Last edited by hbghlyj at 2022-2-14 08:54:00这份力学讲义的52页
Figure 17: A particle moves from a point $P$ at time $t$ to a point $Q$ at time $t+\delta t$, with the angle subtended at the origin changing by a small amount $\delta \theta$, sweeping out a region $O P Q O$, of area $\delta A$ as shown in green. Let $r$ denote the distance $|O P|$, with $r+\delta r$ the distance $|O Q|, P^{\prime}$ the point on $O Q$ such that $\left|O P^{\prime}\right|=r$ and $Q^{\prime}$ the point on OP, such that $\left|O Q^{\prime}\right|=r+\delta r$.
Proof of K2: Kepler's second law is a simple consequence of conservation of angular momentum. Recall from conservation of angular momentum we have $r^{2} \dot{\theta}=h=$ constant.
A straight line from the Sun to a planet is simply the position vector $\mathbf{r}(t)$. In a small time interval $\delta t$, as shown in Figure 17) the planet sweeps out a region $O P Q O$ that is approximately triangular.
In the first instance, with the points $O, P, P^{\prime}, Q, Q^{\prime}$ as given in Figure 17 , we consider $\delta r=$ $|O Q|-|O P| \geq 0$, with $r=|O P|$.
Noting the area of the circular sector, of radius $s$, subtending an angle $\psi$ between its radii, is given by $\psi s^{2} / 2$, we have
$$
\frac{1}{2} r^{2} \delta \theta \leq \delta A \leq \frac{1}{2}(r+\delta r)^{2} \delta \theta .
$$
Dividing by $\delta t$ and taking $\delta t \rightarrow 0$, so that $\delta r \rightarrow 0, \delta \theta \rightarrow 0$, we have
$$
\dot{A}=\frac{1}{2} r^{2} \dot{\theta}=\frac{1}{2} h=\text { constant }\tag{6.47}
$$
which also holds by an analogous argument when $\delta r<0$.$\blacksquare$Aside Being a consequence only of conservation of angular momentum, Kepler's second law holds for any central force (even non-conservative ones).
Proof of K3: Recall that the area of an ellipse with semi-major axis $a$ and semi-minor axis $b$ is
$$
A=\pi a b .\tag{6.48}
$$
We know from K2 that this area is swept out at a constant rate $\dot{A}=\frac{1}{2} h$. Integrating this over one period we obtain
$$
A=\int \mathrm{d} A=\frac{1}{2} h \int \mathrm{d} t=\frac{1}{2} h T .\tag{6.49}
$$
Thus the square of the period $T$ is
$$
T^{2}=\frac{4 A^{2}}{h^{2}}=\frac{4 \pi^{2} a^{2} b^{2}}{h^{2}}=\frac{4 \pi^{2}}{G_{N} M_{S}} \cdot \frac{b^{2}}{a r_{0}} \cdot a^{3},\tag{6.50}
$$
where in the last step we have substituted $h^{2}=\kappa r_{0} / m=G_{N} M_{S} r_{0}$ using (6.20), where $\kappa=G_{N} m M_{S}$, $m$ is the mass of the planet and $M_{S}$ is the mass of the Sun.
Recall from Eqn (6.24) for the equation of a Kepler problem ellipse, we have
$$
a=\frac{r_{0}}{\left(1-e^{2}\right)}, \quad b=\frac{r_{0}}{\left(1-e^{2}\right)^{1 / 2}},\tag{6.51}
$$and thus
$$
\frac{b^{2}}{a r_{0}}=1
$$
Hence
$$
T^{2}=\frac{4 \pi^{2}}{G_{N} M_{S}} a^{3},\tag{6.52}
$$
which is precisely Kepler's third law.$\blacksquare$
1. Conservation of angular momentum. The angular momentum vector is defined as the cross-product
$$
\mathbf{M}=m(\mathbf{r} \times \dot{\mathbf{r}}),
$$
where $m$ is the mass of the planet. Differentiating, one finds
$$
\dot{\mathbf{M}}=m(\dot{\mathbf{r}} \times \dot{\mathbf{r}})+m(\mathbf{r} \times \ddot{\mathbf{r}})=\mathbf{0},
$$
whenever the force field is central (i.e. $m \ddot{\mathbf{r}}=\mathbf{F}(\mathbf{r})$ is proportional to the radius-vector $\bf r$). Thus, in a central force field, the angular momentum vector is conserved. In particular, the direction of $\mathbf{M}$ is conserved, that is, the motion of the planet occurs in the plane spanned by the radius-vector $\mathbf{r}(0)$ and the velocity vector $\dot{\mathbf{r}}(0)$ at the initial moment $t=0$. Thereby Kepler's problem in 3D is reduced to Kepler's problem in 2D. 2. Kepler's 2nd law describes relative times the celestial body spends in various parts of the trajectory: In equal times, the radius-vector of the body sweeps equal areas. In other words, the "sectorial velocity" is constant. This is true for motion in any central force field and follows from the the conservation of the angular momentum vector $\mathbf{M}=m(\mathbf{r} \times \dot{\mathbf{r}})$. In particular, the length $\mu:=m|\mathbf{r} \times \dot{\mathbf{r}}|$ of the angular momentum vector is conserved, and so is the sectorial velocity $|\mathbf{r} \times \dot{\mathbf{r}}| / 2=\mu / 2 m$. 3. The cone and the orbits. In the 3-dimensional space with coordinates $(x, y, r)$,
consider the cone given by the equation
$$r^2 = x^2 + y^2.$$
A purely geometric theorem below recasts Kepler’s 1st law as the claim that in Kepler’s 2D problem with the center of attraction located at the origin $(x, y) = (0, 0)$,
trajectories are the projections to the $(x, y)$-plane of plane sections of this cone. 定理 圓錐截线到$(x, y)$-平面的投影是一条二次曲線，其焦點是圓錐的頂點，準線是圓錐的截面和$(x, y)$-平面的交線，偏心率等于截面和$(x, y)$-平面之間的角度的正切。
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Fig. 1 An orbit’s focus, directrix, and eccentricity證明$$P'O=PP'⇒{P'O\over P'Q}={PP'\over P'Q}=\tan(∠PQP^′)$$
4. Kepler's 1st law by differentiation. Suppose that a point is moving on the surface of the cone $r^{2}=x^{2}+y^{2}$ in such a way that the projection of the point to the plane obeys the equation of motion $\ddot{\mathbf{r}}=-k \mathbf{r} /|\mathbf{r}|^{3}$. We represent the position of the point on the cone by its radius-vector $\mathbf{R}$ in space with respect to the origin shifted by distance $l$ along the axis of the cone:
$$
\mathbf{R}:=\mathbf{r}+r \mathbf{e}-l \mathbf{e}, \text { where } \mathbf{e}=(0,0,1) .
$$
We choose the shift to be different for different trajectories and determined by the value of sectorial velocity: $l=\mu^{2} / {km}^{2}$. Since it is conserved, the same value of $l$ serves all points of one trajectory.
Proposition. In Kepler's problem on the plane, each trajectory (with the given value of sectorial velocity), when lifted to the cone $r^{2}=x^{2}+y^{2}$, obeys the equation of motion:
$$
\ddot{\mathbf{R}}=-k \frac{\mathbf{R}}{r^{3}}
$$
Fig. 2 Fictitious angular momentumProof. Differentiating $\mathbf{R}$ with respect to time, we find:
$$
\dot{\mathbf{R}}=\dot{\mathbf{r}}+\frac{\dot{\mathbf{r}} \cdot \mathbf{r}}{r} \mathbf{e}
$$
Differentiating again, we get:
$$
\ddot{\mathbf{R}}=\ddot{\mathbf{r}}+\frac{\ddot{\mathbf{r}} \cdot \mathbf{r}}{r} \mathbf{e}+\frac{\dot{\mathbf{r}} \cdot \dot{\mathbf{r}}}{r} \mathbf{e}-\frac{(\dot{\mathbf{r}} \cdot \mathbf{r})^{2}}{r^{3}} \mathbf{e}
$$
Taking in account the equation of motion $\ddot{\mathbf{r}}=-k \mathbf{r} /|\mathbf{r}|^{3}$, we have:
$$
\ddot{\mathbf{R}}=-k \frac{\mathbf{r}}{r^{3}}-\frac{k}{r^{2}} \mathbf{e}+\frac{(\dot{\mathbf{r}} \cdot \dot{\mathbf{r}})(\mathbf{r} \cdot \mathbf{r})-(\dot{\mathbf{r}} \cdot \mathbf{r})^{2}}{r^{3}} \mathbf{e}
$$
Note that
$$
(\dot{\mathbf{r}} \cdot \dot{\mathbf{r}})(\mathbf{r} \cdot \mathbf{r})-(\dot{\mathbf{r}} \cdot \mathbf{r})^{2}=|\mathbf{r} \times \dot{\mathbf{r}}|^{2}=\frac{\mu^{2}}{m^{2}}=k l,
$$
four times the square of the sectorial velocity. Thus, at a fixed value of sectorial velocity, we obtain:
$$
\ddot{\mathbf{R}}=-k \frac{\mathbf{r}+r \mathbf{e}-l \mathbf{e}}{r^{3}}=-k \frac{\mathbf{R}}{r^{3}} .
$$
Corollary 1. The fictitious angular momentum $\mathbf{N}:=m(\mathbf{R} \times \dot{\mathbf{R}})$ is conserved.
Proof. $\dot{\mathbf{N}}=m(\dot{\mathbf{R}} \times \dot{\mathbf{R}})+m(\mathbf{R} \times \ddot{\mathbf{R}})=\mathbf{0}$, since $\ddot{\mathbf{R}}$ is proportional to $\mathbf{R}$.
In particular, the direction of vector $\mathbf{N}$ is conserved, and so the trajectory lies in the section of the cone by the plane passing through the point $l\bf e$ and spanned by vectors $\mathbf{R}(0)$ and $\dot{\mathbf{R}}(0)$ (perpendicular to $\mathbf{N}$ ). We have arrived at
Corollary 2 (Kepler's 1st law). When lifted from the plane to the cone, Keplerian trajectories become plane sections of the cone.
Corollary 3. Keplerian trajectories with a fixed value of sectorial velocity correspond to the sections of the cone by planes passing through the same point $l\bf e$ on the axis, $l=\mu^{2} / {km}^{2}$.
Note that $l$ here is non-negative. To obtain negative values of $l$ one must assume that $k<0$, i.e. that the planet is repelled from the Sun by the central anti-gravity force inversely proportional to the square of the distance (as in Coulomb's law for electric charges of the same sign). A plane crossing the axis of the cone at a point below the plane $r=0$ can intersect the part of the cone above this plane along one of the branches of a hyperbola, namely the branch unlinked with the axis. Thus, as a bonus, we obtain:
Corollary 4. In Kepler's anti-gravity problem, planets move along branches of hyperbolas with the Sun at the focus, namely along those branches which (being separated from the focus by the directrix) don't go around the Sun.
Remark. The equation of motion on the surface of the cone splits into two: the usual Newton equation for radius-vector $\mathbf{r}$ on the plane, and the scalar equation
$$
\ddot{r}=-\frac{k}{r^{2}}+\frac{\mu^{2}}{m^{2} r^{3}} .
$$
The latter is the effective Newton equation in polar coordinates at a fixed value of sectorial velocity. The traditional solution of Kepler’s problem consists in treating the effective equation as a mechanical system with one degree of freedom and integrating it using the effective energy conservation law.
Differentiating $\mathbf{R}$ with respect to time, we find:
$$
\dot{\mathbf{R}}=\dot{\mathbf{r}}+\frac{\dot{\mathbf{r}} \cdot \mathbf{r}}{r} \mathbf{e}
$$
Corollary 4. In Kepler's anti-gravity problem, planets move along branches of hyperbolas with the Sun at the focus, namely along those branches which (being separated from the focus by the directrix) don't go around the Sun.
Repulsive inverse–square interactions
正如庫侖定律中相同符號的電荷相互排斥,還沒接近就會被彈開
If we change the sign on the inverse square force, the only orbits that are possible are hyperbolas. The hyperbola has two branches, one of which is the solution for the attractive force, and the other is the solution for the repulsive force. The figures below show the trajectories for a variety of impact parame- ters for incoming velocities equal to 1, 2, 4 and 8 units respectively. inference.org.uk/teaching/dynamics/tex/orbits.pdf#page=61.svg(166.59 KB, Downloads: 88)
Remark. The equation of motion on the surface of the cone splits into two: the usual Newton equation for radius-vector $\mathbf{r}$ on the plane, and the scalar equation
$$
\ddot{r}=-\frac{k}{r^{2}}+\frac{\mu^{2}}{m^{2} r^{3}} .
$$
The latter is the effective Newton equation in polar coordinates at a fixed value of sectorial velocity. The traditional solution of Kepler’s problem consists in treating the effective equation as a mechanical system with one degree of freedom and integrating it using the effective energy conservation law.
5. Dandelin’s spheres. The definition of an ellipse as a quadratic curve with the eccentricity < 1 competes with another definition of an ellipse as the locus of points on the plane with a fixed sum of distances to two given points called foci. The following famous geometric proof of the fact that closed conic sections are ellipses was invented by French mathematician G. P. Dandelin (1794 – 1847).
Into a conical cup (Figure 3), place two balls so that they, being tangent to the cone, also touch the secting plane (one from above the other from below). The claim is that the points $F$ and $G$ of tangency with the plane are the foci of the conic section. Indeed, given a point $A$ on the conic section, the segments $AF$ and $AG$ are tangent to the respective balls at the points $F$ an $G$ since the balls touch the plane at these points. The segments $AB$ and $AC$ of the generator $OA$ of the cone are also tangent to the balls, since the balls touch the cone at $B$ and $C$ respectively. Since all tangents to the same ball from the same point have the same length, one concludes that $AF = AB, AG = AC,$ and hence $AF + AG = BC$. The latter is the distance along a generator of
the cone between the parallel circles along which the cone touches the balls. This distance does not depend on the position of $A$ on the conic section.
Exercise. Extend Dandelin’s construction to hyperbolic sections of the cone.
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Fig. 3 Dandelin’s spheres6. Osculating paraboloids. Here we reconcile the two definitions of ellipses by locating the second focus of a Keplerian orbit.
Theorem. Into the conical cup, inscribe a paraboloid of revolution so that it touches the secting plane. Then the projection of the point of tangency to the horizontal plane is the second focus of the projected conic section (the vertex of the cone being the first one).
Proof. First, note that a paraboloid of revolution is the locus of points equidistant from its focus lying on the axis of revolution and the directrix plane perpendicular to it. Consequently, the plane tangent to the paraboloid at a given point is the locus of points equidistant from the focus and the foot of the perpendicular dropped from the given point to the directrix. (This shows that the direction normal to the tangent plane makes equal angles with the direction to the focus and the direction of the axis, which implies the famous optical property of the parabolic mirror to reflect all rays parallel to the axis into the focus.) To justify this claim, consider the plane of all points equidistant from the focus and the aforementioned foot of the perpendicular dropped from the given point. Any other point on this plane is farther from the foot (and hence from the focus) than from the directrix, and therefore lies outside the paraboloid. Thus, the plane has only one common point with the paraboloid, and hence touches it at this point.
Lemma. Paraboloids of revolution inscribed into the cone $r^{2}=x^{2}+y^{2}$ have the plane $r=0$ as the directrix, and the center of the circle of tangency with the cone as the focus.
Indeed, at the circle of tangency, the tangent planes make $45^{\circ}$ with the direction of the axis, hence the center of the circle must be the focus, and hence the plane $r=0$ the directrix.
Now let $P$ be a point on the section of the cone by the plane tangent to the paraboloid at $F$ (Figure 4 ), $P^{\prime}$ and $P^{\prime \prime}$ be the projections of $P$ to the horizontal planes through the focus $O^{\prime}$ of the paraboloid and the vertex $O$ of the cone respectively, and likewise $F^{\prime}$ and $F^{\prime \prime}$ be the projections of $F$. Then the right triangles $P P^{\prime} O^{\prime}$ and $F^{\prime \prime} P^{\prime \prime} P$ are congruent, since $P O^{\prime}=P F^{\prime \prime}$ (as was explained before the lemma), and $P^{\prime} O^{\prime}=P^{\prime \prime} O=P^{\prime \prime} P$ (as was already noted in Section 3). Therefore $P^{\prime \prime} F^{\prime \prime}=P^{\prime} P$ and so $P^{\prime \prime} O+P^{\prime \prime} F^{\prime \prime}=P^{\prime} P^{\prime \prime}$, the distance from the focus $O^{\prime}$ of the paraboloid to the directrix plane, and does not depend on the choice of a point $P$ on the conic section.
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Fig. 4 Osculating paraboloid
Corollary. Elliptic Keplerian orbits with a fixed length of their major axis correspond to the sections of the cone by planes tangent to the same paraboloid of revolution inscribed into the cone (and the length is equal to the distance from the focus to the directrix of the paraboloid).
Indeed, the major axis has length equal to the sum of the distances to the foci. We will see that the same condition characterizes orbits with a fixed period of revolution, and a fixed value of total energy.
Exercise. Extend the proof and the corollary to hyperbolic orbits.
7. Kepler's 3rd law. The period $T$ of revolution can be found as the ratio of the area enclosed by the orbit to the sectorial velocity. The latter is $\mu / 2 m$, and the square of it $\mu^{2} / 4 m^{2}$ coincides with $k l / 4$ (in our earlier notation). The area of an ellipse with semiaxes $a \geq b$ is equal to $\pi a b$. Combining, we find a geometric expression for the square of the period:
$$
T^{2}=\frac{4 \pi^{2}}{k} \frac{a^{2} b^{2}}{l} .
$$
The farthest from and closest to the Sun positions of the planet are called the aphelium and perihelium respectively. Denote by $r_{1}$ and $r_{2}$ the respective distances to the Sun, and examine Figure 5A showing the axial cross-section of the cone over the major axis of an elliptic orbit.
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Fig. 5 Arithmetic, geometric, and harmonic means
Proposition. The major (a) and minor $(b)$ semiaxes of an elliptic orbit, and the altitude ($l$) of the corresponding secting plane over the vertex of the cone are respectively the arithmetic, geometric, and harmonic means between the aphelium $\left(r_{1}\right)$ and perihelium $\left(r_{2}\right)$ distances:
$$
a=\frac{r_{1}+r_{2}}{2}, \quad b=\sqrt{r_{1} r_{2}}, \quad l=\frac{2}{1 / r_{1}+1 / r_{2}}
$$
Proof. The first is obvious. The second is standard (see Figure 5B): By the Pythagorean theorem, since the half-distance $f$ between the foci equals $\left(r_{1}-r_{2}\right) / 2$, we have $b^{2}=a^{2}-f^{2}=\left(r_{1}+r_{2}\right)^{2} / 4-\left(r_{1}-r_{2}\right)^{2} / 4=r_{1} r_{2}$. Lastly (see Figure 5A), the area $r_{1} r_{2}$ of the right triangle $K O M$ equals the sum of the areas ${lr}_{1} / 2$ and ${lr}_{2} / 2$ of the triangles into which the angle bisector $O L$ divides the triangle $K O M$. Thus the length $l$ of the bisector equals $2 r_{1} r_{2} /\left(r_{1}+r_{2}\right)$.
We have $a l=b^{2}$ : the geometric mean between two quantities is the geometric mean between their arithmetic and harmonic means. Thus, $b^{2} / l=a$, and
$$
T^{2}=\frac{4 \pi^{2}}{k} a^{3} .
$$
This is Kepler's 3rd law: The squares of the periods are proportional to the cubes of the orbits' major semiaxes. 8. The total energy. The sum of the kinetic and potential energy
$$
E:=m \frac{|\dot{\mathbf{r}}|^{2}}{2}-\frac{m k}{|\mathbf{r}|},
$$
is conserved, as it is readily verified by differentiation.
On the other hand, at the aphelium and perihelium, where the velocity vector $\dot{\mathbf{r}}$ is perpendicular to $\mathbf{r}$, we have (from conservation of sectorial velocity) $|\mathbf{r}||\dot{\mathbf{r}}|=\mu / m$. Excluding the velocities, we obtain the following quadratic equation for $r=r_{1}$ and $r=r_{2}$
$$
\frac{\mu^{2}}{2 m}\left(\frac{1}{r}\right)^{2}-m k\left(\frac{1}{r}\right)-E=0
$$
By Vieta theorem, we have
$$
\frac{2}{l}=\frac{1}{r_{1}}+\frac{1}{r_{2}}=\frac{2 k m^{2}}{\mu^{2}}, \text { and } r_{1}+r_{2}=-\frac{m k}{E} .
$$
The first equality shows (again) that at a fixed value of sectorial velocity the secting planes hit the axis of the cone at the same point. The second equality implies that elliptic orbits have negative total energy, and that the value of $E / m$ is determined by the major semiaxis of the orbit.
Corollary. Orbits with a fixed value of the total energy (as well as orbits with a fixed period of revolution) correspond to sections of the cone by planes tangent to the same paraboloid of revolution inscribed into the cone.
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Fig. 6 The effect of the angular defect
9. The gravity of curvature. The Keplerian dynamics lifted from the plane to the cone $x^{2}+y^{2}=r^{2}$ can be considered as Lagrangian dynamics on the cone itself with potential energy $-m k / r$, where however the distance $r$ to the vertex of the cone, as well as the kinetic energy, are dictated by the metric on the cone obtained from the degenerate metric $(d x)^{2}+(d y)^{2}+0(d r)^{2}$ in 3-space, induced by the projection of the space to the plane. Consider now how the dynamics changes when the metric on the cone is induced from a more general metric of the form $(d x)^{2}+(d y)^{2}+\varepsilon(d r)^{2}$, Euclidean for $\varepsilon>0$ and Minkovsky for $\varepsilon<0$ (but $>-1$ so that the cone remains Euclidean). When cut along a generator, the cone can be developed isometrically to the plane. The development covers a sector which for $\varepsilon \neq 0$ differs from full angle $2 \pi$ by the angular defect $\alpha$ (easily computed as $\alpha=2 \pi(\sqrt{1+\varepsilon}-1)$. The ratio $\alpha / 2 \pi$ can be interpreted as the amount of the Gaussian curvature of the cone accumulated at the vertex. This curvature has the following effect on the Keplerian dynamics (see Figure 6): the usual Keplerian orbits on the development plane jump from one edge of the cut to the other and proceed along the trajectory rotated through the angle $\alpha$. This results in the rotation of the perihelium by $\alpha$ radians per revolution - somewhat similar to the rotation of the perihelium of Mercury explained by the general theory of relativity.
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Fig. 7 Two beads on the strings10. An exercise on hidden symmetries. The integrability of Kepler’s problem is
based on the law of conservation of angular momentum, which, in full agreement
with E. Noether’s theorem, is due to the rotational symmetry (isotropy) of space. In
the following problem, the isotropy is explicitly broken:
Two identical beads slide without friction along two perpendicular strings (Figure 7) and interact with each other gravitationally, i.e. by central attracting force
proportional to the inverse square of the distance between the beads. Describe the
motion of this system (assuming that each bead crosses the other bead’s string without collision).
References
V. I. Arnold. Huygens and Barrow, Newton and Hooke: Pioneers in mathematical analysis and catastrophe theory from evolvents to quasicrystals. Birkh ̈auser, Basel, 1990.
V. I. Arnold, V. V. Kozlov, A. I. Neishtadt. Mathematical aspects of classical and celestial mechanics.
Dynamical systems – III (Encyclopaedia of Mathematical Sciences), Springer, 1987.
D. L. Goodstein & J. R. Goodstein. Feynmans Lost Lecture, The Motion of Planets Around the Sun. W.
W. Norton & Co., New York, 1996.
J. Milnor. On the geometry of the Kepler problem. Amer. Math. Monthly, 90 (1983), 353-365.
In antiquity, the great mathematician and astronomer Ptolemy had laid out
a system for computing where planets would be found in the sky. It was
based on the theory that the sun and the planets moved around the earth in
epicycles, a pattern of circles which are themselves spinning in circles.
It should be noted that his computations actually worked quite well to
predict the night sky. However, the computations were quite messy. In the
16th century, Copernicus pointed out that the calculations could be
simplified if instead it was assumed that the earth moved around the sun.
This theory had been considered and rejected by Ptolemy. While Copernicus
was correct that it simplified the calculations, it still required the use
of epicycles. Johannes Kepler discovered that epicycles could be
entirely avoided if instead of the planets were assumed to move in
elliptical orbits instead of circular orbits.
Kepler's three laws of
planetary motion are:
Planetary orbits are ellipses with the sun at one focus.
Planets do not move with a constant speed, but the line segment
joining the sun and a planet will sweep out equal areas in equal
times.
The square of the period of the orbit of a planet is proportional to
the cube of the semi-major axis of the orbit.
This was an amazing feat given that Kepler did not actually have the
positions of the planets but was working from the data accumulated by
Tycho Brahe on where different planets appeared in the sky on different
days. Kepler published his first two laws in 1609, and the third law in
1619. The laws were not immediately accepted. Part of this was religious
resistance to the idea that the earth moved about the sun. Galileo's
conviction for heresy in supporting a Copernican rather than Ptolemaic
view was in 1633. But part of the objections were because there seemed
no reason for the use of ellipses rather than perfect circles. Kepler's
theories slowly gained acceptance during the 17th century.
In 1687, Newton published his Principia in which he demonstrated
that Kepler's three laws were all consequences of the single assumption
that planetary motion is governed by an inverse square attraction
(gravity) toward the sun. This both sealed the acceptance of Kepler's laws
in the scientific community, and established Newton's theory of gravity,
since it explained all three laws in one simple assumption. While Newton
was well-known in the scientific community prior to the publication of the
Principia, afterward he became much more generally famous, in much
the same way Einstein would be viewed in the 20th century.
The derivation of Kepler's laws from Newton's theory of gravity is
generally considered one of the greatest intellectual achievements in
history.
Newton's Principia does not use calculus, but it does make
extensive use of the infinitesimal techniques Newton used to develop the
calculus. It seems fitting that as the capstone on the calculus sequence,
we go through the development of Kepler's laws from Newton's theory. We
will use modern calculus and differential equations techniques rather than
the infinitesimal arguments Newton used. In fact, we will show that
writing the equations for planetary motion based on Newton's theory of
gravity leads to a non-linear second order system of differential
equations. Along the way, we will deduce Kepler's second law. We will then
use our standard trick for converting parametric
systems to a single equation along with an inspired change of variables to
reduce the non-linear system to a single linear, constant coefficient,
differential equation. We will solve this using the techniques of chapter
2. This solution can be recognized as a conic section, establishing
Kepler's first law. Finally, we will use Kepler's second law in
combination with a formula for area of an ellipse to establish Kepler's
third law.
Acceleration in polar coordinates
It will be most convenient to work in polar coordinates, with the sun at
the origin and the axes oriented so aphelion, the point where the orbit is
farthest from the sun, is along $\theta=0$. We write the position of
the orbiting planet is the standard form
$$\text{position} = \langle r\cos(\theta),r\sin(\theta) \rangle.$$
I use $\langle a,b \rangle$ for the vector because we will soon have
enough $()$'s that
we it will be helpful to have different separators for vectors.
Here $r=r(t)$ and $\theta=\theta(t)$ are assumed to be functions changing
with time, since the planet is moving. To find the velocity vector, we
differentiate position with respect to time. We use the product rule to
get
$$\text{velocity} = \langle r'\cos(\theta)-r\theta'\sin(\theta),
r'\sin(\theta)+r\theta'\cos(\theta) \rangle.$$
Differentiating again we get acceleration.
$$
\begin{align}
\text{acceleration} &= \langle r''\cos(\theta) - 2r'\theta'\sin(\theta) -
r(\theta')^2\cos(\theta) - r\theta''\sin(\theta),
r''\sin(\theta) + 2r'\theta'\cos(\theta) - r(\theta')^2\sin(\theta) +
r\theta''\cos(\theta) \rangle \\
&= \langle
(r'' - r(\theta')^2\cos(\theta) + (2r'\theta'+r\theta'')(-\sin(\theta)),
(r'' - r(\theta')^2\sin(\theta) + (2r'\theta'+r\theta'')(\cos(\theta))
\rangle \\
&=(r''-r(\theta')^2)\langle\cos(\theta),\sin(\theta)\rangle\quad
+ \quad(2r'\theta' + r\theta'')\langle-\sin(\theta),\cos(\theta)\rangle
\end{align}
$$
Note that our acceleration vector naturally splits into two pieces. Since
the vector $\langle\cos(\theta),\sin(\theta)\rangle$ lies along the radius
from the
origin, we call
$$a_r = r'' - r(\theta')^2$$
the radial component of accelartion. The vector
$\langle-\sin(\theta),\cos(\theta)\rangle$ is normal to the radial vector
and we call
$$a_{\theta} = 2r'\theta' + r\theta''$$
the transverse component of acceleration.
Newton's Law of Gravity
Newton's law of gravity states that the attractive force between two
masses is
$$\frac{GMm}{r^2}$$
where $M$ is the mass of one object, $m$ the mass of the other object, $r$
the distance between the two objects, and $G$ is the gravitational
constant, $G\approx 6.67 \times 10^{-11} m^3/(kg\cdot sec^2).$ We let $M$
be the mass of the sun and $m$ be the mass of the planet. Since $F=ma$,
the $m$ cancels out and we find
the planet undergoes an acceleration of magnitude $GM/r^2$ toward the sun.
Since the acceleration is toward the sun, that means we have $a_r =
-GM/r^2$. On the other hand, $a_{\theta}=0$ since the acceleration is
purely toward the sun and so there is no transverse component. Using our
formulas for $a_r$ and $a_{\theta}$ from the previous section gives
us a system of two non-linear differential equations for $r$ and $\theta$,
$$
\begin{align}
r''-r(\theta')^2 &= -\frac{GM}{r^2} \\
2r'\theta'+r\theta'' &= 0
\end{align}
$$
Conservation of Angular Momentum
From the second equation we can derive Kepler's second law. The key trick
is to observe
$$
\begin{align}
\frac{d}{dt}(r^2\theta') &= 2rr'\theta' + r^2\theta'' \\
&= r(2r'\theta'+r\theta'') \\
&=ra_{\theta} \\
&=0
\end{align}
$$
So $r^2\theta' = c$ for some constant $c$. Physicists refer to this as the
law of conservation of angular momentum. In a little bit, we will find it
very useful to have
$$
\theta' = \frac{c}{r^2}
$$
to be able to replace $\theta'$ terms by $r$ terms. But first, we will
use this result to establish Kepler's second law.
Kepler's Second Law
In calculus you (should have) learned that the area swept out by a polar
curve is
$$
\text{Area} = \int \frac12 r^2 d\theta.
$$
Treating $r$ and $\theta$ as functions of $t$, so $d\theta = \theta'dt$,
and using the result from the previous section that $r^2\theta' = c$, we
get that the area swept out by the line joining the sun (at the origin) to
the planet between times $t_0$ and $t_1$ is
$$
\begin{align}
\text{Area} &= \int_{t_0}^{t_1} \frac12 r^2\theta' dt \\
&= \int_{t_0}^{t_1} \frac{c}{2} dt \\
&=\frac{c(t_1-t_0)}{2}.
\end{align}
$$
Hence the area swept out depends only on the length of time, or in
Kepler's terms, the planet sweeps out equal areas in equal times.
Writing An Equation For The Orbit
We now turn our attention to the key equation,
$$r'' - r(\theta')^2 = -\frac{GM}{r^2}.$$
This equation involves $r$ and $\theta$ as functions of time $t$. If we
want to plot the orbit as a whole, instead of determining the
coordinates of the planet at time $t$, it would be better to have an
equation for $r$ as a function of $\theta$. We have our standard trick
to write
$$
\frac{dr}{d\theta} = \frac{r'}{\theta'}.
$$
But this doesn't actually help much since our equations aren't just simple
formulas for $r'$ and $\theta'$. However, there is an inspired change of
variables that makes everything work out neatly. Instead of working with
$r$, we will use $u = 1/r$. Then $\displaystyle u' = -r'/r^2$.
And then
$$
\begin{align}
\frac{du}{d\theta} &= \frac{u'}{\theta'} \\
&= \frac{-r'/r^2}{c/r^2} \\
&= -\frac{r'}{c}
\end{align}
$$
where we used the formula for $\theta'$ we found in the section on
"Conservation of Angular Momentum."
Now since we have a second order equation, we will need to compute the
second derivative of $u$ with respect to $\theta$.
$$
\begin{align}
\frac{d^2u}{d\theta^2} &= \frac{d}{d\theta}\left(\frac{du}{d\theta}\right)
\\
&= \frac{d/dt(du/d\theta)}{d\theta/dt} \\
&= \frac{d/dt(-r'/c)}{\theta'} \\
&= \frac{-r''/c}{c/r^2} \\
&= -\frac{r^2r''}{c^2}.
\end{align}
$$
Now return to our equation
$$
r'' - r(\theta')^2 = -\frac{GM}{r^2}.
$$
Making our standard substitution for $\theta'$ we get
$$
r'' - r\left(\frac{c}{r^2}\right)^2 = -\frac{GM}{r^2}.
$$
Multiply through by $\displaystyle -\frac{r^2}{c^2}$ to obtain
$$
-\frac{r^2r''}{c^2} + \frac{1}{r} = \frac{GM}{c^2}.
$$
Now substitute in our formulas for $u$ and $\displaystyle
\frac{d^2u}{d\theta^2}$ to get
$$
\frac{d^2u}{d\theta^2} + u = \frac{GM}{c^2}.
$$
So we have found a nice constant coefficient linear differential equation.
And we can solve this using the techniques of chapter 2 to obtain
$$
u = \frac{GM}{c^2} - A\cos(\theta - \theta_0),
$$
where $A$ is an arbitrary constant. Changing back to $r = 1/u$ we get
$$
r = \frac{1}{GM/c^2 - A\cos(\theta - \theta_0)}.
$$
Remember that we defined our coordinate system so aphelion occurs at
$\theta = 0$. Since $r$ will be largest when the denominator is smallest,
and $\cos$ takes its maximum value at 0, we must have $\theta_0=0$ so the
orbit of the planet in polar coordinates is
$$
r = \frac{1}{GM/c^2 - A\cos(\theta)}.
$$
Kepler's First Law
Isaac Newton would have recognized the equation above as the general
equation for a conic section in polar coordinates with a focus at the
origin and major axis along the $x$-axis. And that
establishes Kepler's first law, that the orbit of the planets is an
ellipse with the sun at one focus. Now the ellipse is not the only conic
section. But it is the only conic section with a closed curve, so a planet
that repeats the same orbit must follow an ellipse.
Since polar forms of conic sections has drifted out of the curriculum at
most schools, you may be curious about how to recognize the formula above
describes a conic section. The algebra is straightforward, but rather
messy. We start with
$$
r = \frac{1}{GM/c^2 - A\cos(\theta)}.
$$
We will assume $A < GM/c^2$, which corresponds to an ellipse ($A >
GM/c^2$ is a hyperbola while $A = GM/c^2$ is a parabola). We will find
it convenient to first multiply the fraction on the right top and bottom
by $c^2/(GM)$ to get
$$
r = \frac{c^2/(GM)}{1 - (Ac^2/(GM))\cos(\theta)}.
$$
We can then prepare the equation to switch back to rectangular coordinates
as follows.
$$
\begin{align}
r - (Ac^2/(GM))r\cos(\theta) &= c^2/(GM) \\
r &= c^2/(GM) + (Ac^2/(GM))r\cos(\theta) \\
r^2 &= \left(GM/c^2 + (Ac^2/(GM))r\cos(\theta)\right)^2
\end{align}
$$
Now we can substitute $r^2 = x^2 + y^2$ and $r\cos(\theta)=x$ to get
$$
x^2+y^2 = \left(c^2/(GM) + (Ac^2/(GM))x\right)^2.
$$
This is a quadratic expression, and any second order equation in $x$ and
$y$ corresponds to a conic section (and vice-versa). You can square the
left-hand-side and move terms over to the right and complete the square to
get the standard textbook presentation
$$
\left(\frac{x-f}{a}\right)^2 +
\left(\frac{y}{b}\right)^2 = 1,
$$
but this gets rather messy and I've decided trying to follow the
manipulations is more likely to confuse than enlighten, so we'll stop
here. You can look at problems 19-21 which sketch out the algebra.
The ratio $Ac^2/(GM)$ is called the eccentricity of the orbit. If the
eccentricity is less than 1, the orbit is an ellipse. If the eccentricity
is greater than 1 it is a hyperbola, and if the eccentricity is exactly 1
it is a parabola. Now while it is possible for an extra-solar object to
pass through the solar system with such a high velocity that it follows a
hyperbolic trajectory, the odds against such an event are literally
"astronomical." What has been observed is that a comet which has a very
eccentric orbit gets disturbed by the gravity of a planet to increase its
speed (and hence its eccentricity) to move from an eccentricity of just
under 1 to just over. Since the new trajectory has eccentricity so close
to 1, astronomers typically call these parabolic trajectories, even though
they are technically hyperbolic.
Newton understood
that his work should also apply to comets, but had difficulty getting
predictions that matched with observation. The trouble is that cometary
orbits typically get affected by the gravity of the planets. Newton's
friend, Edmund Halley, worked out those details, which led to his ability
to predict the return of the comet that now bears his name.
To make it look more
like the standard textbook presentation of an ellipse, we can write
$$
\begin{align}
x^2 + y^2 &= (GM/c^2)^2 + 2Ax + (Ac^2/(GM))^2x^2 \\
(1-(Ac^2/(GM))^2)x^2 - 2Ax + y^2 &= (GM/c^2)^2 \\
(1-(Ac^2/(GM))^2)\left(x^2 - \left(\frac{2A}{1-(Ac^2/(GM))^2)}\right)x +
\frac{A^2}{(1 - (Ac^2/(GM))^2)^2}\right) + y^2
&= (GM/c^2)^2 + \frac{A^2}{1-(Ac^2/(GM))^2} \\
(1-(Ac^2/(GM))^2)\left(x - \frac{A}{(1 - (Ac^2/(GM))^2)}\right)^2 + y^2
&= (GM/c^2)^2 + \frac{A^2}{1-(Ac^2/(GM))^2} \\
\alpha(x-f)^2 + y^2 = \beta
\end{align}
$$
where $\alpha=(1-(Ac^2/(GM))^2)$,
$\displaystyle f=\frac{A}{(1 - (Ac^2/(GM))^2)}$, and
$\beta=\displaystyle (GM/c^2)^2 + \frac{A^2}{1-(Ac^2/(GM))^2}$. Since
$0 < \alpha$, this is the
textbook form of an ellipse with major axis along the
$x$-axis (since $\alpha < 1$) and center at $(f,0)$ (recall
that the center of
an ellipse is midway between the two foci).
Area Of An Orbit
Kepler's third law relates the period, the time it takes the planet to
complete an orbit, to the semimajor axis of the ellipse. Kepler's second
law relates time to the area swept out, and we also know how to find the
area of an ellipse given the major and minor axes. We will use this to
find two formulas for the area of an orbit, and then use those to deduce
Kepler's third law.
The standard formula for area of an ellipse is $\pi ab$ where $a$ is the
semi-major axis and $b$ is the semi-minor axis. If we worked out the full
formula for the ellipse in the previous section we could get $a$ and $b$
from the equation, but there is an easier way, that involves developing a
new and pretty formula for area in terms of perihelion and aphelion.
Perihelion is the closest distance from the orbit to the sun (from the
ellipse to the focus), while aphelion is the farthest distance. The
diagram above, which I modified from a
Science
Buddies diagram, is not
accurate in that the Sun should properly be much closer to the left edge
for the indicated orbit, but it gets across the relationships between the
major and minor axes and perihelion and aphelion. In particular, if we let
$f$ be the focal distance, the distance from the sun at a focus to the
center of the ellipse, $a$ be the semi-major axis, and $b$ the semi-minor
axis, then perihelion is $a-f$ while aphelion is $a+f$, so the semi-major
axis is the arithmetic mean of perihelion and aphelion,
$$
a = \frac{\text{perihelion} + \text{aphelion}}{2}.
$$
To get the semi-minor axis, we recall that the defining property of an
ellipse is that the sum of the distances from any point on the ellipse to
the two foci is a constant, which is $2a$, the major axis. So by
symmetry, the distance from the Sun to the ellipse along the hypoteneuse
in the diagram is $a$, while the sides of the right triangle are $f$ and
$b$. Then we have
$$
\begin{align}
b &= \sqrt{a^2 - f^2} \\
&=\sqrt{(a-f)(a+f)} \\
&=\sqrt{\text{perihelion}\times\text{aphelion}}.
\end{align}
$$
So the semi-minor axis, $b$, is the geometric mean of the perihelion and
aphelion. Finally, the area is $\pi$ times the product of the arithmetic
and geometric means of the perihelion and aphelion,
$$
\text{Area} = \pi\frac{\text{perihelion} + \text{aphelion}}{2}
\sqrt{\text{perihelion} \times \text{aphelion}}.
$$
So now we need to find the perihelion and the aphelion. Fortunately, we
can do this easily from the polar form we derived above,
$$
r = \frac{1}{GM/c^2 - A\cos(\theta)}.
$$
Since $r$ is the distance to the sun, perihelion is just the minimum value
of $r$, while aphelion is the maximum value of $r$. Since a fraction is
smallest when the denominator is largest, to find perihelion we just need
to maximize $GM/c^2 - A\cos(\theta)$. But since we know $-1 \le
\cos(\theta) \le 1$, we can see that the maximum value will be
$GM/c^2 + A$ when $\cos(\theta) = -1$, so perihelion will be
$$
\text{perihelion} = \frac{1}{GM/c^2 + A}.
$$
We similarly find
$$
\text{aphelion} = \frac{1}{GM/c^2 - A}.
$$
Then we plug these into our area formula to get
$$
\begin{align}
\text{Area} &= \pi
\frac12\left(\frac{1}{GM/c^2 + A}+\frac{1}{GM/c^2 - A}\right)
\sqrt{\left(\frac{1}{GM/c^2 + A}\right)\left(\frac{1}{GM/c^2 - A}\right)}
\\
&=\pi\left(\frac{GM/c^2}{(GM/c^2)^2 - A^2}\right)
\frac{1}{\sqrt{(GM/c^2)^2 - A^2}} \\
&=\pi \frac{GM/c^2}{\left( (GM/c^2)^2 - A^2 \right)^{3/2}}
\end{align}
$$
Kepler's Third Law
From the above we can compute
$$
\begin{align}
\text{semi-major axis} &= \frac{\text{perihelion} + \text{aphelion}}2 \\
&= \frac{1}{2}\left(\frac{1}{GM/c^2 + A}+\frac{1}{GM/c^2 - A}\right)
\\
&=\frac{GM/c^2}{(GM/c^2)^2 - A^2}.
\end{align}
$$
We want to relate this to the orbital period $T$, the time it takes the
planet to complete an orbit. In the section "Kepler's First Law," we
deduced that the area swept out in a time $t$ was $ct/2$. So in time $T$,
we sweep out area $cT/2$. But this must be the area of the entire orbit,
which we computed in previous section. So
$$ \begin{align}
\frac{cT}{2} &= \pi \frac{GM/c^2}{\left((GM/c^2)^2 - A^2 \right)^{3/2}} \\
T &= 2\pi \frac{GM/c^3}{\left( (GM/c^2)^2 - A^2 \right)^{3/2}} \\
T^2 &= 4\pi^2 \frac{(GM)^2/c^6}{\left( (GM/c^2)^2 - A^2 \right)^3} \\
T^2 &= \frac{4\pi^2}{GM}\frac{(GM)^3/c^6}{\left( (GM/c^2)^2 - A^2
\right)^3} \\
T^2 &= \frac{4\pi^2}{GM}\left(\frac{(GM)/c^2}{(GM/c^2)^2 - A^2}\right)^3
\\
T^2 &= \frac{4\pi^2}{GM}(\text{semi-major axis})^3.
\end{align}
$$
So the orbital period squared is proportional to the semi-major axis
cubed, with the constant of proportionality $\displaystyle
\frac{4\pi^2}{GM}$.
Note that we have not only established Kepler's third law, we've also
worked out the specific constant of proportionality in terms of
physical constants. This enables us to "weigh" the sun.
An accurate determination of the semi-major axis of
the earth's orbit
was made based on measurements taken during the transits of Venus
in 1761 and 1769. An accurate measurement for $G$ was carried out by Henry
Cavendish in 1798 using a torsion balance. Since the orbital period of the
earth is 1 year, from this information we are able to measure the mass of
the Sun.
You may use $G \approx 6.67 \times 10^{-11} m^3 kg^{-1} sec^{-2}$ for the
gravitational constant in the following problems.
According to Newton's Law of Gravity, the acceleration due to gravity
at the earth's surface is $\displaystyle g = \frac{GM}{r^2}$ where $M$ is
the mass of the earth and $r$ is the radius of the earth. Using
$g \approx 9.81 m/sec^2$ and $r \approx 6.37 \times 10^6 m$, approximate
the mass of the earth.
We are unable to stand on the surface of the sun to measure the
acceleration of its gravity. But we can find the mass using Kepler's
third law, that $\displaystyle \frac{T^2}{a^3} = \frac{4\pi^2}{GM}$, where
$T$ is the orbital period, $a$ is the semi-major axis, and $M$ is the mass
of the sun. For earth's orbit, $T \approx 3.156 \times 10^7 sec$ and
$a \approx 1.496 \times 10^{11} m$. What is the approximate mass of the
sun?
Europa orbits Jupiter with semi-major axis about $6.709 \times 10^8 m$
and period $3.068 \times 10^5 sec$. What is the approximate mass of
Jupiter?
Phobos orbits Mars with semi-major axis about $9.376 \times 10^6 m$
and period $2.755 \times 10^4 sec$. What is the approximate mass of
Mars?
For the next problems you may find it easier to use the standard
gravitational parameter for the sun,
$\mu = GM \approx 1.327 \times 10^{20} m^3/sec^2$
rather than having to mulitply $G$ and $M$ each time.
The European Space Agency recently landed the Philae probe on Comet
67P/Churyumov-Gerasimenko (67P/C-G). Comet 67P/C-G is a short period comet
with an orbital period of approximately $2.03 \times 10^8 sec$. It has a
perihelion of approximately $1.86 \times 10^{11} m$. What is its aphelion?
One group of short period comets typically have orbits that take them
from aphelion at about the distance of Jupiter from the Sun to the inner
solar system for perihelion. This gives them a typical semi-major axis of
around $10^{11} m$ to $5 \times 10^{11} m$. What would the periods be for
such comets?
Observing that the fossil record shows major extinction events tend to
happen about every 25 million years, it has been theorized that perhaps
there is a red or brown dwarf star orbiting the sun, called
Nemesis, that has an orbit that disturbs trans-Neptunian bodies
every 25 million years or so, leading to much higher chances of cometary
impacts on the inner planets. What would the semi-major axis be for an
orbit with a period of 25 million years (remember that $\mu$ is given
above in units of seconds, not years)? Note: astronomical surveys have
looked for such an object and not found anything, so the Nemesis
hypothesis is currently considered incorrect.
You may find it interesting to compare your answers to problems 9 and 10
to the mean speed of the Earth in orbit, which is about $30,000 m/sec$.
Note that the Earth's orbit is close to circular (eccentricity 0.0167) so
the speed only varies a few percent over the year.
Speed is the magnitude of the velocity vector. Show that at perihelion
and aphelion, speed is $c/r$ where $c$ is the angular velocity. Hint:
there is something special about perihelion and aphelion that will
simplify the calculations.
Halley's comet has a period of about $2.38 \times 10^9 sec$, a
perihelion of about $8.766 \times 10^{10} m$, and aphelion of about $5.25
\times 10^{12} m$ (you could actually deduce this last value from the
first two using Kepler's third law). What is the speed of Halley's comet
at perihelion? Note: it is usually easiest to find $c$ by remembering the
area of the orbit is $\displaystyle \frac{cT}{2}$ from our derivation of
Kepler's second law.
Halley's comet has a period of about $2.38 \times 10^9 sec$, a
perihelion of about $8.766 \times 10^{10} m$, and aphelion of about $5.25
\times 10^{12} m$ (you could actually deduce this last value from the
first two using Kepler's third law). What is the speed of Halley's comet
at aphelion? Note: it is usually easiest to find $c$ by remembering the
area of the orbit is $\displaystyle \frac{cT}{2}$ from our derivation of
Kepler's second law.
Suppose a comet is observed with a perihelion of $1.30 \times 10^{11}
m$ and moving with a speed of $4.50 \times 10^4 m/sec$ at perihelion.
What is the period of the orbit of the comet? Hint: recall that orbits
take the form $\displaystyle r = \frac{1}{GM/c^2 - A\cos(\theta)}.$ With
this data it is easiest to solve for $c$ using the speed data and then
solve for $A$. Then use these values to find the area of the orbit and
from that the period.
Suppose a comet is observed with a perihelion of $7.50 \times 10^{10}
m$ and moving with a speed of $5.00 \times 10^4 m/sec$ at perihelion.
What is the period of the orbit of the comet? Hint: recall that orbits
take the form $\displaystyle r = \frac{1}{GM/c^2 - A\cos(\theta)}.$ With
this data it is easiest to solve for $c$ using the speed data and then
solve for $A$. Then use these values to find the area of the orbit and
from that the period.
Suppose a comet has a perihelion of $5.50 \times 10^{10} m.$ What
speed would the comet need to have at perihelion to be on a "parabolic"
orbit and escape the solar system?
Suppose a comet has a perihelion of $8.00 \times 10^{10} m.$ What
speed would the comet need to have at perihelion to be on a "parabolic"
orbit and escape the solar system?
Suppose Bennett's Law of Gravity says that the gravitational force
between two masses is given by $\displaystyle -\frac{GMm}{r}$ (where the
denominator is $r$ instead of $r^2$). Show the Kepler's second law still
holds for this model.
With Bennett's Law of Gravity from the previous problem, the key
equation in "Writing an Equation for the Orbit" becomes
$$
r'' - r(\theta')^2 = - \frac{GM}{r}.
$$
Show that if you start from here and make the substitution $u = 1/r$ and
write the equation as a differential equation for $u$ as a function of
$\theta$, you get a non-linear equation.
More generally, show that the law of gravitational force
$\displaystyle -\frac{GM}{r^n}$ leads to a linear equation for $u$ as a
function of $\theta$ only when $n=2$ or $n=3$.
From Kepler's first law, we know that if we have an equation for
$u = 1/r$ as a function of $\theta$, then the solution must have the
form $u = B - A\cos(\theta)$, since that is the general form for an
ellipse with a focus at the origin and aphelion at $\theta = 0$.
Show that this is a solution for the law of gravitational force
$\displaystyle -\frac{GM}{r^n}$ only when $n=2$.
By definition, an ellipse consists of all points with the property
that the sum of
the distances from the point to two foci is a fixed quantity. Pick the
coordinate system so the foci are
$(f,0)$ and $(-f,0)$ and the sum of the distances be $2a$, and let
$b^2 = a^2 - f^2$. Show the equation of the ellipse is
$$ \left(\frac{x}{a}\right)^2 + \left(\frac{y}{b}\right)^2 = 1$$
Show that the polar equation $\displaystyle r =
\frac{ep}{1-e\cos(\theta)}$ becomes
$$\left(\frac{x-f}{a}\right)^2 + \left(\frac{y}{b}\right)^2 = 1
$$
with $\displaystyle f = \frac{e^2p}{1-e^2}$,
$\displaystyle a = \frac{ep}{1-e^2}$, and
$\displaystyle b = \frac{ep}{\sqrt{1-e^2}}$.
Show that $b^2 + f^2 = a^2$ for the values of $a$, $b$, and $f$ in
problem 20. Note: problems 19-21 show that
$$ r = \frac{ep}{1-e\cos(\theta)}$$
is the equation of an ellipse with a focus at the origin, since you now
have the equation in problem 20 is that of the ellipse from problem 19,
only translated by $f$, the distance from the focus to the origin.
Note: Because these problems use rounded values for the data, the
answers you get will often differ slightly from the actual values you will
find in a reference work.
1. $5.97 \times 10^{24} kg$.
3. $1.899 \times 10^{27} kg$.
5. $8.49 \times 10^{11} m$.
7. $1.3 \times 10^{16} m$.
9. Approximately 54,500 m/sec.
11. $1.24 \times 10^{10} sec$ (which is about 393 years).
13. Approximately 69,500 m/sec.
15. Since the transverse component of acceleration is still 0, the proof
of the second law in the text works exactly the same for Bennett's law of
gravity as it does for Newton's.
17. Our key equation is
$$
r'' - r(\theta')^2 = - \frac{GM}{r^n}.
$$
Using the conservation of angular momentum, $\displaystyle \theta' =
\frac{c}{r^2}$, we get
\begin{align*}
r'' - r(\theta')^2 &= -\frac{GM}{r^n} \\
r'' - r\left(\frac{c}{r^2}\right)^2 &= -\frac{GM}{r^n} \\
r'' - \left(\frac{c^2}{r^3}\right) &= -\frac{GM}{r^n} \\
-\frac{r^2r''}{c^2} + \frac{1}{r} &= \left(\frac{GM}{c^2}\right)r^{2-n}
\end{align*}
Then making the substitution $u=1/r$, which leads to $\displaystyle
\frac{d^2u}{d\theta^2} = -\frac{r^2r''}{c^2}$ as shown in the text we get
$$
\frac{d^2u}{d\theta^2} + u = \left(\frac{GM}{c^2}\right)u^{n-2}.
$$
For this to be a linear equation, we can only have $u$ to the zeroth or
first power, so we only get a linear equation if $n=2$ or $n=3$.
19. We want points $(x,y)$ where the sum of the distances to $(f,0)$
and $(-f,0)$ is $2a$. From the distance formula we thus get
$$
\sqrt{(x-f)^2+y^2} + \sqrt{(x+f)^2+y^2} = 2a.
$$
To simplify this equation we isolate one of the square roots and square
both sides, then isolate the remaining square root and square both sides
again as follows.
\begin{align*}
\sqrt{(x-f)^2+y^2} + \sqrt{(x+f)^2+y^2} &= 2a \\
\sqrt{(x-f)^2+y^2} &= 2a - \sqrt{(x+f)^2+y^2} \\
(x-f)^2+y^2 &= 4a^2 - 4a\sqrt{(x+f)^2+y^2} + (x+f)^2 + y^2 \\
x^2 - 2xf + f^2 +y^2 &= 4a^2 - 4a\sqrt{(x+f)^2+y^2} + x^2+2xf+f^2 + y^2 \\
-2xf &= 4a^2 - 4a\sqrt{(x+f)^2+y^2} + 2xf \\
4a\sqrt{(x+f)^2+y^2} &= 4a^2+4xf \\
a\sqrt{(x+f)^2+y^2} &= a^2+xf \\
a^2\left((x+f)^2 + y^2\right) &= a^4 +2a^2xf + x^2f^2 \\
a^2x^2 + 2a^2xf + a^2f^2 + a^2y^2 &= a^2 + 2a^2xf + x^2f^2 \\
(a^2-f^2)x^2 + a^2y^2 &= a^4-a^2f^2 \\
(a^2-f^2)x^2 + a^2y^2 &= a^2(a^2-f^2) \\
\left(\frac{x^2}{a^2}\right) + \left(\frac{y^2}{a^2 - f^2}\right) &=1.
\end{align*}
21. We compute
\begin{align*}
b^2 + f^2 &= \frac{e^2p^2}{1-e^2} + \frac{e^4p^2}{(1-e^2)^2} \\
&= \frac{(1-e^2)e^2p^2}{(1-e^2)^2} + \frac{e^4p^2}{(1-e^2)^2} \\
&= \frac{e^2p^2 - e^4p^2 + e^4p^2}{(1-e^2)^2} \\
&= \frac{e^2p^2}{(1-e^2)^2} \\
&= \left(\frac{ep}{1-e^2}\right)^2 \\
&= a^2.
\end{align*}
青青子衿
Posted at 2025-4-8 09:37:02
