5.2 Central Forces

We will now examine a very special type of force. These forces have many useful properties which make solving problems involving them a lot simpler.

Definition 5.2.

A central force is a force which acts along the radial direction and only depends on the radial distance $r$ . Central forces thus have the form

\boldsymbol{F}=F(r)\hat{\boldsymbol{r}}.

(5.2)

If $F(r)<0$ , the force is attractive, and if $F(r)>0$ , the force is repulsive.

We have seen one example of a central force already, the universal law of gravitation. Another example is Coulomb’s law.

5.2.1 Properties of Central Forces

Motion under a central force obeys the following rules:

•

The motion is confined to a plane.
•

The angular momentum is conserved ( $\frac{\operatorname{\mathrm{d}}\!\boldsymbol{L}}{\operatorname{\mathrm{d}}\!t}=0$ ).
•

The position vector sweeps out equal area in equal time.

Let’s prove that angular momentum is conserved under the influence of a central force. Looking at the derivative, we get

$\displaystyle\frac{\operatorname{\mathrm{d}}\!\boldsymbol{L}}{\operatorname{% \mathrm{d}}\!t}$	$\displaystyle=\frac{\operatorname{\mathrm{d}}\!}{\operatorname{\mathrm{d}}\!t}% (\boldsymbol{r}\times\boldsymbol{p})$	(5.3)
	$\displaystyle=\frac{\operatorname{\mathrm{d}}\!\boldsymbol{r}}{\operatorname{% \mathrm{d}}\!t}\times\boldsymbol{p}+\boldsymbol{r}\times\frac{\operatorname{% \mathrm{d}}\!\boldsymbol{p}}{\operatorname{\mathrm{d}}\!t}$	(5.4)
	$\displaystyle=\boldsymbol{v}\times m\boldsymbol{v}+\boldsymbol{r}\times F(R)% \hat{\boldsymbol{R}}$	(5.5)
	$\displaystyle=0.$	(5.6)

Therefore, $\boldsymbol{L}$ is constant over time, it is conserved.

By definition, $\boldsymbol{L}$ is perpendicular to $\boldsymbol{r}$ . We can also see this by taking the dot product of $\boldsymbol{L}$ with the position vector:

\boldsymbol{r}\cdot\boldsymbol{L}=\boldsymbol{r}\cdot(\boldsymbol{r}\times% \boldsymbol{p})=\boldsymbol{p}\cdot(\boldsymbol{r}\times\boldsymbol{r})=0.

(5.7)

Since $\boldsymbol{L}$ is constant, this implies that $\boldsymbol{r}$ is confined to the plane perpendicular to $\boldsymbol{L}$ .

Because of this, it is convenient to use cylindrical coordinates with the plane $z=0$ as the plane of motion. Using equation LABEL: (note that $\dot{z}=0$ ) and the cross products between unit vectors, we get

$\displaystyle\boldsymbol{L}$	$\displaystyle=m\boldsymbol{r}\times\boldsymbol{v}$	(5.8)
	$\displaystyle=mR\hat{\boldsymbol{R}}\times(\dot{R}\hat{\boldsymbol{R}}+R\dot{% \theta}\hat{\boldsymbol{\theta}})$	(5.9)
	$\displaystyle=mR^{2}\dot{\theta}\hat{\boldsymbol{k}}$	(5.10)
	$\displaystyle=mh\hat{\boldsymbol{k}}.$	(5.11)

In the last line we have defined $h=R^{2}\dot{\theta}$ . This is the specific angular momentum (angular momentum normalised by mass), and is useful in some problems.

5.2.2 Kepler’s Second Law

Finally, we will prove the last fact about motion under central forces, the law of equal areas, which is a more general version of Kepler’s second law of planetary motion. Consider an object on a curved path at two points separated by a small time interval $\operatorname{\mathrm{d}}\!t$ .

The shaded arc is approximately a triangle, so the area get closer and closer to $A=\frac{1}{2}R^{2}\operatorname{\mathrm{d}}\!\theta$ as $\operatorname{\mathrm{d}}\!t\to 0$ . Therefore, we have

\frac{\operatorname{\mathrm{d}}\!A}{\operatorname{\mathrm{d}}\!t}=\frac{1}{2}R% ^{2}\frac{\operatorname{\mathrm{d}}\!\theta}{\operatorname{\mathrm{d}}\!t}=% \frac{1}{2}h=\text{const}.

(5.12)

So the rate of area swept out over time is constant.

5.2.3 Motion Under a Central Force

What does motion under a central force look like? By Newton’s second law, we know that

a_{R}=\ddot{R}-R\dot{\theta}^{2}=\frac{F(R)}{m}.

(5.13)

Substituting in the specific angular momentum, we get

\ddot{R}-\frac{h^{2}}{R^{3}}=\frac{F(R)}{m}.

(5.14)

This is the differential equation we need to solve for any central force problem.

Example 5.1.

Prove Kepler’s third law: the square of a planet’s orbital period is proportional to the cube of the semi-major axis of its orbit.

Planets move according to the universal law of gravitation, which is

\boldsymbol{F}=-\frac{GMm}{R^{2}}\hat{\boldsymbol{R}}.

(5.15)

We will make an approximation and assume that the sun is fixed at the origin and the planets orbit in circular orbits. In reality, the a given planet and the sun orbit their common centre of mass in elliptical paths, but the difference is very small making this approximation makes the problem much simpler. Since $R=\text{const}$ , then $\dot{R}=\ddot{R}=0$ . So $a_{R}=-R\dot{\theta}^{2}$ . Since angular momentum $L=mR^{2}\dot{\theta}$ is conserved, $\dot{\theta}=\omega$ is a constant. Thus the equation of motion 5.13 becomes

	$\displaystyle-\frac{GMm}{R^{2}}$	$\displaystyle=-mR\omega^{2}$		(5.16)
	$\displaystyle GM$	$\displaystyle=R^{3}\omega^{2}.$		(5.17)

The period of the orbit is equal to the displacement divided by the speed, in this case $T=\frac{2\pi}{\omega}$ , so rearranging for $R^{3}$ we get

R^{3}=\frac{GM}{4\pi^{2}}T^{2}.

(5.18)

Example 5.2.

Consider a central force of the form $\boldsymbol{F}=-\frac{mk^{2}}{r^{3}}\hat{\boldsymbol{r}}$ . If a particle under the influence of this force starts at a distance $r=r_{0}$ from the origin with no radial speed ( $v_{r,0}=0$ ), what will its motion look like for different values of $k$ ?

Substituting in the force into equation 5.13, we get

	$\displaystyle\ddot{r}-\frac{h}{r^{3}}=-\frac{k^{2}}{r^{3}}$		(5.19)
	$\displaystyle\ddot{r}+\frac{k^{2}-h^{2}}{r^{3}}=0.$		(5.20)

For $k^{2}=h^{2}$ , we get $\ddot{r}=0$ , which can be integrated twice to give $r=r_{0}+v_{r,0}t$ . Since $v_{r,0}=0$ , we have $r=r_{0}$ which is circular motion. The angular speed is given by equation 5.10, $\omega=\frac{L}{mr_{0}^{2}}=\frac{mvr_{0}}{mr_{0}^{2}}=\frac{v}{r_{0}}$ . This is just our familiar result from uniform circular motion, which makes sense because if we plug $k^{2}=h^{2}$ into the form of the force above, we get

\boldsymbol{F}=-\frac{mr^{4}\omega^{2}}{r^{3}}\hat{\boldsymbol{r}}=-mr\omega^{% 2}\hat{\boldsymbol{r}},

(5.21)

which is just the formula for centripetal force.

What about the cases where $k^{2}\neq h^{2}$ ? Unfortunately, the differential equation is not solvable analytically for any other case. Luckily, there is another way to solve it! What we have to do is parameterise the orbit in terms of the angular displacement $\theta$ instead of $t$ by making use of the relation $h=r^{2}\dot{\theta}$ . We will also make the substition $u=1/r$ , which is a very useful substition when solving problems with central forces of the form $F(r)=\frac{\alpha}{r^{n}}$ . Therefore, the aim is to replace all time derivatives of $r$ with derivatives of $u$ with respect to $\theta$ . Using the chain rule, we get

$\displaystyle\dot{r}$	$\displaystyle=\frac{\operatorname{\mathrm{d}}\!r}{\operatorname{\mathrm{d}}\!u% }\dot{u}=\frac{\operatorname{\mathrm{d}}\!r}{\operatorname{\mathrm{d}}\!u}% \frac{\operatorname{\mathrm{d}}\!u}{\operatorname{\mathrm{d}}\!\theta}\dot{\theta}$	(5.22)
	$\displaystyle=-r^{2}\dot{\theta}\frac{\operatorname{\mathrm{d}}\!u}{% \operatorname{\mathrm{d}}\!\theta}$	(5.23)
	$\displaystyle=-h\frac{\operatorname{\mathrm{d}}\!u}{\operatorname{\mathrm{d}}% \!\theta}.$	(5.24)

For $\ddot{r}$ , we get

$\displaystyle\ddot{r}$	$\displaystyle=\frac{\operatorname{\mathrm{d}}\!}{\operatorname{\mathrm{d}}\!t}% \left(-h\frac{\operatorname{\mathrm{d}}\!u}{\operatorname{\mathrm{d}}\!\theta}% \right)=-h\frac{\operatorname{\mathrm{d}}\!^{2}u}{\operatorname{\mathrm{d}}\!% \theta^{2}}\dot{\theta}$	(5.25)
	$\displaystyle=-\frac{h^{2}}{r^{2}}\frac{\operatorname{\mathrm{d}}\!^{2}u}{% \operatorname{\mathrm{d}}\!\theta^{2}}$	(5.26)
	$\displaystyle=-h^{2}u^{2}\frac{\operatorname{\mathrm{d}}\!^{2}u}{\operatorname% {\mathrm{d}}\!\theta^{2}}.$	(5.27)

Now, the equation of motion becomes

	$\displaystyle-h^{2}u^{2}\frac{\operatorname{\mathrm{d}}\!^{2}u}{\operatorname{% \mathrm{d}}\!\theta^{2}}+(k^{2}-h^{2})u^{3}=0$		(5.28)
	$\displaystyle\frac{\operatorname{\mathrm{d}}\!^{2}u}{\operatorname{\mathrm{d}}% \!\theta^{2}}-\frac{k^{2}-h^{2}}{h^{2}}u=0.$		(5.29)

Letting $\beta^{2}=\lvert\frac{k^{2}-h^{2}}{h^{2}}\rvert$ , if $k^{2}>h^{2}$ then the differential equation becomes

\frac{\operatorname{\mathrm{d}}\!^{2}u}{\operatorname{\mathrm{d}}\!\theta^{2}}% -\beta^{2}u=0.

(5.30)

The general solution is $u=Ae^{\beta\theta}+Be^{-\beta\theta}$ .

For the opposite case $k^{2}<h^{2}$ , the negative sign gets cancelled out and we get

\frac{\operatorname{\mathrm{d}}\!^{2}u}{\operatorname{\mathrm{d}}\!\theta^{2}}% +\beta^{2}u=0.

(5.31)

This has the general solution $u=A\cos(\beta\theta)+B\sin(\beta\theta)$ .