Dynamics of a Material Point

Abstract

A positional force is said to be central with center O if its force law is

$$\mathbf{F} = f(r)\frac{\mathbf{r}} {r},\quad r = \vert \mathbf{r}\vert,$$

(14.1)

where r is the position vector relative to O.

These keywords were added by machine and not by the authors. This process is experimental and the keywords may be updated as the learning algorithm improves.

1 Central Forces

Definition 14.1.

A positional force is said to be central with center O if its force law is

$$\mathbf{F} = f(r)\frac{\mathbf{r}} {r},\quad r = \vert \mathbf{r}\vert,$$

(14.1)

where r is the position vector relative to O.

The elementary work dW of F in the elementary displacement dr is

$$\mathrm{d}W = \mathbf{F} \cdot \mathrm{d}\mathbf{r} = f(r)\frac{\mathbf{r}} {r} \cdot \mathrm{d}\mathbf{r} = f(r)\mathrm{d}r.$$

Then, taking into account (13.41), we can state that a central force is conservative and its potential energy is given by

$$U = -\int \nolimits \nolimits f(r)\mathrm{d}r.$$

(14.2)

A first example of central force is the elastic force for which (14.1) and (14.2) become

$$\mathbf{F} = -kr\frac{\mathbf{r}} {r},\quad U = \frac{1} {2}k{r}^{2},$$

(14.3)

respectively. Another important example of central force is the Newtonian force for which

$$\mathbf{F} = -\frac{k} {{r}^{2}} \frac{\mathbf{r}} {r},\quad U = -\frac{k} {r}.$$

(14.4)

A Newtonian force is said to be attractive if k > 0, repulsive if k < 0.

The torque with respect to the pole O of any central force vanishes. Consequently, the angular momentum K _O is constant:

$${ \mathbf{K}}_{O} ={ \mathbf{K}}_{O}(0),$$

(14.5)

where K _O (0) is the initial value of K _O . On the basis of this result we can prove the following theorem.

Theorem 14.1.

The trajectory of a material point P subject to a central force lies in a plane containing the initial position and velocity of P and orthogonal to the initial angular momentum K _O(0) (Laplace’s plane). Further, in this plane the areal velocity of P is constant (Kepler’s second law).

Proof.

Since K _O is constant, we have that

$${ \mathbf{K}}_{O} = \mathbf{r} \times m\dot{\mathbf{r}} = \mathbf{r}(0) \times m\dot{\mathbf{r}}(0) ={ \mathbf{K}}_{O}(0).$$

(14.6)

Consequently, we can state that r and $\dot{\mathbf{r}}$ belong to the plane π orthogonal to K _O and containing r(0) and $\dot{\mathbf{r}}(0)$. Let Oxyz be a Cartesian system of coordinates such that Oxy ≡ π. In these coordinates, we have that r = (x, y, 0), $\dot{\mathbf{r}} = (\dot{x},\dot{y},0)$, and K _O (0) = (0, 0, K _z (0)). Then (14.6) implies that

$$x\dot{y} -\dot{ x}y = {K}_{z}(0)/m,$$

(14.7)

and the theorem is proved when we recall (11.21). □

The dynamic equation of a point mass P subject to a central force is

$$m\ddot{\mathbf{r}} = f(r)\frac{\mathbf{r}} {r}.$$

(14.8)

If we introduce polar coordinates (r, φ) with center O into the Laplace plane π and we recall formulae (11.19), which give the components of the acceleration in polar coordinates, from (14.8) we obtain the following system of ordinary differential equations:

$$\begin{array}{rcl} m(\ddot{r} - r\dot{{\varphi }}^{2})& =& f(r),\end{array}$$

(14.9)

$$\begin{array}{rcl} \frac{\mathrm{d}} {\mathrm{d}t}\left ({r}^{2}\dot{\varphi }\right )& =& 0\end{array}$$

(14.10)

in the unknowns r(t) and φ(t). Equation (14.10) again states that the areal velocity is constant:

$$\frac{1} {2}{r}^{2}\dot{\varphi } = \frac{1} {2m}{K}_{z}(0) \equiv \frac{1} {2}c.$$

(14.11)

Introducing (14.11) into (14.9), we obtain the following equation in the unknown function r(t):

$$\ddot{r} = \frac{1} {m}\left (f(r) + \frac{m{c}^{2}} {{r}^{3}} \right ).$$

(14.12)

Finally, as the central forces are conservative, the total energy [see (13.43)]

$$\frac{1} {2}m\dot{{\mathbf{r}}}^{2} + U(r) = E$$

(14.13)

assumes a constant value E depending on the initial data. In polar coordinates, we have that $\dot{{\mathbf{r}}}^{2} =\dot{ {r}}^{2} + {r}^{2}\dot{{\varphi }}^{2}$ [see (11.15)] and taking into account (14.11), (14.13) becomes

$$\frac{1} {2}m\left (\dot{{r}}^{2} + \frac{{c}^{2}} {{r}^{2}}\right ) + U(r) = E.$$

(14.14)

In view of (14.11), (14.13) gives

$$\dot{{r}}^{2} = \frac{2} {m}(E - {U}_{\mathrm{eff}}(r)),$$

(14.15)

where we have introduced the effective potential

$${U}_{\mathrm{eff}}(r) = U(r) + \frac{m{c}^{2}} {2{r}^{2}}.$$

(14.16)

Equation (14.11) shows that, if the initial data are chosen in such a way that c = 0, then φ(t) = φ(0) for any t > 0 and the trajectory lies on a straight line containing O. When c≠0, the function $\dot{\varphi }(t)$ has a constant sign, so that φ(t) is always increasing or decreasing, according to the initial data. Consequently, there exists the inverse function

$$t = t(\varphi ),$$

(14.17)

and the trajectory, instead of being expressed by the parametric equations r(t), φ(t), can be put in the polar form r(φ). To find a differential equation giving the function r(φ), it is sufficient to note that the formulae

$$\begin{array}{rcl} \dot{r}& =& \frac{\mathrm{d}r} {\mathrm{d}\varphi }\dot{\varphi } = \frac{c} {{r}^{2}} \frac{\mathrm{d}r} {\mathrm{d}\varphi } = -c \frac{\mathrm{d}} {\mathrm{d}\varphi }\left (\frac{1} {r}\right ),\end{array}$$

(14.18)

$$\begin{array}{rcl} \ddot{r}& =& \frac{\mathrm{d}\dot{r}} {\mathrm{d}\varphi }\dot{\varphi } = \frac{c} {{r}^{2}} \frac{\mathrm{d}\dot{r}} {\mathrm{d}\varphi } = -\frac{{c}^{2}} {{r}^{2}} \frac{{\mathrm{d}}^{2}} {\mathrm{d}{\varphi }^{2}}\left (\frac{1} {r}\right ),\end{array}$$

(14.19)

allow us to write (14.9) and (14.14), respectively, in the form (Binet)

$$\begin{array}{rcl} \frac{{\mathrm{d}}^{2}u} {\mathrm{d}{\varphi }^{2}} + u& =& - \frac{1} {m{c}^{2}{u}^{2}}f\left (\frac{1} {u}\right ),\ \end{array}$$

(14.20)

$$\begin{array}{rcl} \frac{1} {2}{\left (\frac{\mathrm{d}u} {\mathrm{d}\varphi }\right )}^{2}& =& \frac{E} {m{c}^{2}} -\frac{1} {2}{u}^{2} - \frac{1} {m{c}^{2}}U\left (\frac{1} {u}\right ),\end{array}$$

(14.21)

where

$$u = \frac{1} {r(\varphi )}.$$

We highlight that (14.12) and (14.20) contain the constant quantity c. In other words, these equations require a knowledge of the areal velocity or, equivalently, of the angular momentum about an axis orthogonal to the plane in which the orbit lies [see (14.11)] and containing the center of the force. This plane is obtained by giving the initial data r(0) and $\dot{\varphi }(0)$. We can always assume that the initial datum φ(0) is equal to zero by a convenient choice of the polar axis. On the other hand, to integrate (14.12), we must also assign the initial datum $\dot{r}(0)$. The data r(0), $\dot{\varphi }(0)$, and $\dot{r}(0)$ are also necessary for integrating (14.20) since $(\mathrm{d}r/\mathrm{d}\varphi )(0) =\dot{ r}(0)/\dot{\varphi }(0)$ [see (14.18)].

Weierstrass’ analysis (Chap. 10) can be applied to (14.12) and (14.14) to determine the qualitative behavior of the function r(t) as well as the phase portrait in the space $(r,\dot{r})$. In contrast, if we are interested in determining the qualitative behavior of the function r(φ) and the corresponding phase portrait in the space (r, dr ∕ dφ), then we apply Weierstrass’s analysis to equations (14.20) and (14.21) since (14.21) is a first integral of (14.20). We explicitly note that the preceding phase portrait depends on the choice of the constant c. The different curves of the phase portrait in the space $(r,\dot{r})$ are obtained by assigning the initial data r(0), $\dot{r}(0)$, whereas the curve of the phase portrait in the space (r, dr ∕ dφ) is determined by the initial data (r(0), (dr ∕ dφ)(0)). The notebook Weierstrass provides examples of these phase portraits.

Regarding the central forces we recall the following important theorem.

Theorem 14.2 (Bernard).

The elastic forces and the Newtonian forces are the only central forces for which all bounded orbits are closed.

2 Newtonian Forces

Before analyzing the motion of a material point subject to a Newtonian force, we remember one possible way to define a conic. In the plane π we fix a point O at a distance D from a straight line a. Finally, we denote by OP and d the distances of any point P ∈ π from O and a, respectively. Then, a conic is the locus Γ of the points P such that the eccentricity of Γ, defined by the ratio

$$e = \frac{OP} {d},$$

is constant. The point O and the straight line a are respectively called the focus and the directrix of the conic Γ. Applying the preceding definition of a conic to both cases of Fig. 14.1 and using polar coordinates in the plane π, we obtain the relations

$$\begin{array}{l} e = \frac{r} {D - r\cos (\varphi - {\varphi }_{0})}, \\ e = \frac{r} {r\cos (\varphi - {\varphi }_{0}) - D}, \end{array}$$

which lead us to the following equations of a conic:

$$ \begin{array}{rcl} r& =& \frac{eD} {1 + e\cos (\varphi - {\varphi }_{0})},\end{array}$$

(14.22)

$$ \begin{array}{rcl} r& =& \frac{eD} {-1 + e\cos (\varphi - {\varphi }_{0})}.\end{array}$$

(14.23)

Fig. 14.1

Conic

For a material point subject to a Newtonian force (14.4), (14.20) yields

$$\frac{{\mathrm{d}}^{2}u} {\mathrm{d}{\varphi }^{2}} + u = \frac{1} {p},$$

(14.24)

where

$$p = \frac{m{c}^{2}} {k}.$$

(14.25)

The general integral of the linear equation (14.24), with constant coefficients, is

$$u = \frac{1} {p} + R\cos (\varphi - {\varphi }_{0}),$$

where R > 0 and φ₀ are arbitrary constants depending on the initial data. Going back to the variable r and introducing the quantity

$$e = \vert p\vert R,$$

(14.26)

we obtain an equation of the orbit

$$\begin{array}{rcl} r& =& \frac{p} {1 + e\cos (\varphi - {\varphi }_{0})},\;\mathrm{if}\;k > 0,\end{array}$$

(14.27)

$$\begin{array}{rcl} r& =& \frac{p} {-1 + e\cos (\varphi - {\varphi }_{0})},\;\mathrm{if}\;k < 0,\end{array}$$

(14.28)

for attractive (k > 0) or repulsive (k < 0) Newtonian forces. In view of (14.22) and (14.23), we conclude that a material point under the influence of a Newtonian force moves, with constant areal velocity, along a conic having a focus at the center O.

To recognize the meaning of the constants p and φ₀ in (14.27) and (14.28), we begin by noting that these relations do not depend on the sign of the difference (φ − φ₀). Consequently, the axis φ = φ₀ is a symmetry axis s of the conic. Further, for $\varphi = {\varphi }_{0} \pm \pi /2$, we have r = p, so that the parameter of the conicp is equal to the length of the segment OQ, where Q is the intersection point between the conic and the straight line orthogonal at point O to the axis s (Fig. 14.2). Finally, when 0 ≤ e < 1, and φ = φ₀, r assumes its lowest value

$$ {r}_{\mathrm{min}} = \frac{p} {1 + e}, $$

(14.29)

whereas its highest value, assumed for φ = φ₀ ± π, is

$$ {r}_{\mathrm{max}} = \frac{p} {1 - e}. $$

(14.30)

The values (14.29) and (14.30) correspond to the perihelion and aphelion, respectively. In conclusion, when 0 ≤ e < 1, we have that r _min ≤ r ≤ r _max, and the conic is an ellipsis. In particular, it becomes a circumference when e = 0. If e = 1, then we have

$$ {r}_{\mathrm{min}} = \frac{p} {2},\quad {r}_{\mathrm{max}} {=\lim }_{\varphi \rightarrow {\varphi }_{0}+\pi }r = \infty, $$

(14.31)

and the conic is a parabola.

Fig. 14.2

Axis and parameter of the conic

Finally, when e > 1, the perihelion is still given by (14.29) but there exist two values φ = φ₀ ± α for which r = ∞. The two straight lines with these angular coefficients are asymptotes of the conic, which is a branch of a hyperbola with its focus at O.

We can summarize the preceding considerations as follows: a material point subject to an attractive Newtonian central force moves along a conic that may be an ellipse, a parabola, or a branch of a hyperbola. In this last case, the focus O is internal to the branch of the hyperbola. If the force is repulsive, then the trajectory is still a branch of the hyperbola and O is an external focus. In any case, during the motion the areal velocity is constant.

It is important to relate the eccentricity of the conic to the total energy E. In view of (14.14), the conservation of energy (14.18) assumes the form

$$\frac{1} {2}m{c}^{2}\left [{\left ( \frac{\mathrm{d}} {\mathrm{d}\varphi }\left (\frac{1} {r}\right )\right )}^{2} + \frac{1} {{r}^{2}}\right ] -\frac{k} {r} = E.$$

Taking into account (14.25)–(14.27), after some calculations, we can prove the result

$$E = \frac{k} {2p}({e}^{2} - 1) = \frac{{k}^{2}} {2m{c}^{2}}({e}^{2} - 1),$$

which implies that

$$e = \sqrt{1 + \frac{2m{c}^{2 } } {{k}^{2}} E}.$$

(14.32)

This formula yields the following table:

Table 1

relating energy and eccentricity.

We conclude this section with two important remarks.

Remark 14.1.

After determining the orbit r(φ) of a material point subject to a Newtonian force, we must know φ(t) to complete the analysis of the motion. We limit ourselves to the case of closed orbits for which 0 ≤ e ≤ 1. Since the areal velocity is constant, the function φ(t) is a solution of the equation [see (14.11) and (14.27)]

$$\dot{\varphi } = \frac{c} {{r}^{2}(\varphi )} = \frac{c} {{p}^{2}}{(1 + e\cos (\varphi - {\varphi }_{0}))}^{2}.$$

(14.33)

By changing the polar axis, we can always suppose $\varphi (0) = {\varphi }_{0} = 0$. Then, separating the variables in the preceding equation, we obtain the implicit form of the solution

$${\int \nolimits \nolimits }_{0}^{\varphi } \frac{\mathrm{d}\varphi } {{(1 + e\cos \varphi )}^{2}} = \frac{c} {{p}^{2}}t.$$

(14.34)

To find for elliptic orbits an approximate solution of (14.33), we note that if, in particular, the orbit is a circumference (e = 1), then the angular velocity has the constant value a ≡ c ∕ p ² since r is constant. For elliptic orbits, which are similar to circular orbits, it is reasonable to search for an approximate solution of (14.33) in the form

$$\varphi (t) = at + e{\varphi }_{1}(t) + {e}^{2}{\varphi }_{ 2}(t) + \cdots \,.$$

(14.35)

Introducing (14.35) into (14.33) we obtain

$$a + e\dot{{\varphi }}_{1} + {e}^{2}\dot{{\varphi }}_{ 2} + \cdots = a{(1 + e\cos (a + e\dot{{\varphi }}_{1} + {e}^{2}\dot{{\varphi }}_{ 2} + \cdots \,))}^{2}.$$

(14.36)

Applying Taylor’s expansion to the right-hand side of (14.36) and collecting the terms that multiply the same power of the eccentricity, we obtain the following sequence of equations:

$$\begin{array}{rcl} \dot{{\varphi }}_{0}& =& a,\end{array}$$

(14.37)

$$\begin{array}{rcl} \dot{{\varphi }}_{1}& =& 2a\cos (at),\end{array}$$

(14.38)

$$\begin{array}{rcl} \dot{{\varphi }}_{2}& =& {a\cos }^{2}(at) - 2a\sin (at){\varphi }_{ 1}(t),\end{array}$$

(14.39)

$$\begin{array}{rcl} & & \cdots \cdots \\ \end{array}$$

Finally, solving the preceding equations with the initial data φ _i (0) = 0, i = 0, 1, 2, …, we obtain

$$\varphi (t) = at + 2e\sin (at) + {e}^{2}\left (-\frac{3} {2}at + \frac{5} {4}\sin (2at)\right ) + \cdots \,.$$

(14.40)

Remark 14.2.

A material point P with mass m, subject to an attractive Newtonian force, initially occupies the position (r ₀, φ₀). We want to determine for which initial velocity v ₀ of P the orbit is a parabola or a hyperbola, i.e., is open. The lowest value of these velocities is called the escape velocity. We have already shown that open orbits are possible if and only if the total energy verifies the condition

$$E = \frac{1} {2}m{v}_{0}^{2} -\frac{k} {r} \geq 0.$$

Consequently, the escape velocity is given by

$${v}_{0} = \sqrt{ \frac{2k} {m{r}_{0}}}.$$

(14.42)

3 Two-Body Problem

Let S be an isolated system of two material points P ₁ and P ₂ subject to the action of the mutual gravitational force. If m ₁ and m ₂ denote their masses, the motion of S is described by the equations

$$\begin{array}{rcl}{ m}_{1}\ddot{{\mathbf{r}}}_{1}& =& -h \frac{{m}_{1}{m}_{2}} {\vert {\mathbf{r}}_{1} -{\mathbf{r}}_{2}{\vert }^{3}}({\mathbf{r}}_{1} -{\mathbf{r}}_{2}),\ \end{array}$$

(14.43)

$$\begin{array}{rcl}{ m}_{2}\ddot{{\mathbf{r}}}_{2}& =& -h \frac{{m}_{1}{m}_{2}} {\vert {\mathbf{r}}_{2} -{\mathbf{r}}_{1}{\vert }^{3}}({\mathbf{r}}_{2} -{\mathbf{r}}_{1}),\end{array}$$

(14.44)

where the position vector r _i , i = 1, 2, is relative to the origin O of an inertial frame I (Fig. 14.3). Further, the position vector r _G relative to O of the center of mass G of S yields

$$({m}_{1} + {m}_{2}){\mathbf{r}}_{G} = {m}_{1}{\mathbf{r}}_{1} + {m}_{2}{\mathbf{r}}_{2},$$

(14.45)

whereas the position vector r of P ₂ relative to P ₁ is given by

$$ \mathbf{r} ={ \mathbf{r}}_{2} -{\mathbf{r}}_{1}. $$

(14.46)

Solving (14.45) and (14.46) with respect to r ₁ and r ₂, we obtain

$$ \begin{array}{rcl}{ \mathbf{r}}_{1}& =&{ \mathbf{r}}_{G} + \frac{{m}_{2}} {{m}_{1} + {m}_{2}}\mathbf{r},\\ \end{array} $$

$$ \begin{array}{rcl}{ \mathbf{r}}_{2}& =&{ \mathbf{r}}_{G} - \frac{{m}_{1}} {{m}_{1} + {m}_{2}}\mathbf{r}.\\ \end{array} $$

These equations show that the two-body problem can be solved by determining the vector functions r _G (t) and r(t). Since the system S is isolated, we can state that the center of mass moves uniformly on a straight line; therefore, the function r _G (t) is known when the initial data r ₁(0), $\dot{{\mathbf{r}}}_{1}(0)$, r ₂(0), $\dot{{\mathbf{r}}}_{2}(0)$ are given [see (14.45)]. Regarding the function r(t), we note that, dividing (14.43) by m ₁ and (14.43) by m ₂ and subtracting the second equation from the first one, we obtain the equation

$$ {m}_{2}\ddot{\mathbf{r}} = -h\frac{{m}_{2}({m}_{1} + {m}_{2})} {\vert \mathbf{r}{\vert }^{3}} \mathbf{r}, $$

(14.49)

which can equivalently be written in the form

$$ \mu \ddot{\mathbf{r}} = -h\frac{{m}_{1}{m}_{2}} {\vert \mathbf{r}{\vert }^{3}} \mathbf{r}, $$

(14.50)

where we have introduced the reduced mass

$$ \mu = \frac{{m}_{1}{m}_{2}} {{m}_{1} + {m}_{2}}. $$

(14.51)

Fig. 14.3

Two-body system

Both Eqs. (14.49) and (14.50) show that r(t) can be determined by solving a problem of a material point subject to a Newtonian force. In other words, if we introduce the variables r _G and r, the two-body problem is reduced to the problem of a material point moving under the action of a Newtonian force.

Exercise 14.1.

Let P ₁ x ₁ ^′ x ₂ ^′ x ₃ ^′ be a noninertial frame of reference with the origin at the point P ₁ and axes parallel to the axes of an inertial frame Ox ₁ x ₂ x ₃. Prove that Newton’s equation relative to the material point P ₂ in the frame P ₁ x ₁ ^′ x ₂ ^′ x ₃ ^′ coincides with (14.49), provided that we take into account the fictitious forces.

4 Kepler’s Laws

In this section, we study the motion of the planets about the Sun. First, we note that the dimension of the planets is much smaller compared with their distances from the Sun. Consequently, it is reasonable to model the planetary system with a system of material points moving under the influence of the mutual gravitational interaction. Such a problem is too complex to be solved. However, the solar mass is much greater than the mass of any planet. This circumstance suggests adopting a simplified model to analyze the motion of the planets. This model is justified by the following considerations.

The motion of a planet P _i is determined by the gravitational action of the Sun and the other planets. Consequently, in an inertial frame I, the motion of the planets P _i of mass m _i is a solution of the equation

$${m}_{i}\ddot{{\mathbf{r}}}_{i} = -h\frac{M{m}_{i}} {\vert {\mathbf{r}}_{i}{\vert }^{3}}{ \mathbf{r}}_{i} -{\sum \nolimits }_{j\neq i}h\frac{{m}_{i}{m}_{j}} {\vert \mathbf{r}{\vert }_{ij}^{3}}{ \mathbf{r}}_{ij},$$

(14.52)

where h is the gravitational constant, M is the solar mass, r _i is the position vector of the planet P _i with respect to the Sun, and ${\mathbf{r}}_{ij} ={ \mathbf{r}}_{i} -{\mathbf{r}}_{j}$ is the position vector of P _i relative to the planet P _j . Since the solar mass M ≫ m _i , for any planet, while the distances | r _i  | and | r _ij  | are of the same order of magnitude, in a first approximation, we can neglect the action of the other planets on P _i with respect to the prevailing action of the Sun. In this approximation we are faced with a two-body problem: the Sun and the single planet. Consequently, to describe the Sun–planet system, we can resort to (14.49), which, owing to the condition M ≫ m _i , reduces to the equation

$${m}_{i}\ddot{{\mathbf{r}}_{i}} = -h\frac{{m}_{i}M} {\vert {\mathbf{r}}_{i}{\vert }^{3}}{ \mathbf{r}}_{i},$$

(14.53)

according to which, in a first approximation, the planet moves about the Sun as a material point subject to a Newtonian force. We are in a position to prove the following Kepler’s laws:

• 1. The orbit of a planet is an ellipse having a focus occupied by the center of the Sun.

• 2. The orbit is described by a constant areal velocity.

• 3. The ratio between the square of the revolution period T _i and the cube of the greatest semiaxis a _i of the ellipse is given by

  $$\frac{{T}_{i}^{2}} {{a}_{i}^{3}} = \frac{4{\pi }^{2}} {hM},$$

  (14.54)

  i.e., it does not depend on the planet.

The first two laws are contained in the results we have already proved for the motion of a point under the influence of a Newtonian force. To prove the third law, we start by recalling that the area bounded by the ellipse described by the planet is given by the formula πa _i b _i , where b _i is the smallest semiaxis of the ellipse. By the second law, we have that [see (14.11)]

$$\pi {a}_{i}{b}_{i} = \frac{1} {2}{c}_{i}{T}_{i}.$$

On the other hand, it can be proved that ${p}_{i} = {b}_{i}^{2}/{a}_{i}$, so that

$${p}_{i} = \frac{{c}_{i}^{2}{T}_{i}^{2}} {4{\pi }^{2}{a}_{i}^{3}}.$$

Comparing this result with (14.25), we obtain the formula

$$\frac{{T}_{i}^{2}} {{a}_{i}^{3}} = \frac{4{\pi }^{2}{m}_{i}} {{k}_{i}}$$

which, recalling that for Newtonian forces k _i  = hm _i M, implies the Kepler’s third law.

The preceding considerations show that Kepler’s laws are not exact laws since they are obtained by neglecting the mutual actions of the other planets on the motion of a single planet. Taking into account this influence is a very difficult task since it requires a more sophisticated formulation of the problem. This more accurate analysis shows that the orbits of the planets are more complex since they are open curves that, roughly, can be regarded as ellipses slowly rotating about the focus occupied by the Sun.

5 Scattering by Newtonian Forces

Let P be a material point of mass m moving under the action of a repulsive Newtonian force with center at O. If we suppose its energy E to be positive, the trajectory of P will be a branch of a hyperbola γ that does not contain the center O. We denote by ψ the angle between the two asymptotes of γ and define as a scattering angle the angle $\alpha = \pi - \psi$. This is the angle formed by the velocity v _∞ ⁽ⁱ⁾ of P when it comes from infinity and the velocity v _∞ ^(e) when P comes back to infinity (Fig. 14.4). Finally, we call a shock parameter s the distance of O from the asymptote corresponding to the velocity v _∞ ⁽ⁱ⁾.

Fig. 14.4

Scattering of a particle

In the polar coordinates (r, φ), the equation of γ is

$$r = \frac{\vert p\vert } {e\cos \varphi - 1},$$

and it gives for the angle ψ the value

$$\cos \frac{\psi } {2} = \frac{1} {e}.$$

Further, we also have that

$$\sin \frac{\alpha } {2} =\sin \left (\frac{\pi } {2} -\frac{\psi } {2} \right ) =\cos \frac{\psi } {2} = \frac{1} {e}$$

and

$$\tan \frac{\alpha } {2} = \frac{1} {\sqrt{{e}^{2 } - 1}}.$$

(14.55)

On the other hand, the conservation of total energy and angular momentum is expressed by the following formulae:

$$\begin{array}{rcl} E& =& \frac{1} {2}{v}^{2} -\frac{k} {r} = \frac{1} {2}m{\left ({v}_{\infty }^{(i)}\right )}^{2},\end{array}$$

(14.56)

$$\begin{array}{rcl} \vert {\mathbf{K}}_{O}\vert & =& \lim\limits_{r->\infty }\vert {\mathbf{K}}_{O}\vert = m\,s\,{v}_{\infty }^{(i)} = s\sqrt{2mE},\end{array}$$

(14.57)

which allow us to give (14.32) the equivalent form

$$e = \sqrt{1 +{ \left (\frac{2sE} {k} \right )}^{2}}.$$

(14.58)

Using this relation and (14.57), we finally obtain that

$$s = \frac{k} {2E}\cot \frac{\alpha } {2}.$$

(14.59)

The preceding formula is important for the following reasons. Rutherford proposed the nuclear atomic model according to which an atom is formed by a small nucleus, containing the entire positive charge of the atom, and electrons rotating along elliptic orbits with a focus occupied by the nucleus. This model was in competition with the model proposed by Thomson in which the positive charge was continuously distributed in a cloud having the dimension of the atom, whereas the electrons were contained in fixed positions inside this cloud. Rutherford understood that, to resolve the dispute regarding the correct model, it was sufficient to throw charged particles (α particles) against the atoms contained in a thin gold leaf and to evaluate the scattering angles of those particles. In fact, Thomson’s model, owing to the assumed charge distribution inside the atom, implied small scattering angles. In contrast, the concentrated positive charge of Rutherford’s model implied high scattering angles, especially for small shock parameters. The experimental results of Rutherford’s experience were in full agreement with Eq. (14.59), which, in particular, implies a back scattering for very small shock parameters.

6 Vertical Motion of a Heavy Particle in Air

Let P be a mass point acted upon by its weight m g and the air resistance

$$\mathbf{F} = -R(v)\frac{\mathbf{v}} {v},$$

(14.60)

where m is the mass of P, g the gravitational acceleration, and v the length of the velocity v. To suggest a reasonable form of the function R(v), we recall that a mass point is a schematic model of a small body B. When B is falling in the air, the motion is influenced by the form of B, even if B is small. Consequently, the air resistance must include something resembling those characteristics of B that we discarded in adopting the model of a point particle. A good description of the phenomenon we are considering can be obtained assuming that the air resistance has the following form:

$$R(v) = \mu A\alpha f(v),$$

(14.61)

where μ is the mass density of the air, A is the area of the projection of B on a plane orthogonal to v, and α is a form coefficient that depends on the profile of B. Finally, f(v) is an increasing function confirming the conditions

$$f(0) = 0, \lim\limits_{v\rightarrow \infty }f(v) = \infty.$$

(14.62)

Taking into account the preceding considerations, we can state that the equation governing the motion of P is

$$m\dot{\mathbf{v}} = m\mathbf{g} - R(v)\frac{\mathbf{v}} {v}.$$

(14.63)

Since in this section we limit our attention to falls along the vertical a, we begin by proving that if the initial velocity v ₀ is directed along a, then the whole trajectory lies on a vertical straight line. Let Oxyz be a frame of reference with the origin O at the initial position of P and the axis Oz vertical and downward directed. In this frame, the components along Ox and Oy of (14.63) are written as

$$\begin{array}{rcl} \ddot{x}& =& -\frac{R(v)} {mv} \dot{x},\ \end{array}$$

(14.64)

$$\begin{array}{rcl} \ddot{y}& =& -\frac{R(v)} {mv} \dot{y}.\end{array}$$

(14.65)

Multiplying (14.64) by $\dot{x}$ and (14.65) by $\dot{y}$, we obtain

$$\begin{array}{rcl} \frac{\mathrm{d}} {\mathrm{d}t}\dot{{x}}^{2}& =& -2\frac{R(v)} {mv} \dot{{x}}^{2} \leq 0,\end{array}$$

(14.66)

$$\begin{array}{rcl} \frac{\mathrm{d}} {\mathrm{d}t}\dot{{y}}^{2}& =& -2\frac{R(v)} {mv} \dot{{y}}^{2} \leq 0,\end{array}$$

(14.67)

so that $\dot{{x}}^{2}(t)$ and $\dot{{y}}^{2}(t)$ are not increasing functions of time. Since initially $\dot{x}(0) =\dot{ y}(0) = 0$, we can state that $\dot{x}(t)$ and $\dot{y}(t)$ vanish identically. But in addition, $x(0) = y(0) = 0$, and then $x(t) = y(t) = 0$, for any value of time, and P moves vertically. It remains to analyze the motion along the vertical axis Oz. To this end, we denote by k the vertical downward-oriented unit vector and note that

$$-\frac{\mathbf{v}} {v} = - \frac{\dot{z}} {\vert \dot{z}\vert }\mathbf{k} = \mp \mathbf{k},$$

(14.68)

where we must take the − sign if ż > 0 (downward motion) and the + sign if ż < 0 (upward motion). Finally, to evaluate the vertical motion of P, we must integrate the equation

$$\ddot{z} = mg\left (1 \mp \frac{R(\vert \dot{z}\vert )} {mg} \right )$$

(14.69)

with the initial conditions

$$z(0) = 0,\quad \dot{z}(0) =\dot{ {z}}_{0}.$$

(14.70)

We must consider the following possibilities:

$$\mathrm{(a)}\;\dot{{z}}_{0} < 0,\quad \mathrm{(b)}\;\dot{{z}}_{0} \geq 0.$$

(14.71)

In case (a), we must take the + sign in (14.69) at least up to the instant t ^∗ for which ż(t ^∗) = 0. In the time interval [0, t ^∗], we have that $\ddot{z} > 0$, ż < 0, so that

$$\frac{\mathrm{d}\dot{{z}}^{2}} {\mathrm{d}t} = 2\dot{z}\ddot{z} < 0,$$

and the motion is decelerated. Integrating (14.69) in the interval [0, t ^∗] we obtain the formula

$${\int \nolimits \nolimits }_{\dot{{z}}_{0}}^{0} \frac{\mathrm{d}\dot{z}} {1 + \frac{R(\vert \dot{z}\vert )} {mg} } = g{t}^{{_\ast}},$$

(14.72)

and t ^∗ is finite since the function under the integral is bounded in the interval [ż ₀, 0].

In case (b), the motion is initially progressive, so that we take the − sign in (14.69) at least up to an eventual instant t _∗ in which ż(t _∗) = 0:

$$\ddot{z} = mg\left (1 -\frac{R(\dot{z})} {mg} \right ).$$

(14.73)

On the other hand, it is evident that, under hypotheses (14.62) on f(v), there is one and only one value V such that

$$mg = R(V ).$$

(14.74)

Consider the following three cases:

$$(i)\,\dot{z}(0) = V,\;\;(ii)\,\dot{z}(0) > V,\;\;(iii)\,\dot{z}(0) < V.$$

(14.75)

It is plain to prove that ż(t) = V is a solution of (14.73) verifying the initial condition (i). Owing to the uniqueness theorem, this is the only solution satisfying datum (i). In case (ii) we have that ż > V for any value of t. In fact, if there is an instant t _∗ in which ż(t _∗) = V, then, taking this instant as the initial time, we should always have ż(t) = V, against the hypothesis ż ₀ > V. Further, if ż(t) > V for any t, then from (14.73) we have $\ddot{z}(t) < 0$; consequently, $\dot{z}(t)\ddot{z}(t) < 0$ and ż ²(t) is always decreasing. In conclusion, the motion is progressive and decelerated, and

$$\lim \limits_{t\rightarrow \infty }\dot{z}(t) = V.$$

(14.76)

Reasoning in the same way, we conclude that in case (iii) the motion is progressive and accelerated, and (14.76) still holds.

As an application of the previous analysis, let us consider the case of a parachutist of mass m for which we require a prefixed limit velocity V, i.e., an impact velocity with the ground that does not cause damage to the parachutist. From (14.61) we obtain the cross section A of the parachute in order to attain the requested limit velocity:

$$A = \frac{mg} {\alpha \mu f(V )}.$$

7 Curvilinear Motion of a Heavy Particle in Air

In this section, a qualitative analysis of Eq. (14.63) is presented in the general case in which the direction of the initial velocity v ₀ is arbitrary. Let Oxyz be a frame of reference with the origin O at the initial position of P and an upward directed vertical axis Oz. Further, we suppose that the coordinate plane Oxz contains v ₀, and Ox is directed in such a way as to have $\dot{{x}}_{0} > 0$. Since in this frame $\dot{{y}}_{0} = 0$, and the component of Eq. (14.63) along Oy is still (14.65), we have that $\dot{y}(t) = 0$, at any instant, and we can state that the trajectory lies in the vertical plane containing the initial position and velocity.

From (14.64) and the initial condition $\dot{x}(0) =\dot{ {x}}_{0} > 0$ it follows that the function $\vert \dot{x}(t)\vert$ is not increasing [see (14.66)]. Moreover, there is no instant t ^∗ such that $\dot{x}({t}^{{_\ast}}) = 0$. In fact, if such an instant existed, we could consider the initial value problem given by (14.64) and the initial datum $\dot{x}({t}^{{_\ast}}) = 0$. Since the only solution of this problem is given by $\dot{x}(t) = 0$, the motion should take place along the vertical axis Oz, and this conclusion should be in contradiction with the condition $\dot{x}(0) =\dot{ {x}}_{0} > 0$. In conclusion, $\dot{x}(t)$ is a positive nonincreasing function for any t ≥ 0 so that

$$ \begin{array}{rcl} \lim\limits_{t\rightarrow \infty }\dot{x}(t)& =& \dot{{x}}_{\infty } <\dot{ {x}}_{0},\end{array}$$

(14.77)

$$ \begin{array}{rcl} \lim\limits_{t\rightarrow \infty }\ddot{x}(t)& =& 0.\end{array}$$

(14.78)

In the limit t → ∞, (14.64) and (14.78) lead us to the result

$$ \lim\limits_{t\rightarrow \infty }\dot{x}(t) = 0.$$

(14.79)

We have already proved that the trajectory γ of the material point P is contained in the vertical plane Oxz. To discover important properties of γ, we introduce the unit vector $\mathbf{t} = \mathbf{v}/v$ tangent to γ and denote by α the angle that t forms with the Ox-axis. Then, projecting (14.63) onto the Ox-axis and along t, we have the system

$$\begin{array}{rcl} m \frac{\mathrm{d}} {\mathrm{d}t}(v\cos \alpha )& =& -R(v)\cos \alpha,\end{array}$$

(14.80)

$$\begin{array}{rcl} m\dot{v}& =& -mg\sin \alpha - R(v)\end{array}$$

(14.81)

in the unknowns α(t) and v(t). Introducing into (14.80) the value of $\dot{v}$ deduced from (14.81), we obtain the equation

$$\frac{\mathrm{d}} {\mathrm{d}t}(\cos \alpha ) = \frac{g} {2v}\sin 2\alpha,$$

(14.82)

which can also be written as

$$\dot{\alpha }(t) = -\frac{g\cos \alpha } {v}.$$

(14.83)

Since we refer to the case $\dot{x}(0) > 0$, at the initial instant t = 0 the angle α satisfies the condition

$$-\frac{\pi } {2} < {\alpha }_{0} < \frac{\pi } {2}.$$

Then from (14.82) we have that

$$\begin{array}{rcl} \dot{\alpha }(0) < 0.& &\end{array}$$

(14.84)

Consequently, we can state that α(t), starting from any initial value belonging to the interval $(-\pi /2,\pi /2)$, decreases at least up to an eventual instant t ^∗ in which the right-hand side of (14.83) vanishes, that is, when (Fig. 14.5)

$$\alpha ({t}^{{_\ast}}) = \mp \frac{\pi } {2}.$$

The value t ^∗ is infinite since, in the opposite case, if we adopt the initial datum $\dot{x}({t}^{{_\ast}}) = v({t}^{{_\ast}})\cos \alpha ({t}^{{_\ast}}) = 0$, the corresponding motion should always take place along a vertical line, against the hypothesis that for t = 0 the velocity has a component along the Ox-axis. In conclusion, we can state that the trajectory is downward concave [α(t) decreases] and has a vertical asymptote (Fig. 14.6).

Fig. 14.5

Behavior of function α(t)

Fig. 14.6

Trajectories in air

Finally, we want to prove that, if v(t) is a regular function of t at infinity, then

$$\lim\limits_{t\rightarrow \infty }v(t) = V,$$

(14.85)

where V is the unique solution of (14.74). In fact, the regularity of the positive function v(t) implies one of the following possibilities:

$$\lim \limits_{t\rightarrow \infty }v(t) = +\infty,{\quad \lim }_{t\rightarrow \infty }v(t) = {v}_{\infty } < +\infty.$$

(14.86)

In the first case, since v(t) is positive, either of the following conditions holds:

$$\lim\limits_{t\rightarrow \infty }\dot{v}(t) = +\infty,{\quad \lim }_{t\rightarrow \infty }\dot{v}(t) = a > 0.$$

On the other hand, when $\lim\nolimits_{t\rightarrow \infty }v(t) = +\infty$, from (14.81) we obtain $\lim\limits_{t\rightarrow \infty }\dot{v}$ $(t) = -\infty$, and then (14.86)₁ is false.

When (14.86)₂ is verified, we have

$$\lim\limits_{t\rightarrow \infty }\dot{v(t)} = 0,$$

so that, by (14.81), we prove (14.85).

We wish to conclude this section with some considerations about the mathematical problem we are faced with. Equations (14.83) and (14.81) can also be written in the form

$$\begin{array}{rcl} \dot{\alpha }(t)& =& -\frac{g\cos \alpha } {v},\ \end{array}$$

(14.87)

$$\begin{array}{rcl} \frac{\mathrm{d}v} {\mathrm{d}\alpha }\dot{\alpha }(t)& =& -g\sin \alpha -\frac{R(v)} {m},\end{array}$$

(14.88)

from which we deduce the equation

$$\frac{\mathrm{d}v} {\mathrm{d}\alpha } = v\tan \alpha + \frac{vR(v)} {mg\cos \alpha }$$

(14.89)

in the unknown v(α). Its solution,

$$v = F(\alpha,{\alpha }_{0},{v}_{0}),$$

(14.90)

for given initial data α₀ and v ₀, defines a curve in the plane (α, v). When this function is known, it is possible to integrate the equation of motion. In fact, in view of (14.87), we have that

$$\begin{array}{rcl} v\cos \alpha & =& \dot{x} =\dot{ \alpha }\frac{\mathrm{d}x} {\mathrm{d}\alpha } = -\frac{g\cos \alpha } {v} \frac{\mathrm{d}x} {\mathrm{d}\alpha }, \\ v\sin \alpha & =& \dot{z} =\dot{ \alpha }\frac{\mathrm{d}z} {\mathrm{d}\alpha } = -\frac{g\cos \alpha } {v} \frac{\mathrm{d}z} {\mathrm{d}\alpha }\end{array}$$

(.)

In turn, this system implies

$$\begin{array}{rcl} \frac{\mathrm{d}x} {\mathrm{d}\alpha }& =& -\frac{{v}^{2}} {g} = -\frac{{F}^{2}(\alpha,{\alpha }_{0},{v}_{0})} {g}, \\ \frac{\mathrm{d}z} {\mathrm{d}\alpha }& =& -\frac{{v}^{2}} {g} \tan \alpha = -\frac{{F}^{2}(\alpha,{\alpha }_{0},{v}_{0})} {g} \tan \alpha, \\ \end{array}$$

so that

$$\begin{array}{rcl} x& =& -\frac{1} {g}{\int \nolimits \nolimits }_{{\alpha }_{0}}^{\alpha }{F}^{2}(\xi,{\alpha }_{ 0},{v}_{0})\mathrm{d}\xi \equiv G(\alpha,{\alpha }_{0},{v}_{0}),\end{array}$$

(14.91)

$$\begin{array}{rcl} z& =& -\frac{1} {g}{\int \nolimits \nolimits }_{{\alpha }_{0}}^{\alpha }{F}^{2}(\xi,{\alpha }_{ 0},{v}_{0})\tan \alpha \,\mathrm{d}\xi \equiv H(\alpha,{\alpha }_{0},{v}_{0}).\end{array}$$

(14.92)

These relations, which give the parametric equations of the trajectory for any choice of the initial data, represent the solution of the fundamental ballistic problem since they make it possible to determine the initial angle α₀ and the initial velocity v ₀ to hit a given target.

8 Terrestrial Dynamics

By terrestrial dynamics we mean the dynamics relative to a frame of reference R _e  = (O ^′, x ₁, x ₂, x ₃) at rest with respect to the Earth. Before writing the equation of motion of a material point in the frame R _e  = (O ^′, x ₁, x ₂, x ₃), we must determine how R _e  = (O ^′, x ₁, x ₂, x ₃) moves relative to an inertial frame I = (Oξ₁, ξ₂, ξ₃). We take the frame I = (Oξ₁, ξ₂, ξ₃) with its origin at the center O of the Sun, axes oriented toward fixed stars, and containing the terrestrial orbit in the coordinate plane Oξ₁, ξ₂ (Fig. 14.7). It is well known that the motion of the Earth relative to I is very complex. However, we reach a sufficiently accurate description of terrestrial dynamics supposing that the Earth describes an elliptic orbit about the Sun while it rotates uniformly about the terrestrial axis a.

Fig. 14.7

Terrestrial frame

Then we consider the frame R _e ^′ = (O ^′, ξ₁ ^′, ξ₂ ^′, ξ₃ ^′) with its origin at the center O ^′ of the Earth, axes Oξ₃ ^′ ≡ a, and the other axes with fixed orientation relative to the axes of I. It is evident that the motion of R _e ^′ relative to I is translational but not uniform. Finally, we consider a frame R _e  = (O ^′, x ₁, x ₂, x ₃) with Ox ₃ ≡ a and the other two axes at rest with respect to the Earth. Because the dynamical phenomena we want to describe take place in time intervals small compared with the solar year, we can assume that the rigid motion of R _e ^′ relative to I is translational and uniform. In other words, R _e ^′ can be supposed to be inertial, at least for time intervals much shorter than the solar year. Finally, the motion of R _e relative to R _e ^′ is a uniform rotation about the terrestrial axis a. Now we are in a position to determine the fictitious forces acting on a material point P moving relative to the frame R _e :

$$-m{\mathbf{a}}_{\tau } = m{\omega }_{T}^{2}\,\overrightarrow{QP},\quad - m{\mathbf{a}}_{ c} = -2m{\omega }_{T} \times \dot{{\mathbf{r}}}^{{\prime}},$$

(14.93)

where m is the mass of P, ω _T the terrestrial angular velocity, Q the projection of P onto the rotation axis a, and $\dot{{\mathbf{r}}}^{{\prime}}$ the velocity of P relative to R _e . Consequently, the equation governing the motion of P relative to the frame R _e is

$$m\ddot{{\mathbf{r}}}^{{\prime}} ={ \mathbf{F}}^{{_\ast}} + m\mathbf{A} + m{\omega }_{ T}^{2}\,\overrightarrow{QP} - 2m{\omega }_{ T} \times \dot{{\mathbf{r}}}^{{\prime}},$$

(14.94)

where

$$m\mathbf{A} = -h\frac{{M}_{T}m} {\vert \mathbf{r}{\vert }^{{\prime}3}}{ \mathbf{r}}^{{\prime}}$$

(14.95)

is the gravitational force acting on P due to the Earth and F ^∗ is the nongravitational force.

We define the weight p of P as the opposite of the force F _e we need to apply to P, so that it remains at rest in the terrestrial frame R _e . From (14.94) we have that

$${\mathbf{F}}_{e} + m\mathbf{A} + m{\omega }_{T}^{2}\,\overrightarrow{QP} = \mathbf{0},$$

and then the weight of P is given by

$$\mathbf{p} = m\left (\mathbf{A} + {\omega }_{T}^{2}\,\overrightarrow{QP}\right ) \equiv m\mathbf{g},$$

(14.96)

where g is the gravitational acceleration. The preceding equation shows that the weight of P is obtained by adding the gravitational force m A, which is exerted on P by the Earth, and the centrifugal force $m{\omega }_{T}^{2}\,\overrightarrow{QP}$. Supposing that the Earth is spherical and recalling that the first force is always directed toward the center O ^′ of the Earth, whereas the second force is orthogonal to the rotation axis of the Earth, we can state that the weight reaches its minimum at the equator and its maximum at the poles. Further, at the poles and at the equator, the direction of p coincides with the vertical to the surface of the Earth. Finally, we remark that the direction of g does not depend on the mass of P. This fundamental property follows from the identification of the masses m appearing in (14.95) and in the centrifugal force $m{\omega }_{T}^{2}\,\overrightarrow{QP}$. If we define as gravitational mass m _g the mass appearing in the gravitational force (14.95) and as inertial mass the mass m _i appearing in Newton’s equation of motion, instead of (14.96), we obtain

$$\mathbf{g} = \mathbf{A} + \frac{{m}_{i}} {{m}_{g}}{\omega }_{T}^{2}\overrightarrow{QP}.$$

(14.97)

In other words, if we distinguish the two masses respectively appearing in the universal attraction law and in Newton’s law of motion, then, due to the presence of the ratio m _i  ∕ m _g , the acceleration g depends on the nature of the body. However, E$\ddot{\mathrm{o}}$tv$\ddot{\mathrm{o}}$s and Zeeman showed, with very high experimental accuracy, that g does not depend on the ratio m _i  ∕ m _g . Consequently, if we adopt the same unit of measure for m _i and m _g , we can assume that ${m}_{i}/{m}_{g} = 1$. This result plays a fundamental role in general relativity.

With the introduction of the weight (14.96) and in the absence of other forces, (14.94) becomes

$$\ddot{{\mathbf{r}}}^{{\prime}} = \mathbf{g} - 2m{\omega }_{ T} \times \dot{{\mathbf{r}}}^{{\prime}}.$$

(14.98)

To consider a first application of (14.98), we choose a frame of reference R _e as in Fig. 14.8 and write (14.98) in a convenient nondimensional form. Let us introduce a reference length L and a reference time interval T. Since the rotation of the Earth produces much smaller effects on the falls of heavy bodies than does the weight, we can take L and T in such a way that $L/{T}^{2} = g$. On the other hand, we cannot use the same time interval T as a measure of ω since a turn of the Earth about its axis requires 24 h, whereas a heavy body falls in a few minutes or seconds depending on the throw height. Consequently, we introduce a new reference time interval, T _T . If we use the same letters to denote dimensional and nondimensional quantities, (14.98) can also be written as

$$ \ddot{{\mathbf{r}}}^{{\prime}} = -\mathbf{k} + 2\epsilon {\omega }_{ T} \times \dot{{\mathbf{r}}}^{{\prime}}, $$

(14.99)

where k is the unit vector along Oz (Fig. 14.8) and ε is the nondimensional parameter

$$\epsilon = \frac{T} {{T}_{T}}.$$

For a falling time T = 60 s and T = 86, 400 s (24 h) we have ε ≃ 0. 0007. In other words, we are faced with a problem to which we can apply Poincaré’s method. Since in the frame R _e k = (0, 0, 1) and ω _T  = (0, ω _T cosγ, ω _T sinγ), (14.99) gives the system

$$ \begin{array}{rcl} \ddot{x}& =& -2\epsilon {\omega }_{T}(\dot{z}\cos \gamma -\dot{ y}\sin \gamma ), \\ \ddot{y}& =& -2\epsilon {\omega }_{T}\dot{x}\sin \gamma, \\ \ddot{z}& =& -1 + 2\epsilon {\omega }_{T}\dot{x}\cos \gamma, \\ \end{array} $$

which is equivalent to the first-order system

$$ \begin{array}{rcl} \dot{x}& =& u, \\ \dot{y}& =& v, \\ \dot{z}& =& w, \\ \dot{u}& =& -2\epsilon {\omega }_{T}(w\cos \gamma - v\sin \gamma ), \\ \dot{v}& =& -2\epsilon {\omega }_{T}u\sin \gamma, \\ \dot{w}& =& -1 + 2\epsilon {\omega }_{T}u\cos \gamma \end{array}$$

(.)

Fig. 14.8

Terrestrial frame

The first-order Poincaré expansion of the solution of the preceding system, obtained using the notebook Poincare and referred to by x(t), y(t), and z(t), is

$$\begin{array}{rcl} x& =& \frac{1} {3}\epsilon {t}^{3}\omega \cos \gamma, \\ y& =& 0, \\ z& =& 1 -\frac{{t}^{2}} {2} \end{array}$$

(.)

This approximate solution shows that the rotation of the Earth, while a heavy body is falling, gives rise to an eastward deviation.

9 Simple Pendulum

We call a heavy particle P with mass m, constrained to moving along a smooth circumference γ lying in a vertical plane π (Fig. 14.9), a simple pendulum.

Fig. 14.9

Simple pendulum

The equation of motion of P is

$$m\ddot{\mathbf{r}} = m\mathbf{g} + \Phi,$$

(14.100)

where g is the gravitational acceleration and Φ the reactive force exerted by the constraint. Projecting (14.100) onto the unit vector t tangent to γ and recalling that the curvilinear abscissa s = lφ, where l is the radius of γ, and Φ ⋅t = 0 [see (13.51)], we obtain a second-order differential equation

$$\ddot{\varphi } = -\frac{g} {l} \sin \varphi $$

(14.101)

in the unknown φ(t). This equation admits one and only one solution when the initial data φ(0) and $\dot{\varphi (0)}$ are given. Since there is no solution in a finite form of the preceding equation, we resort to a qualitative analysis of (14.101). Due to the form of (14.101), we can resort to the method discussed in Sect. 10.8 to obtain a phase portrait (Fig. 14.10).

Fig. 14.10

Phase portrait of simple pendulum

The conservation of the total mechanical energy is written as

$$\frac{1} {2}m{l}^{2}\dot{{\varphi }}^{2} - mgl\cos \varphi = E,$$

so that we have

$$\dot{{\varphi }}^{2} = \frac{2} {m{l}^{2}}(E + mgl\cos \varphi ).$$

(14.102)

Since the potential energy $U(\varphi ) = -mgl\cos \varphi$ is periodic, it is sufficient to analyze the motion for φ ∈ [ − π, π].

The motion is possible if E + mglcosφ ≥ 0. When $E = -mgl$, point P occupies the position φ = 0 with velocity equal to zero [see (14.102)]. This position is a stable equilibrium position since the potential energy has a minimum at φ = 0. For − mgl < E < mgl the motion is periodic between the two simple zeros of the equation

$$E + mgl\cos \varphi = 0,$$

and the corresponding period of the motion is given by the formula

$$T = 2{\int \nolimits \nolimits }_{{\varphi }_{1}(E)}^{{\varphi }_{2}(E)} \frac{\mathrm{d}\varphi } {\sqrt{ \frac{2} {m{l}^{2}}(E + mgl\cos \varphi )}},$$

(14.103)

which is easily obtained by (14.102). For E = mgl we have again an equilibrium position φ =  ∓ π that is unstable since in this position the potential energy has a maximum. In view of (14.102), the level curve corresponding to this value of the energy is

$$\dot{{\varphi }}^{2} = 2\frac{g} {l} (1 +\cos \varphi ).$$

(14.104)

This level curve contains three trajectories. One of them corresponds to the already mentioned unstable equilibrium position. Since in this position the first derivative of potential energy vanishes, any motion with initial condition − π < φ(0) < π and velocity such that the total energy assumes the value E = mgl tends to the position ± π without reaching it. This circumstance is highlighted by the dashed part of the level curve.

10 Rotating Simple Pendulum

In this section we consider a simple pendulum P moving along a smooth vertical circumference γ that uniformly rotates about a vertical diameter a with angular velocity ω. We study the motion of P relative to a frame of reference R = Cxyz, having its origin at the center C of γ, Cz ≡ a, and the plane Cyz containing γ (Fig. 14.11). Since R is not an inertial frame, the equation governing the motion of P is

$$ m\ddot{{\mathbf{r}}}^{{\prime}} = m\mathbf{g} + m{\omega }^{2}\overrightarrow{QP} - 2m\omega \times \dot{{\mathbf{r}}}^{{\prime}} + \Phi, $$

(14.105)

where Q is the orthogonal projection of P on a, t the unit vector tangent to γ, and Φ the reactive force satisfying the condition Φ ⋅t = 0. Denoting by l the radius of γ, recalling the relation s = lφ, where s is the curvilinear abscissa on γ, and projecting (14.105) along t, we obtain the second-order differential equation

$$ m\ddot{\varphi } =\sin \varphi \left ({\omega }^{2}\cos \varphi -\frac{g} {l} \right ). $$

(14.106)

There is no closed solution of this equation, and then we again resort to the qualitative analysis of Sect. 10.8. Multiplying (14.106) by $\dot{\varphi }$, we obtain

$$ \dot{{\varphi }}^{2} = \frac{2} {m{l}^{2}}\left (E + mgl\cos \varphi -{\frac{m{\omega }^{2}{l}^{2}} {2} \cos }^{2}\varphi \right ). $$

(14.107)

To determine the behavior of the solutions of (14.106), we analyze the potential

$$ U(\varphi ) = -mgl\cos \varphi +{ \frac{m{\omega }^{2}{l}^{2}} {2} \cos }^{2}\varphi $$

(14.108)

in the interval [ − π, π]. First, we have

$$ U(0) = ml\left (\frac{{\omega }^{2}l} {2} - g\right ), $$

(14.109)

Fig. 14.11

Rotating simple pendulum

and

$$\begin{array}{rcl}{ U}^{{\prime}}(\varphi ) = ml\sin \varphi (g - {\omega }^{2}l\cos \varphi ).& &\end{array}$$

(14.110)

Therefore, if

$$\frac{g} {{\omega }^{2}l} > 1,$$

then U ^′(φ) = 0 if and only if φ = 0, ± π. In this hypothesis, we can state that the sign of U ^′(φ) [see (14.110)] depends on the sign of sinφ. Consequently, U(φ) decreases for φ < 0, increases for φ > 0, and has a minimum when φ = 0 and a maximum for φ =  ± π (Fig. 14.12).

Fig. 14.12

Phase portrait for (g ∕ ω² l) > 1

In contrast, if

$$\begin{array}{rcl} \frac{g} {{\omega }^{2}l} < 1,& & \\ \end{array}$$

then U ^′(φ) = 0 when $\varphi = 0,\pm \pi,\pm \arccos \left ( \frac{g} {{\omega }^{2}l}\right )$. Moreover, φ = 0, ± π are maxima of U(φ), whereas the angles ± arccos(g ∕ ω² l) correspond to minima. The relative phase portrait is shown in Fig. 14.13. This analysis shows that the phase portrait undergoes a profound change when the ratio g ∕ ω² l changes in a small neighborhood of 1. For this reason it is called a bifurcation value.

Fig. 14.13

Phase portrait for (g ∕ ω² l) < 1

11 Foucault’s Pendulum

Let P be a heavy point moving on the surface S of a smooth sphere with radius l whose center C is fixed with respect to a terrestrial frame of reference R. We wish to study the small oscillations of this spherical pendulum relative to a terrestrial observer R = Oxyz. The axis Oz is chosen to coincide with the vertical of the place containing the center C of S, and the axes Ox and Oy are horizontal (Figs. 14.14 and 14.15). The equation governing the motion of P relative to a terrestrial observer R = Oxyz is

$$ \ddot{\mathbf{r}} = \mathbf{g} - 2{\omega }_{T} \times \dot{\mathbf{r}} + \Phi, $$

(14.111)

where ω _T is the angular velocity of the Earth and Φ is the reactive force exerted on P by the spherical constraint S with the equation

$$ f(x,y,z) \equiv {x}^{2} + {y}^{2} + {(z - l)}^{2} - {l}^{2} = 0. $$

(14.112)

The sphere S is smooth so that Φ = λ ∇ f, where λ is an unknown function of (x, y, z). Since also ω _T  = (ω _T sinα, 0, ω _T cosα), the components of (14.111) along the axes of R are

$$ \begin{array}{rcl} \ddot{x}& =& 2{\omega }_{T}\dot{y}\cos \alpha + 2x\lambda,\\ \end{array} $$

$$ \begin{array}{rcl} \ddot{y}& =& 2{\omega }_{T}\dot{z}\sin \alpha - 2{\omega }_{T}\dot{x}\cos \alpha + 2y\lambda,\\ \end{array} $$

$$ \begin{array}{rcl} \ddot{z}& =& -g - 2{\omega }_{T}\,\dot{y}\,\sin\,\alpha - 2(l - z)\lambda.\\ \end{array} $$

We are interested in the small oscillations of the spherical pendulum about the position φ = 0. This circumstance simplifies the study of Eqs. (14.113)–(14.115), as we can see by a nondimensional analysis. Introduce as the length of reference the radius l of the sphere S, and denote by $\overline{\varphi }$ the angle corresponding to the largest oscillation; then we have that (Fig. 14.16)

$$ x,y \simeq l\overline{\varphi },\quad z \simeq l{\overline{\varphi }}^{2}. $$

(14.116)

Fig. 14.14

Spherical pendulum

Fig. 14.15

Spherical pendulum and terrestrial frame

Fig. 14.16

Order of magnitude of x, y, and z

Now we introduce the two reference times

$$T = \sqrt{ \frac{l} {g}},\quad {T}_{T},$$

(14.117)

where T is approximately equal to the oscillation period of a simple pendulum and T _T is equal to the number of seconds in a day. Writing (14.115) in nondimensional form and using the same symbols for nondimensional and dimensional quantities, we have

$$\frac{l{\overline{\varphi }}^{2}} {{T}^{2}} \ddot{z} = -g - 2 \frac{l\overline{\varphi }} {T\,{T}_{T}}{\omega }_{T}\dot{y}\sin \alpha - 2l\lambda + l{\overline{\varphi }}^{2}z\lambda,$$

that is,

$$\lambda \simeq -\frac{g} {2l}.$$

(14.118)

Operating in the same way with (14.115) and taking into account (14.118), we obtain the approximate system

$$\begin{array}{rcl} \ddot{x}& =& 2{\omega }_{T}\dot{y}\cos \alpha -\frac{g} {l} x,\end{array}$$

(14.119)

$$\begin{array}{rcl} \ddot{y}& =& -2{\omega }_{T}\dot{x}\cos \alpha -\frac{g} {l} y,\end{array}$$

(14.120)

which in vector form is written as

$$\ddot{{\mathbf{r}}}_{\perp } = -\frac{g} {l}{ \mathbf{r}}_{\perp }- 2{\omega }^{{_\ast}}\times \dot{{\mathbf{r}}}_{ \perp },$$

(14.121)

where r _⊥ is the vector obtained projecting the position vector r of P onto the horizontal plane Oxy and ω^∗ = (0, 0, ω _T cosα).

To solve (14.121), we introduce a new frame of reference R ^′ = (Ox ^′ y ^′ z ^′) with the same origin of R, Oz ^′ ≡ Oz, and rotating about Oz with angular velocity  − ω^∗. In the transformation R → R ^′ we have [see (12.38), (12.39), and (12.44)]

$$\begin{array}{rcl} \mathbf{r}& =&{ \mathbf{r}}^{{\prime}}, \\ \dot{\mathbf{r}}& =& \dot{{\mathbf{r}}}^{{\prime}}-{\omega }^{{_\ast}}\times {\mathbf{r}}^{{\prime}}, \\ \ddot{\mathbf{r}}& =& \ddot{{\mathbf{r}}}^{{\prime}} +{ \mathbf{a}}_{ \tau } - 2{\omega }^{{_\ast}}\times \dot{{\mathbf{r}}}^{{\prime}} =\ddot{{ \mathbf{r}}}^{{\prime}}- {\omega }^{{_\ast}2}{\mathbf{r}}^{{\prime}}- 2{\omega }^{{_\ast}}\times \dot{{\mathbf{r}}}^{{\prime}}\end{array}$$

(.)

Introducing the preceding relations into (14.121) and noting that

$$2{\omega }^{{_\ast}}\times ({\omega }^{{_\ast}}\times {\mathbf{r}}^{{\prime}}) = -2{\omega }^{{_\ast}2}{\mathbf{r}}^{{\prime}} + 2({\omega }^{{_\ast}}\cdot {\mathbf{r}}^{{\prime}}){\omega }^{{_\ast}} = -2{\omega }^{{_\ast}2}{\mathbf{r}}^{{\prime}}$$

since ω^∗ ⋅r ^′ = 0, (14.121) becomes

$$\ddot{{\mathbf{r}}}^{{\prime}} = -\left (\frac{g} {l} + {\omega }^{{_\ast}2}\right ){\mathbf{r}}^{{\prime}}.$$

(14.122)

This equation shows that in the frame R ^′, the pendulum oscillates with a period $2\pi /\sqrt{\frac{g} {l} + {\omega }^{{_\ast}2}}$, whereas its plane of oscillation rotates with angular velocity − ω^∗ about the vertical axis Oz.

12 Exercises

• 1. Neglecting the resistance of the air, prove that the velocity of a point particle projected upward in a direction inclined at 60 to the horizontal is half of its initial velocity when it arrives at its greatest height.

• 2. Neglecting the resistance of the air, find the greatest distance that a stone can be thrown with initial velocity v ₀.

• 3. A simple pendulum executes small oscillations in a resisting medium with damping $-h\dot{\varphi }$. Determine the qualitative behavior of the motion and the loss of energy after n oscillations.

• 4. Supposing the Earth to be spherical, with what velocity must a projectile be fired from the Earth’s surface for its trajectory to be an ellipse with major axis a?

• 5. A bomb is dropped from an airplane flying horizontally at height h and with speed v. Assuming a linear law of resistance − mgk v, show that, if k is small, the fall time is approximately

  $$\sqrt{\frac{2h} {g}} \left (1 + \frac{1} {6}gk\sqrt{\frac{2h} {g}} \right ).$$

• 6. A particle moves in a plane under the influence of the central force

  $$f(r) = \frac{b} {{r}^{2}} + \frac{c} {{r}^{4}},$$

  where a and b are positive constants and r is the distance from the center. Analyze qualitatively the motion, prove the existence of a circular orbit, and determine its radius.

• 7. Repeat the analysis required in the preceding exercise for a particle moving under the action of the central force

  $$f(r) = k{r}^{-2}{\mathrm{e}}^{-{r}^{2} }.$$

• 8. A point moves under the action of a Newtonian attractive force. Resorting to the stability theorems of Chap. 10 relative to the equilibrium stability, verify that the elliptic orbits are stable. Are the motions along these orbits stable?

• 9. A particle moves in a plane attracted to a fixed center by a central force

  $$f(r) = -\frac{k} {{r}^{3}},$$

  where k is a positive constant. Find the equation of the orbits.

• 10. Let γ be a vertical parabola with equation y = x ². Determine the motion of a particle P of mass m constrained to moving smoothly on γ under the influence of its weight and of an elastic force $\mathbf{F} = -k\overrightarrow{QP}$, where k is a positive constant and Q is the orthogonal projection of P onto the axis of the parabola.