## What is central force?

Central force is the force that acts along the center of mass of one body to the center of mass of another body. Its direction from or away from a point on a body to another is called the center of the force. Such forces include gravitational force between two massive objects or electrostatic forces between two charged particles. So from atoms to planets, the central forces play an important role in shaping our daily lives. One of the remarkable features of central forces is that they are conservative, i.e. the total energy of a central force system can always be expressed as a gradient of a potential energy function.

## Circular motion

The central force problem is easier dealt with in polar coordinates, due to its circular symmetry, so we proceed in polar coordinates. The position vector in polar coordinates is given by

$\overrightarrow{R}\left(t\right)=r\hat{r}$

$r$ is the radius and $\hat{r}$ is the unit vector directed along the radius.

The radial unit vector and angular unit vector is given in terms of cartesian coordinates is given by

$\hat{r}=\mathrm{cos}\Theta \hat{i}+\mathrm{sin}\Theta \hat{j}......\left(1\right)\phantom{\rule{0ex}{0ex}}\hat{\Theta}=-\mathrm{sin}\Theta \hat{i}+\mathrm{cos}\Theta \hat{j}......\left(2\right)\phantom{\rule{0ex}{0ex}}$

The time derivative of the radial unit vector is given by

$\frac{d\hat{r}}{dt}=\frac{d}{dt}\left(\mathrm{cos}\Theta \hat{i}+\mathrm{sin}\Theta \hat{j}\right)\phantom{\rule{0ex}{0ex}}\frac{d\hat{r}}{dt}=-\mathrm{sin}\Theta \frac{d\Theta}{dt}\hat{i}+\mathrm{cos}\Theta \frac{d\Theta}{dt}\hat{j}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\text{from(2)}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\frac{d\hat{r}}{dt}=\dot{\Theta}\hat{\Theta}......\left(3\right)$

and the time derivative of angular coordinates is

$\frac{d\hat{\Theta}}{dt}=\frac{d}{dt}\left(-\mathrm{sin}\Theta \hat{i}+\mathrm{cos}\Theta \hat{j}\right)\phantom{\rule{0ex}{0ex}}\frac{d\hat{\Theta}}{dt}=-\mathrm{cos}\Theta \frac{d\Theta}{dt}\hat{i}-\mathrm{sin}\Theta \frac{d\Theta}{dt}\hat{j}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\text{from(1)}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\frac{d\hat{\Theta}}{dt}=-\dot{\Theta}\hat{r}......\left(4\right)$

The velocity in polar coordinates is

$\overrightarrow{v}=\frac{d}{dt}\left(r\hat{r}\right)\phantom{\rule{0ex}{0ex}}\overrightarrow{v}=\frac{dr}{dt}\hat{r}+r\frac{d\hat{r}}{dt}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\text{from(3)}\phantom{\rule{0ex}{0ex}}\overrightarrow{v}=\dot{r}\hat{r}+r\dot{\Theta}\hat{\Theta}......\left(5\right)$

and the acceleration in polar coordinates is

$\overrightarrow{a}=\frac{d}{dt}\overrightarrow{v}\phantom{\rule{0ex}{0ex}}\overrightarrow{a}=\frac{d}{dt}\left(\dot{r}\hat{r}+r\dot{\Theta}\hat{\Theta}\right)\phantom{\rule{0ex}{0ex}}\overrightarrow{a}=\frac{d\dot{r}}{dt}\hat{r}+\dot{r}\frac{d\hat{r}}{dt}+\frac{d}{dt}r\dot{\Theta}\hat{\Theta}\phantom{\rule{0ex}{0ex}}\overrightarrow{a}=\frac{d\dot{r}}{dt}\hat{r}+\dot{r}\frac{d\hat{r}}{dt}+\frac{dr}{dt}\dot{\Theta}\hat{\Theta}+r\frac{d\dot{\Theta}}{dt}\hat{\Theta}+r\dot{\Theta}\frac{d\hat{\Theta}}{dt}\phantom{\rule{0ex}{0ex}}\overrightarrow{a}=\ddot{r}\hat{r}+\dot{r}\dot{\Theta}\hat{\Theta}+\dot{r}\dot{\Theta}\hat{\Theta}+r\ddot{\Theta}\hat{\Theta}+r\dot{\Theta}\left(-\dot{\Theta}\hat{r}\right)\phantom{\rule{0ex}{0ex}}\overrightarrow{a}=\ddot{r}\hat{r}+\dot{r}\dot{\Theta}\hat{\Theta}+\dot{r}\dot{\Theta}\hat{\Theta}+r\ddot{\Theta}\hat{\Theta}-r{\dot{\Theta}}^{2}\hat{r}\phantom{\rule{0ex}{0ex}}\overrightarrow{a}=\left(\ddot{r}-r{\dot{\Theta}}^{2}\right)\hat{r}+\left(r\ddot{\Theta}+2\dot{r}\dot{\Theta}\right)\hat{\Theta}.......\left(6\right)$

## Central force motion as one body problem

The central force problem can be reduced to one body problem.

Then we have,

Suppose in an isolated system that consists of two particles with masses m_{1} and m_{2} interacting via central force. The particles have relative position vector given by,

$\overrightarrow{r}=\overrightarrow{{r}_{1}}-\overrightarrow{{r}_{2}}$

Now let the central force between them be f(r). Newton's second law dictates that the equation of motion of two particles that are given by,

${m}_{1}{\overrightarrow{r}}_{1}=f\left(r\right)\hat{r}......\left(7\right)\phantom{\rule{0ex}{0ex}}{m}_{2}{\overrightarrow{r}}_{2}=-f\left(r\right)\hat{r}......\left(8\right)$

The equations of motion of the two particles are coupled together by the relative position vector $\overrightarrow{r}$, the problem can be simplified if we eliminate the position vectors of the two particles with their relative position vector. To this end, note that the center of mass of the particles is given by,

$\overrightarrow{{r}_{CM}}=\frac{{m}_{1}{\overrightarrow{r}}_{1}+{m}_{2}{\overrightarrow{r}}_{2}}{{m}_{1}+{m}_{2}}$

Now from the conservation of linear momentum we know that if there is no external force on the system of particles, then the acceleration of their center of mass is zero, i.e.

${\ddot{\overrightarrow{r}}}_{CM}=0$

From the equation of motion, we get

$\ddot{{\overrightarrow{r}}_{1}}=\frac{f\left(r\right)}{{m}_{1}}\hat{r}\phantom{\rule{0ex}{0ex}}\ddot{{\overrightarrow{r}}_{2}}=-\frac{f\left(r\right)}{{m}_{2}}\hat{r}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\text{subtractingthesetwoequationsgives,}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\ddot{{\overrightarrow{r}}_{1}}-\ddot{{\overrightarrow{r}}_{2}}=\left(\frac{1}{{m}_{1}}+\frac{1}{{m}_{2}}\right)f\left(r\right)\hat{r}$

To further simplify the algebra we can write,

$\frac{1}{\mu}=\frac{1}{{m}_{1}}+\frac{1}{{m}_{2}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\text{where\mu isthereducedmass}$

Now we have

$\mu \ddot{\overrightarrow{r}}=f\left(r\right)\hat{r}......\left(9\right)$

Using equation (9) the two-body central force problem characterized by equations (7) and (8) is now reduced to essentially one particle problem. This method although elegant can be generalized in the case of more than two particles or objects.

## Features of central force

The exact solution of the equation of motion (9) depends upon the form of force function f(r), but still, some of the features of central force motion can be inferred directly from equation (9). For this analysis the principle of conservation of angular momentum and conservation of energy proves helpful.

### Conservation of angular momentum

The motion in central force is constrained to a plane, in other words, the orbits are planar in central force motion. To see this, start with the definition of angular momentum,

$\overrightarrow{L}=\overrightarrow{r}\times \mu \dot{\overrightarrow{r}}......\left(10\right)$

From (10) we can see that the position vector is always perpendicular to angular momentum. Now since angular momentum is fixed, so is the plane of the position vector.

Now from equations (6) and (9), we get

$\mu \left[\left(\ddot{r}-r{\dot{\Theta}}^{2}\right)\hat{r}+\left(r\ddot{\Theta}+2\dot{r}\dot{\Theta}\right)\hat{\Theta}\right]=f\left(r\right)\hat{r}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\text{whichgives}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mu \left(\ddot{r}-r{\dot{\Theta}}^{2}\right)=f\left(r\right)......\left(11\right)\phantom{\rule{0ex}{0ex}}\mu \left(r\ddot{\Theta}+2\dot{r}\dot{\Theta}\right)=0.......\left(12\right)$

As a consequence of equation (12), there is no tangential component of central force so the angular momentum remains constant.

### Law of equal areas

As discussed above, the angular momentum in central force motion is constant, which means that it can be set equal to a constant, i.e.

$\mu {r}^{2}\dot{\Theta}=L......\left(13\right)$

Now in polar coordinates, the differential area of any elliptical orbit is given by,

$dA=\frac{{r}^{2}d\theta}{2}\phantom{\rule{0ex}{0ex}}\frac{dA}{dt}=\frac{{r}^{2}}{2}\frac{d\theta}{dt}......\left(14\right)$

Using (13) in (14) gives,

$\frac{dA}{dt}=\frac{L}{2\mu}$

Since L and µ are constants, the area covered per unit time dA/dt is also a constant. So the area swept by bodies moving in central force orbits is constant.

### Conservation of energy

The total energy of a system is the sum of Kinetic Energy (KE) and Potential Energy (U(r)), i.e.

$E=KE+U\left(r\right)\phantom{\rule{0ex}{0ex}}E=\frac{1}{2}\mu {v}^{2}+U\left(r\right)\phantom{\rule{0ex}{0ex}}E=\frac{1}{2}\mu \left({\dot{r}}^{2}+{r}^{2}{\dot{\Theta}}^{2}\right)+U\left(r\right)......\left(15\right)$

Using equation (13), we get

$E=\frac{1}{2}\mu {\dot{r}}^{2}+\frac{1}{2}\frac{{L}^{2}}{\mu {r}^{2}}+U\left(r\right)$

We can set the last two terms on RHS equal to an effective potential.

$\frac{1}{2}\frac{{L}^{2}}{\mu {r}^{2}}+U\left(r\right)={U}_{eff}\left(r\right)$

So we have,

$E=\frac{1}{2}\mu {\dot{r}}^{2}+{U}_{eff}\left(r\right)$

The effective potential is also known as centrifugal potential. The term $\frac{1}{2}\frac{{L}^{2}}{\mu {r}^{2}}$ is not true potential energy related to a force, but a potential energy term arising to angular part of kinetic energy.

## Central force motion of a free particle

Consider two non-interacting particles moving towards each other with masses ${\text{m}}_{1}{\text{andm}}_{2}$ and velocities ${\text{v}}_{1}{\text{andv}}_{2}$ respectively. The reduced mass of the system is then

$\mu =\frac{{m}_{1}{m}_{2}}{{m}_{1}+{m}_{2}}$ ..............(16)

And the relative velocity between them is

${v}_{0}={v}_{1}-{v}_{2}$

Assuming that there is no interactive potential between the particles, then, the effective potential is

${U}_{\mathrm{eff}}=\frac{{L}^{2}}{2\mu {r}^{2}}$

So, the total energy of the system is

$E=\frac{1}{2}\mu {{v}_{0}}^{2}+\frac{{L}^{2}}{2\mu {r}^{2}}$

Since the angular momentum is conserved, we can obtain an expression for it by considering the situation when both of them are nearest and passing by, at which, the separation is b, so the angular momentum is,

$L=\mu b{v}_{0}$

This gives the effective potential of the system to be,

${U}_{\mathrm{eff}}=\frac{1}{2}\mu {{{v}_{0}}^{2}}^{}{b}^{2}{r}^{2}$

Since the total energy of the system remains constant i.e. E, the kinetic energy can be written as

$E=\mathrm{KE}+{U}_{\mathrm{eff}}\phantom{\rule{0ex}{0ex}}\mathrm{KE}=E-{U}_{\mathrm{eff}}$

Now as we know, the kinetic energy of the system cannot be negative, the motion remains constrained to regions where $E-{U}_{\mathrm{eff}}\ge 0$. Initially, the distance r between the particle is very large and hence the effective potential is almost zero and the kinetic energy is maximum. As the particles start approaching each other, the kinetic energy starts to decrease and attains a minimum at the turning point

$\frac{1}{2}\mu {{v}_{0}}^{2}={U}_{eff}\left(b\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\frac{1}{2}\mu {{v}_{0}}^{2}=\frac{1}{2}\mu {{\nu}^{2}}_{0}{b}^{2}{r}^{2}$

which gives

${r}_{t}=b$

where ${r}_{t}$ is the turning point.

## Context and Applications

This topic is significant in the professional exams for both undergraduate and graduate courses, especially for

- Bachelor of Technology (Mechanical)
- Master of Technology (Mechanical)
- Bachelor of Science (Physics)
- Master of Science (Physics)

## Practice Problems

1. If two identical objects of masses 10 kg each are interacting through central force, find the reduced mass of the equivalent one body system.

- 10 kg
- 20 kg
- 5 kg
- 15 kg

**Answer**- c

**Explanation:** The expression for reduced mass is $\frac{1}{\mu}=\frac{1}{{m}_{1}}+\frac{1}{{m}_{2}}$, plugging in the values answers 5 kg.

2. Find the direction of central force between two bodies whose center of masses are located at ${\overrightarrow{r}}_{1}=2\hat{i}+10\hat{i}+15\hat{i}\text{and}{\overrightarrow{r}}_{2}=7\hat{i}+1\hat{i}+5\hat{i}$.

- $2\hat{i}+19\hat{i}-10\hat{k}$
- $2\hat{i}+4\hat{i}+10\hat{k}$
- $5\hat{i}+10\hat{i}-1\hat{k}$
- $5\hat{i}-9\hat{i}-10\hat{k}$

**Answer**- d

**Explanation:** The direction of central force is given by $\overrightarrow{r}={\overrightarrow{r}}_{2}-{\overrightarrow{r}}_{1}$, so the answer is ${\overrightarrow{r}}_{2}-{\overrightarrow{r}}_{1}=5\hat{i}-9\hat{i}-10\hat{k}$

3. What is the conservation angular momentum a consequence of?

- Law of equal area
- Circular motion
- Conservation of energy
- Conservation of linear momentum

**Answer**- a

**Explanation:** Since in central force motion the rate of the swept area is given by $\frac{dA}{dt}=\frac{L}{2\mu}$, a constant rate of the sweeping area would imply a constant angular momentum. Hence the correct answer is a.

4. In what case does the effective potential play a role?

- Gravitational force
- Inertia
- Circular motion
- Linear motion

**Answer**- c

**Explanation:** The effective potential comes into play when there is circular motion due to central force. In such a case the kinetic energy due to rotation is given by $\frac{1}{2}\mu \left({r}^{2}{\dot{\Theta}}^{2}\right)$. This in terms of angular momentum L can be written as $\frac{{L}^{2}}{2\mu {r}^{2}}$which is also known as effective potential.

5. Which of the following forces is not a central force.

- Gravitational force
- Electromagnetic force]
- Frictional force
- None of the above

**Answer-** c

**Explanation:** Frictional force only acts between contact surfaces and not between the center of masses of objects. Hence it is not a central force.

## Common Mistakes

A common mistake is thinking about the effect of centrifugal potential, i.e. centrifugal force to be a real force. Instead, the centrifugal force is an effect due to the non-inertial reference frame. This force can be eliminated by transforming to an inertial reference frame.

## Related Concepts

- Gravitational force
- Three-body problem
- Circular motion

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