Motion under central force

(i) Central Force

In any force field, if the force acting on a particle has the following characteristics:

1. The line of action of the force passes through a fixed point called the centre of force.
2. The magnitude of the force depends only upon the distance of the particle from the centre of force.

$$ \vec{F}(\vec{r}) = F(r)\,\hat{r} \tag{1} $$

where \(\vec{r}\) is the position vector of the particle, \(\hat{r}\) is the unit vector along \(\vec{r}\), and \(F(r)\) is the magnitude of the central force, which is a function of the distance \(r\) from the centre of force.

Examples

1. Gravitational field:

2. Electrostatic force:

(ii) Motion of a Particle Under a Central Force Is Always in a Plane

For a particle moving under a central force,

$$ \vec{F} = F(r)\hat{r}. $$

Taking the vector product of \(\vec{r}\) with the force,

$$ \vec{r}\times\vec{F} = F(r)\vec{r}\times\hat{r} $$ $$ \vec{r}\times\vec{F} = 0.   ∵\vec{A}\times\vec{A}=0 $$ $$ m[\vec{r}\times \frac{d^2\vec{r}}{dt^2}] = 0.  ∵\vec{F} = m\frac{d^2\vec{r}}{dt^2}. $$

Now,

$$ ∵\frac{d}{dt} \left( \vec{r}\times \frac{d\vec{r}}{dt} \right) = \frac{d\vec{r}}{dt} \times \frac{d\vec{r}}{dt} + \vec{r}\times \frac{d^2\vec{r}}{dt^2}=0 $$

Since

$$ \frac{d}{dt} \left( \vec{r}\times\frac{d\vec{r}}{dt} \right) = 0. $$

Hence,

$$ \vec{r}\times\frac{d\vec{r}}{dt} = \text{constant} \tag{2} $$

The constant vector is perpendicular to the plane containing \(\vec{r}\) and \(\dot{\vec{r}}\). Therefore, this plane remains fixed throughout the motion.

Hence, the motion of a particle under a central force is always planar.

(iii) Equation of Motion of a Particle Under a Central Force

The equation of motion of a particle moving in a plane under a central force can be expressed conveniently in terms of polar coordinates \((r,\theta)\).

Let the Cartesian coordinates of the particle at any instant be \((x,y)\) and its polar coordinates be \((r,\theta)\).

$$ x = r\cos\theta, \qquad y = r\sin\theta. $$

Differentiating with respect to time,

$$ \frac{dx}{dt} = \frac{dr}{dt}\cos\theta - r\sin\theta\frac{d\theta}{dt}, $$

$$ \frac{dy}{dt} = \frac{dr}{dt}\sin\theta + r\cos\theta\frac{d\theta}{dt}. $$

Differentiating again with respect to time,

$$ \frac{d^2x}{dt^2} = \left( \frac{d^2r}{dt^2} - r\left(\frac{d\theta}{dt}\right)^2 \right)\cos\theta - \left( 2\frac{dr}{dt}\frac{d\theta}{dt} + r\frac{d^2\theta}{dt^2} \right)\sin\theta, $$

$$ \frac{d^2y}{dt^2} = \left( \frac{d^2r}{dt^2} - r\left(\frac{d\theta}{dt}\right)^2 \right)\sin\theta + \left( 2\frac{dr}{dt}\frac{d\theta}{dt} + r\frac{d^2\theta}{dt^2} \right)\cos\theta. $$

Therefore, the acceleration vector is

$$ \vec{a} = \left[ \frac{d^2x}{dt^2}\hat{i} + \frac{d^2y}{dt^2}\hat{j} \right]$$

Thus

$$ \vec{a} = \left[ \frac{d^2r}{dt^2} - r\left(\frac{d\theta}{dt}\right)^2 \right] (\hat{i}\cos\theta+\hat{j}\sin\theta) + \left[ r\frac{d^2\theta}{dt^2} + 2\frac{dr}{dt}\frac{d\theta}{dt} \right] (-\hat{i}\sin\theta+\hat{j}\cos\theta). $$ $$ \vec{a} = \left[ \frac{d^2r}{dt^2} - r\left(\frac{d\theta}{dt}\right)^2 \right]\hat{r} + \left[ r\frac{d^2\theta}{dt^2} + 2\frac{dr}{dt}\frac{d\theta}{dt} \right]\hat{\theta} $$

Where, radial direction is

$$ \hat{r} = \hat{i}\cos\theta + \hat{j}\sin\theta, $$

and angular direction is

$$ \hat{\theta} = -\hat{i}\sin\theta + \hat{j}\cos\theta. $$

According to defination of central force and Newton’s second law,

$$ \vec{F} = F(r)\hat{r}.    and    \vec{F} = m\vec{a}.$$

Therefore,

$$ F(r)\hat{r} = m\vec{a}. $$ $$ F(r)\hat{r} = m \left[ \frac{d^2r}{dt^2} - r\left(\frac{d\theta}{dt}\right)^2 \right]\hat{r} + m\left[ r\frac{d^2\theta}{dt^2} + 2\frac{dr}{dt}\frac{d\theta}{dt} \right]\hat{\theta} $$

Comparing the coefficients of \(\hat{r}\) and \(\hat{\theta}\), the radial and angular components of acceleration are

$$ a_r = m\left[ \frac{d^2r}{dt^2} - r\left(\frac{d\theta}{dt}\right)^2\right]=F(r) \tag{3} $$

$$ a_\theta = r\frac{d^2\theta}{dt^2} + 2\frac{dr}{dt}\frac{d\theta}{dt}=0 \tag{4} $$

Thus, the motion equations are

$$ \frac{d^2r}{dt^2} - r\left(\frac{d\theta}{dt}\right)^2 = \frac{F(r)}{m} \tag{5} $$ $$ r\frac{d^2\theta}{dt^2} + 2\frac{dr}{dt}\frac{d\theta}{dt} = 0 \tag{6} $$

Equations (5) and (6) are the equations of motion of a particle under a central force.

(iv) Law of Conservation of Angular Momentum

Starting from equation (6),

$$ r\frac{d^2\theta}{dt^2} + 2\frac{dr}{dt}\frac{d\theta}{dt} = 0. $$

Multiplying by \(mr\),

$$ mr^2\frac{d^2\theta}{dt^2} + 2mr\frac{dr}{dt}\frac{d\theta}{dt} = 0. $$

This can be written as

$$ \frac{d}{dt} \left( mr^2\frac{d\theta}{dt} \right) = 0. $$

Therefore,

$$ mr^2\frac{d\theta}{dt} = \text{constant}. \tag{7} $$

The angular momentum of a particle about the centre of force is

$$ L = |\vec{r}\times\vec{p}| = mr^2\frac{d\theta}{dt}. $$

Hence,

$$ L = mr^2\dot{\theta} = \text{constant} \tag{8} $$

Thus, the angular momentum of a particle moving under a central force remains constant.

Law of Conservation of Angular Momentum:
When a particle moves under a central force, the torque about the centre of force is zero. Consequently, its angular momentum remains constant.

$$ \vec{\tau} = 0 \quad\Longrightarrow\quad \vec{L} = \text{constant} $$