}-\displaystyle {{A\over E}}r+ \displaystyle {{l^{2}\over 2\mu E}}} }\;. $$

Apocenter and pericenter are given by the roots of the quadratic form $(-r^{2} -Ar/E+l^{2} /2\mu E)$:

$$\begin{aligned} r_{\mathrm{A/P}} ={p\over 1\mp \varepsilon } =- {A\over 2E} (1\pm \varepsilon ) \end{aligned}$$

(A.6)

(these are the points where $\mathrm{d} r/\mathrm{d} t=0$). With

$$ r_{\mathrm{P}} r_{\mathrm{A}} ={A^{2}\over 4E^{2}} \bigl (1-\varepsilon ^{2} \bigr )=- {l^{2}\over 2\mu E} $$

we obtain (A.5). This equation can be integrated. With the condition $\varphi (r_{\mathrm{P}} )=0$ one obtains

$$\begin{aligned} \varphi (r)=\mathrm{arccos} \left[ {1\over r_{\mathrm{A}}-r_{\mathrm{P}}} \left( 2 {r_{\mathrm{A}}r_{\mathrm{P}} \over r} -r_{\mathrm{A}} -r_{\mathrm{P}} \right) \right] \;. \end{aligned}$$

(A.7)

As $\varphi (r_{\mathrm{A}} )-\varphi (r_{\mathrm{P}} )=\pi$, one confirms that the pericenter, force center, and apocenter lie on a straight line. Two succeeding pericenter constellations have azimuths differing by $2\pi ,$ i.e. they coincide. There is no precession of the pericenter.

(b) Let $r_{\mathrm{P}}$ and $r_{\mathrm{A}}$ be defined as in (A.6). The new apocenter and pericenter positions, in the perturbed potential, are denoted by $r^{\prime }_{\mathrm{A}}$ and $r^{\prime }_{\mathrm{P}}$, respectively. One has

$$ \bigl (r-r_{\mathrm{P}} \bigr )\bigl (r_{\mathrm{A}} -r\bigr )+ {B\over E} =\bigl (r-r^{\prime }_{\mathrm{P}} \bigr )\bigl (r^{\prime } _{\mathrm{A}} -r\bigr )\;, $$

and therefore

$$\begin{aligned} r^{\prime }_{\mathrm{P}} r^{\prime }_{\mathrm{A}} =r_{\mathrm{P}} r_{\mathrm{A}} - {B\over E} \;. \end{aligned}$$

(A.8)

Equation (A.5) is modified as follows:

$$ {\mathrm{d}\varphi \over \mathrm{d}r} ={1\over r} \sqrt{{r_{\mathrm{P}}r_{\mathrm{A}}\over (r-r^{\prime }_{\mathrm{P}}) (r^{\prime }_{\mathrm{A}}-r)}} =\sqrt{{r_{\mathrm{P}}r_{\mathrm{A}}\over r^{\prime }_{\mathrm{P}}r^{\prime }_{\mathrm{A}}}} \, \sqrt{{r^{\prime }_{\mathrm{P}}r^{\prime }_{\mathrm{A}}\over (r-r^{\prime }_{\mathrm{P}})(r^{\prime }_{\mathrm{A}}-r)}} \;. $$

This equation can be integrated as before under (a):

$$\begin{aligned} \varphi (r)= \sqrt{{r_{\mathrm{P}}r_{\mathrm{A}}\over r^{\prime }_{\mathrm{P}} r^{\prime }_{\mathrm{A}}}} \mathrm{arccos} \left[ {1\over r^{\prime }_{\mathrm{A}}-r^{\prime }_{\mathrm{P}}} \left( 2 {r^{\prime }_{\mathrm{A}}r^{\prime }_{\mathrm{P}}\over r} - r^{\prime }_{\mathrm{A}} -r^{\prime }_{\mathrm{P}} \right) \right] \;. \end{aligned}$$

(A.9)

From (A.8) two successive pericenter configurations differ by

$$\begin{aligned} 2\pi \sqrt{{r_{\mathrm{P}}r_{\mathrm{A}}\over r^{\prime }_{\mathrm{P}} r^{\prime }_{\mathrm{A}}}} ={2\pi l\over \sqrt{l^{2}+2\mu B}} \;. \end{aligned}$$

(A.10)

This difference can be studied numerically, as a function of positive or negative B. Positive B means that the additional potential is repulsive so that, from (A.10), the pericenter will “stay behind”. Negative B means additional attraction and causes the pericenter to “advance”.

4. Rosettelike Orbits.  Study the finite orbits in the attractive potential $U(r)=a/r^{\alpha } ,$ for some values of the exponent $\alpha$ in the neighborhood of $\alpha =1.$

Solution. Use as a starting point the system (1.70′)–(1.71′) of first-order differential equations, written in dimensionless form:

$$\begin{aligned} {\mathrm{d}\varrho \over \mathrm{d}\tau } =\pm \sqrt{2b\varrho ^{-\alpha }-\varrho ^{-2}-2} \,{\buildrel {\scriptstyle \mathrm def}\over =}\, f(\varrho )\;,\quad {\mathrm{d}\varphi \over \mathrm{d}\tau } ={1\over \varrho ^{2}} \;. \end{aligned}$$

(A.11)

From this calculate the second derivatives:

$$ {\mathrm{d}^{2}\varrho \over \mathrm{d}\tau ^{2}}= {\mathrm{d}\over \mathrm{d}\varrho } \left( {\mathrm{d}\varrho \over \mathrm{d}\tau } \right) {\mathrm{d}\varrho \over \mathrm{d}\tau } ={1\over \varrho ^{3}} \bigl (1-b\alpha \varrho ^{2-\alpha } \bigr ) \,{\buildrel {\scriptstyle \mathrm def}\over =}\, g(\varrho )\;,\quad {\mathrm{d}^{2}\varrho \over \mathrm{d}\tau ^{2}} =-{2\over \varrho ^{3}} f(\varrho )\;. $$

Equation (A.11) can be solved approximately by means of simple Taylor series:

$$\begin{aligned} \varrho _{n+1}= & {} \varrho _{n} +hf\bigl (\varrho _{n} \bigr ) +{\textstyle {{1\over 2}}} h^{2} g\bigl (\varrho _{n} \bigr ) +O\bigl (h^{3} \bigr ) \;,\nonumber \\ \varphi _{n+1}= & {} \varphi _{n} +h {1\over \varrho ^{2}_{n}} -h^{2}{1\over \varrho ^{3}_{n}} f\bigl (\varrho _{n} \bigr ) +O \bigl (h^{3} \bigr ) \;, \end{aligned}$$

(A.12)

for the initial conditions $\tau _{0} =0,$ $\varrho (0)=R_{0} ,$ $\varphi (0)=0.$ The step size h for the time variable can be taken to be constant. Thus, if one plots the rosette pointwise, one can follow the temporal evolution of the motion. (In Figs. 1.18, 1.19, 1.20, 1.21 and 1.22 we have chosen h to be variable, instead, taking $h=h_{0} \varrho /R_{0},$ with $h_{0} =0.02.$)

5. Scattering Orbits for a Repulsive Potential.  A particle of fixed momentum ${{\varvec{p}}}$ is scattered in the field of the potential $U(r)=A/r,$ (with $A>0).$ Study the scattering orbits as a function of the impact parameter.

Solution. The orbit is given by

$$\begin{aligned} r=r(\varphi )={p\over 1+\varepsilon \cos (\varphi -\varphi _{0})} \end{aligned}$$

(A.13)

with $\varepsilon >1.$ The energy E must be positive. We choose $\varphi _{0} =0$ and introduce the impact parameter $b=l/\vert {{\varvec{p}}} \vert$ and the quantity $\lambda \,{\buildrel {\scriptstyle \mathrm{def}}\over =}\, A/E$ as a characteristic length of the problem. The equation of the hyperbola (A.13) then reads

$$\begin{aligned} {r(\varphi )\over \lambda } ={2b^{2}/\lambda ^{2}\over 1+\sqrt{1+4b^{2}/\lambda ^{2}}\cos \varphi } \;. \end{aligned}$$

(A.13′)

Introducing Cartesian coordinates (see Sect. 1.7.2), we find that (A.13′) becomes

$$ {4x^{2}\over \lambda ^{2}} - {y^{2}\over b^{2}} =1\;. $$

This hyperbola takes on a symmetric position with respect to the coordinate axes, its asymptotes having the slopes $\mathrm{tan} \varphi _{0}$ and $-\mathrm{tan} \varphi _{0} ,$ respectively, where

$$\begin{aligned} \varphi _{0} =\mathrm{arctan} \left( {2b\over \lambda } \right) \;. \end{aligned}$$

(A.14)

We restrict the discussion to the left-hand branch of the hyperbola. We want the particle always to come in along the same direction, say along the negative x-axis. For a given impact parameter b this is achieved by means of a rotation about the focus on the positive x-axis, viz.

$$\begin{aligned} u= & {} (x-c)\cos \varphi _{0} +y\sin \varphi _{0} \;, \nonumber \\ v= & {} -(x-c)\sin \varphi _{0} +y\cos \varphi _{0} \;, \end{aligned}$$

(A.15)

where $c=\sqrt{1+4b^{2}/\lambda ^{2}}/2$ is the distance of the focus from the origin, and $y=\pm b \sqrt{4x^{2}/\lambda ^{2}-1}.$ For all b,  the particle comes in from $-\infty$ along a direction parallel to the u-axis, with respect to the coordinate system (u, v). Starting from the pericenter $(x_{0} /\lambda =-{1\over 2} ,$ $y_{0} =0$), let y run upwards and downwards and use (A.15) to calculate the corresponding values of x and y (see Fig. 1.32).

6. Temporal Evolution for Rutherford Scattering.  For the example in Sect. 1.29 calculate and plot a few positions of the projectile and target as a function of time, in the laboratory system.

Solution. In the laboratory system the orbits are given, as functions of $\varphi ,$ by (1.90)–(1.92). With $\varphi _{1} =\pi - \varphi ,\,\varphi _{2} =2\pi -\varphi$

$$\begin{aligned} {{\varvec{r}}}_{1}= & {} {{\varvec{r}}}_{\mathrm{S}} + {1\over 2}{{\varvec{r}}} =\sqrt{{E_{\mathrm{r}}\over m}} t(1,0) +{p\over 2} \, {1\over \varepsilon \cos (\varphi -\varphi _{0})-1}(-\cos \varphi ,\,\sin \varphi ) \;,\nonumber \\ {{\varvec{r}}}_{2}= & {} {{\varvec{r}}}_{\mathrm{S}} - {1\over 2} {{\varvec{r}}} =\sqrt{{E_{\mathrm{r}}\over m}} t(1,0) -{p\over 2} \, {1\over \varepsilon \cos (\varphi -\varphi _{0})-1}(\cos \varphi ,\,-\sin \varphi ) \;. \end{aligned}$$

(A.16)

The integral (1.95) that relates the variables t and $\varphi$ is calculated as in Example 2. Noting that here $\varepsilon >1$ and making use of the formulae

$$ \mathrm{arctan} \, x=-{i\over 2} \ln {1+\mathrm{i}x\over 1-\mathrm{i}x} \;,\quad {mp\over l} \sqrt{{E_{\mathrm{r}}\over m}} =\sqrt{\varepsilon ^{2}-1} \;, $$

we find that

$$\begin{aligned} \sqrt{{E_{\mathrm{r}}\over m}} t(\varphi )={p\over 2} \left( {1\over \varepsilon ^{2}-1} \ln {1+u\over 1-u} + {\varepsilon \over \sqrt{\varepsilon ^{2}-1}} \, {\sin (\varphi -\varphi _{0})\over \varepsilon \cos (\varphi -\varphi _{0})-1} \right) \;, \end{aligned}$$

(A.17)

where u stands for the expression

$$ u\equiv \sqrt{{\varepsilon +1\over \varepsilon -1}} \mathrm{tan} {\varphi -\varphi _{0}\over 2} \;. $$

Furthermore, we have

$$\begin{aligned}&\cos \varphi _{0} ={1\over \varepsilon } \;,\quad \sin \varphi _{0} ={\sqrt{\varepsilon ^{2}-1}\over \varepsilon } \;,\\&\mathrm{tan} {\varphi -\varphi _{0}\over 2} = {\sin \varphi -\sin \varphi _{0}\over \cos \varphi +\cos \varphi _{0}} = {\varepsilon \sin \varphi -\sqrt{\varepsilon ^{2}-1}\over 1+\varepsilon \cos \varphi } \;. \end{aligned}$$

Equation (A.17) gives the relation between $\varphi$ and t. Using dimensionless coordinates $(2x/p,\, 2y/p),$ one plots points for equidistant values of $\varphi$ and notes the corresponding value of the dimensionless time variable

$$ \tau \,{\buildrel {\scriptstyle \mathrm{def}}\over =}\, {2\over p} \sqrt{{E_{\mathrm{r}}\over m}} \;. $$

Figure 1.35 shows the example $\varepsilon =0.155,$ $\varphi _{0} = 30^{\circ } .$ Alternatively, one may choose a fixed time interval with respect to $t(\varphi _{0} )=0$ and calculate the corresponding values of $\varphi$ from (A.17).