6.4 Motion Under a Central Force

We’ll now study the motion of a object moving under the influence of a central force; that is, a force whose magnitude at any point $P$ other than the origin depends only on the distance from $P$ to the origin, and whose direction at $P$ is parallel to the line connecting $P$ and the origin, as indicated in Figure 6.4.1 for the case where the direction of the force at every point is toward the origin. Gravitational forces are central forces; for example, as mentioned in Section 4.3, if we assume that Earth is a perfect sphere with constant mass density then Newton’s law of gravitation asserts that the force exerted on an object by Earth’s gravitational field is proportional to the mass of the object and inversely proportional to the square of its distance from the center of Earth, which we take to be the origin.

If the initial position and velocity vectors of an object moving under a central force are parallel, then the subsequent motion is along the line from the origin to the initial position. Here we’ll assume that the initial position and velocity vectors are not parallel; in this case the subsequent motion is in the plane determined by them. For convenience we take this to be the $xy$-plane. We’ll consider the problem of determining the curve traversed by the object. We call this curve the orbit.

Figure 6.4.1.

We can represent a central force in terms of polar coordinates

\[ x=r\cos\theta,\quad y=r\sin\theta \]

\[ {\bf F}(r,\theta)=f(r)(\cos\theta\,{\bf i}+\sin\theta\,{\bf j}). \]

We assume that $f$ is continuous for all $r>0$. The magnitude of ${\bf F}$ at $(x,y)=(r\cos\theta,r\sin\theta)$ is $|f(r)|$, so it depends only on the distance $r$ from the point to the origin the direction of ${\bf F}$ is from the point to the origin if $f(r)<0$, or from the origin to the point if $f(r)>0$. We’ll show that the orbit of an object with mass $m$ moving under this force is given by

\[ r(\theta)={1\over u(\theta)}, \]

where $u$ is solution of the differential equation

\begin{equation} {d^2u\over d\theta^2}+u=-{1\over mh^2u^2}f(1/u), \tag{6.4.1}\end{equation}

and $h$ is a constant defined below.

Newton’s second law of motion (${\bf F}=m{\bf a}$) says that the polar coordinates $r=r(t)$ and $\theta=\theta(t)$ of the particle satisfy the vector differential equation

\begin{equation} m(r\cos\theta\,{\bf i}+r\sin\theta\,{\bf j})''= f(r)(\cos\theta\,{\bf i}+\sin\theta\,{\bf j}). \tag{6.4.2}\end{equation}

To deal with this equation we introduce the unit vectors

\[ {\bf e}_1=\cos\theta\,{\bf i}+\sin\theta\,{\bf j} \mbox{\quad and \quad} {\bf e}_2=-\sin\theta\,{\bf i}+\cos\theta\,{\bf j}. \]

Note that ${\bf e}_1$ points in the direction of increasing $r$ and ${\bf e}_2$ points in the direction of increasing $\theta$ (Figure 6.4.2); moreover,

\begin{equation} {d{\bf e}_1\over d\theta}={\bf e}_2,\quad {d{\bf e}_2\over d\theta}=-{\bf e}_1, \tag{6.4.3}\end{equation}

and

\[ {\bf e}_1\cdot{\bf e}_2=\cos\theta(-\sin\theta)+\sin\theta\cos\theta=0, \]

so ${\bf e}_1$ and ${\bf e}_2$ are perpendicular. Recalling that the single prime $(')$ stands for differentiation with respect to $t$, we see from (6.4.3) and the chain rule that

\begin{equation} {\bf e}_1'=\theta'{\bf e}_2\mbox{\quad and \quad} {\bf e}_2'=-\theta'{\bf e}_1. \tag{6.4.4}\end{equation}

Figure 6.4.2.

Now we can write (6.4.2) as

\begin{equation} m(r{\bf e}_1)''=f(r){\bf e}_1. \tag{6.4.5}\end{equation}

But

\[ (r{\bf e}_1)'=r'{\bf e}_1+r{\bf e}_1'=r'{\bf e}_1+r\theta'{\bf e}_2 \]

(from (6.4.4)), and

\begin{eqnarray*} (r{\bf e}_1)''&=&(r'{\bf e}_1+r\theta'{\bf e}_2)' \\ &=&r''{\bf e}_1+r'{\bf e}_1'+(r\theta''+r'\theta'){\bf e}_2 +r\theta'{\bf e}_2' \\ &=&r''{\bf e}_1+r'\theta'{\bf e}_2+(r\theta''+r'\theta'){\bf e}_2 -r(\theta')^2{\bf e}_1\mbox{\quad (from \eqref{eq:6.4.4})} \\ &=&\left(r''-r(\theta')^2\right){\bf e}_1+(r\theta''+2r'\theta'){\bf e}_2. \end{eqnarray*}

Substituting this into (6.4.5) yields

\[ m\left(r''-r(\theta')^2\right){\bf e}_1+m(r\theta''+2r'\theta'){\bf e}_2=f(r){\bf e}_1. \]

By equating the coefficients of ${\bf e}_1$ and ${\bf e}_2$ on the two sides of this equation we see that

\begin{equation} m\left(r''-r(\theta')^2\right)=f(r) \tag{6.4.6}\end{equation}

and

\[ r\theta''+2r'\theta'=0. \]

Multiplying the last equation by $r$ yields

\[ r^2\theta''+2rr'\theta'=(r^2\theta')'=0, \]

\begin{equation} r^2\theta'=h, \tag{6.4.7}\end{equation}

where $h$ is a constant that we can write in terms of the initial conditions as

\[ h=r^2(0)\theta'(0). \]

Since the initial position and velocity vectors are

\[ r(0){\bf e}_1(0)\mbox{\quad and \quad} r'(0){\bf e}_1(0)+r(0)\theta'(0){\bf e}_2(0), \]

our assumption that these two vectors are not parallel implies that $\theta'(0)\ne0$, so $h\ne0$.

Now let $u=1/r$. Then $u^2=\theta'/h$ (from (6.4.7)) and

\[ r'=-\dst{u'\over u^2}=-h\left(u'\over \theta'\right), \]

which implies that

\begin{equation} r'=-h{du\over d\theta}, \tag{6.4.8}\end{equation}

since

\[ {u'\over\theta'}={du\over dt}\left/{d\theta\over dt}\right.= {du\over d\theta}. \]

Differentiating (6.4.8) with respect to $t$ yields

\[ r''=-h{d\over dt}\left({du\over d\theta}\right) =-h{d^2u\over d\theta^2}\theta', \]

which implies that

\[ r''=-h^2u^2{d^2u\over d\theta^2} \mbox{\quad since\quad} \theta'=hu^2. \]

Substituting from these equalities into (6.4.6) and recalling that $r=1/u$ yields

\[ -m\left(h^2u^2{d^2u\over d\theta^2}+{1\over u}h^2u^4\right) =f(1/u), \]

and dividing through by $-mh^2u^2$ yields (6.4.1).

Eqn. (6.4.7) has the following geometrical interpretation, which is known as Kepler’s Second Law.

Theorem 6.4.1

The position vector of an object moving under a central force sweeps out equal areas in equal times; more precisely$,$ if $\theta(t_1)\le\theta(t_2)$ then the $($signed$)$ area of the sector

\[ \{(x,y)=(r\cos\theta,r\sin\theta)\; :\; 0\le r\le r(\theta),\; \theta(t_1)\le\theta(t_2)\} \]

(Figure 6.4.3) is given by

\[ A={h(t_2-t_1)\over2}, \]

where $h=r^2\theta',$ which we have shown to be constant.

Figure 6.4.3.

Proof Recall from calculus that the area of the shaded sector in Figure 6.4.3 is

\[ A={1\over2}\int_{\theta(t_1)}^{\theta(t_2)} r^2(\theta)\,d\theta, \]

where $r=r(\theta)$ is the polar representation of the orbit. Making the change of variable $\theta=\theta(t)$ yields

\begin{equation} A={1\over2}\int_{t_1}^{t_2} r^2(\theta(t))\theta'(t)\,dt. \tag{6.4.9}\end{equation}

But (6.4.7) and (6.4.9) imply that

\[ A={1\over2}\int_{t_1}^{t_2} h\,dt={h(t_2-t_1)\over2}, \]

which completes the proof.

Motion Under an Inverse Square Law Force

In the special case where $f(r)=-mk/r^2=-mku^2$, so ${\bf F}$ can be interpreted as a gravitational force, (6.4.1) becomes

\begin{equation} {d^2u\over d\theta^2}+u={k\over h^2}. \tag{6.4.10}\end{equation}

The general solution of the complementary equation

\[ {d^2u\over d\theta^2}+u=0 \]

can be written in amplitude–phase form as

\[ u=A\cos(\theta-\phi), \]

where $A\ge0$ and $\phi$ is a phase angle. Since $u_p=k/h^2$ is a particular solution of (6.4.10), the general solution of (6.4.10) is

\[ u=A\cos(\theta-\phi)+{k\over h^2}; \]

hence, the orbit is given by

\[ r=\left(A\cos(\theta-\phi)+{k\over h^2}\right)^{-1}, \]

which we rewrite as

\begin{equation} r={\rho\over 1+e\cos(\theta-\phi)}, \tag{6.4.11}\end{equation}

where

\[ \rho={h^2\over k}\mbox{\quad and \quad} e=A\rho. \]

A curve satisfying (6.4.11) is a conic section with a focus at the origin (Exercise 1). The nonnegative constant $e$ is the eccentricity of the orbit, which is an ellipse if $e<1$ ellipse (a circle if $e=0$), a parabola if $e=1$, or a hyperbola if $e>1$.

Figure 6.4.4.

If the orbit is an ellipse, then the minimum and maximum values of $r$ are

\begin{eqnarray*} r_{\min}&=&{\rho\over1+e}\quad\mbox{(the {\color{blue}\it perihelion distance\/}, attained when $\theta=\phi$)} \\ r_{\max}&=&{\rho\over1-e}\quad\mbox{(the {\color{blue}\it aphelion distance\/}, attained when $\theta=\phi+\pi$)}. \end{eqnarray*}

Figure 6.4.4 shows a typical elliptic orbit. The point $P$ on the orbit where $r=r_{\min}$ is the perigee and the point $A$ where $r=r_{\max}$ is the apogee.

For example, Earth’s orbit around the Sun is approximately an ellipse with $e\approx.017$, $r_{\min}\approx91\times 10^6$ miles, and $r_{\max}\approx 95\times 10^6$ miles. Halley’s comet has a very elongated approximately elliptical orbit around the sun, with $e\approx.967$, $r_{\min}\approx55\times10^6$ miles, and $r_{\max}\approx33\times10^8$ miles. Some comets (the nonrecurring type) have parabolic or hyperbolic orbits.

6.4 Exercises

Find the equation of the curve

\[ r={\rho\over 1+e\cos(\theta-\phi)} \tag*{\rm (A)} \]

in terms of $(X,Y)=\left(r\cos(\theta-\phi),r\sin(\theta-\phi)\right)$, which are rectangular coordinates with respect to the axes shown in Figure 6.4.5. Use your results to verify that (A) is the equation of an ellipse if $0<e<1$, a parabola if $e=1$, or a hyperbola if $e>1$. If $e<1$, leave your answer in the form

\[ {(X-X_0)^2\over a^2}+{(Y-Y_0)^2\over b^2}=1, \]

and show that the area of the ellipse is

\[ A={\pi\rho^2\over(1-e^2)^{3/2}}. \]

Then use Theorem 6.4.1 to show that the time required for the object to traverse the entire orbit is

\[ T={2\pi\rho^2\over h(1-e^2)^{3/2}}. \]

(This is Kepler’s third law; $T$ is called the period of the orbit.)

Figure 6.4.5.

Show answer

If $e=1$, then $Y^2=\rho(\rho-2X)$; if $e\ne1$ $\dst{\left(X+{e\rho\over1-e^2}\right)^2}+\dst{Y^2\over1-e^2}= \dst{\rho^2\over(1-e^2)^2}$ if ;

$e<1$ let $X_0=-\dst{e\rho\over1-e^2}$, $a=\dst{\rho\over1-e^2}$, $b=\dst{\rho\over\sqrt{1-e^2}}$.
Suppose an object with mass $m$ moves in the $xy$-plane under the central force

\[ {\bf F}(r,\theta)=-{mk\over r^2}(\cos\theta\,{\bf i}+\sin\theta\,{\bf j}), \]

where $k$ is a positive constant. As we shown, the orbit of the object is given by

\[ r={\rho\over 1+e\cos(\theta-\phi)}. \]

Determine $\rho$, $e$, and $\phi$ in terms of the initial conditions

\[ r(0)=r_0,\quad r'(0)=r_0', \text{\; and \;} \theta(0)=\theta_0,\quad \theta'(0)=\theta_0'. \]

Assume that the initial position and velocity vectors are not collinear.

Show answer

Let $h=r_0^2\theta_0'$; then $\rho=\dst{h^2\over k}$, $e=\dst{\left[\left({\rho\over r_0}-1\right)^2+\left(\rho r_0'\over h\right)^2\right]^{1/2}}$. If $e=0$, then

$ \theta_0$ is undefined, but also irrelevant if $e\ne0$ then $\phi=\theta_0-\alpha$, where $-\pi\le\alpha<\pi$, $\cos\alpha=\dst{{1\over e}\left({\rho\over r_0}-1\right)}$ and $\sin\alpha=\dst{\rho r_0'\over eh}$.
Suppose we wish to put a satellite with mass $m$ into an elliptical orbit around Earth. Assume that the only force acting on the object is Earth’s gravity, given by

\[ {\bf F}(r,\theta)=-mg\left(R^2\over r^2\right)(\cos\theta\,{\bf i}+\sin\theta\,{\bf j}), \]

where $R$ is Earth’s radius, $g$ is the acceleration due to gravity at Earth’s surface, and $r$ and $\theta$ are polar coordinates in the plane of the orbit, with the origin at Earth’s center.
1. Find the eccentricity required to make the aphelion and perihelion distances equal to $R\gamma_1$ and $R\gamma_2$, respectively, where $1<\gamma_1<\gamma_2$.
2. Find the initial conditions
  
  \[ r(0)=r_0,\quad r'(0)=r_0', \text{\; and \;} \theta(0)=\theta_0,\quad \theta'(0)=\theta_0' \]
  
  required to make the initial point the perigee, and the motion along the orbit in the direction of increasing $\theta$. Hint: Use the results of Exercise 2.
Show answer

(a) $e=\dst{\gamma_2-\gamma_1\over\gamma_1+\gamma_2}$ (b) $r_0=R\gamma_1$, $r_0'=0$, $\theta_0$ arbitrary, $\theta_0'=\dst{\left[2g\gamma_2\over R\gamma_1^3(\gamma_1+\gamma_2)\right]^{1/2}}$
An object with mass $m$ moves in a spiral orbit $r=c\theta^2$ under a central force

\[ {\bf F}(r,\theta)=f(r)(\cos\theta\,{\bf i}+\sin\theta\,{\bf j}). \]

Find $f$.

Show answer

$f(r)=-mh^2\dst\left({6c\over r^4}+{1\over r^3}\right)$
An object with mass $m$ moves in the orbit $r=r_0e^{\gamma\theta}$ under a central force

\[ {\bf F}(r,\theta)=f(r)(\cos\theta\,{\bf i}+\sin\theta\,{\bf j}). \]

Find $f$.

Show answer

$f(r)=-\dst{mh^2(\gamma^2+1)\over r^3}$
Suppose an object with mass $m$ moves under the central force

\[ {\bf F}(r,\theta)=-{mk\over r^3}(\cos\theta\,{\bf i}+\sin\theta\,{\bf j}), \]

with

\[ r(0)=r_0,\quad r'(0)=r_0', \text{\; and \;} \theta(0)=\theta_0,\quad \theta'(0)=\theta_0', \]

where $h=r_0^2\theta_0'\ne0$.
1. Set up a second order initial value problem for $u=1/r$ as a function of $\theta$.
2. Determine $r=r(\theta)$ if (i) $h^2<k$; (ii) $h^2=k$; (iii) $h^2>k$.
Show answer

(a) $\dst{d^2u\over d\theta^2}+\left(1-{k\over h^2}\right)u=0,\; u(\theta_0)=\dst{1\over r_0},\; \dst{du(\theta_0)\over d\theta}=-\dst{r_0'\over h}$. (b) with $\gamma=$

$\left|1-\dst{k\over h^2}\right|^{1/2}$: (i) $r=r_0\left( \cosh\gamma(\theta-\theta_0)-\dst{r_0r_0'\over\gamma h}\sinh\gamma(\theta-\theta_0)\right)^{-1}$ (ii) $r=r_0\left(1 -\dst{r_0r_0'\over h}(\theta-\theta_0)\right)^{-1}$; (iii) $r=r_0\left( \cos\gamma(\theta-\theta_0)-\dst{r_0r_0'\over\gamma h}\sin\gamma(\theta-\theta_0)\right)^{-1}$