Gravitation

1. Introduction

Every object in the universe attracts every other object with a force directed along the line joining their centres. This universal attraction is called gravitation, and it is the force that keeps the planets in orbit around the Sun, holds the Moon in orbit around the Earth, and gives every body its weight near the Earth's surface. This chapter develops Newton's law of gravitation and applies it to the acceleration due to gravity, gravitational field and potential, the motion of satellites, escape velocity, and modern applications such as the Global Positioning System (GPS).

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2. Newton's Law of Gravitation

Statement: Every particle of matter in the universe attracts every other particle with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centres.

Consider two bodies of masses $m_1$ and $m_2$ whose centres are separated by a distance $d$. Let $F$ be the force of attraction between them.

1. The force is directly proportional to the product of the masses: $$F \propto m_1 m_2 \qquad \text{(i)}$$

2. The force is inversely proportional to the square of the separation: $$F \propto \frac{1}{d^2} \qquad \text{(ii)}$$

Combining (i) and (ii):

$$ \begin{aligned} F \propto \frac{m_1 m_2}{d^2} \ \boxed{F = G,\frac{m_1 m_2}{d^2}} \end{aligned} $$

Here $G$ is the constant of proportionality called the universal gravitational constant. Its value is:

$$G = 6.67 \times 10^{-11}\ \text{N m}^2,\text{kg}^{-2}$$

Dimensional formula of $G$: From $G = \dfrac{F d^2}{m_1 m_2}$,

$$ \begin{aligned} [G] &= \frac{[\mathrm{M,L,T^{-2}}][\mathrm{L^2}]}{[\mathrm{M^2}]} \ &= [\mathrm{M^{-1},L^{3},T^{-2}}] \end{aligned} $$

Features of the gravitational force:

• It is always attractive and acts along the line joining the two centres.
• It is a universal force, acting between all pairs of masses.
• It obeys Newton's third law: the two bodies attract each other with equal and opposite forces.
• It is a central force and a conservative force.
• It is independent of the medium between the bodies.

2.1 Acceleration due to gravity ($g$)

Consider a body of mass $m$ resting on the surface of the Earth, where the Earth has mass $M$ and radius $R$. By Newton's law of gravitation, the force of attraction between the Earth and the body is:

$$F = \frac{GMm}{R^2} \qquad \text{(i)}$$

This gravitational pull is the weight of the body. If $g$ is the acceleration due to gravity, then by Newton's second law:

$$F = mg \qquad \text{(ii)}$$

Equating (i) and (ii):

$$ \begin{aligned} mg &= \frac{GMm}{R^2} \ \boxed{g = \frac{GM}{R^2}} \end{aligned} $$

This is the required expression for the acceleration due to gravity. Since the mass $m$ of the body has cancelled out, $g$ is independent of the mass of the falling body: all bodies fall with the same acceleration in the absence of air resistance.

Taking $M = 6.0 \times 10^{24}\ \text{kg}$ and $R = 6.4 \times 10^{6}\ \text{m}$, this gives $g \approx 9.8\ \text{m s}^{-2}$ at the Earth's surface.

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3. Gravitational Field and Field Strength

3.1 Gravitational field

The region of space around a body within which its gravitational force of attraction can be experienced by another body is called its gravitational field.

3.2 Gravitational field strength (intensity)

The gravitational field strength (or gravitational field intensity) at a point in a gravitational field is defined as the gravitational force experienced by a body of unit mass placed at that point.

Consider a body of mass $m$ placed at a distance $r$ from the centre of the Earth (mass $M$, radius $R$). The gravitational force on the body is:

$$F = \frac{GMm}{r^2}$$

The gravitational field strength $E$ is the force per unit mass:

$$ \begin{aligned} E &= \frac{F}{m} \ &= \frac{1}{m}\cdot\frac{GMm}{r^2} \ \boxed{E = \frac{GM}{r^2}} \end{aligned} $$

It is a vector quantity directed towards the centre of the Earth, and its SI unit is $\text{N kg}^{-1}$ (equivalently $\text{m s}^{-2}$).

On the surface of the Earth ($r = R$):

$$ \begin{aligned} E &= \frac{GM}{R^2} \ &= g \end{aligned} $$

Hence, at the Earth's surface the gravitational field strength is numerically equal to the acceleration due to gravity.

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4. Gravitational Potential and Gravitational Potential Energy

4.1 Gravitational potential ($V$)

The gravitational potential at a point in a gravitational field is defined as the amount of work done in bringing a body of unit mass from infinity to that point.

Derivation. Let $P$ be a point at a distance $r$ from the centre of the Earth (mass $M$). Take a unit mass at infinity and bring it towards the Earth. When it is at a point $A$, a distance $x$ from the centre, the gravitational force on it is:

$$ \begin{aligned} F &= \frac{GM(1)}{x^2} \ &= \frac{GM}{x^2} \end{aligned} $$

When the unit mass is displaced a further small distance $dx$ towards the Earth, the small work done by the gravitational force is:

$$dW = F,dx = \frac{GM}{x^2},dx$$

The total work done in bringing the unit mass from infinity to the point $P$ (from $x = \infty$ to $x = r$) is:

$$ \begin{aligned} W &= \int_{\infty}^{r} \frac{GM}{x^2}\ \ dx &= GM\left[\frac{-1}{x}\right]_{\infty}^{r} \ &= GM\left(-\frac{1}{r} + \frac{1}{\infty}\right) \ W &= -\frac{GM}{r} \end{aligned} $$

Since the gravitational potential is this work done per unit mass:

$$\boxed{V = -\frac{GM}{r}}$$

The gravitational potential is negative because the force is attractive and work is done by the field as the mass moves in from infinity. Its SI unit is $\text{J kg}^{-1}$.