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1. Introduction

Gravitation is the force that governs the largest structures in the universe. It holds planets in their orbits around the Sun, the Moon around the Earth, and galaxies together. It also causes the fall of an apple to the ground. Isaac Newton united these celestial and terrestrial phenomena in his universal law of gravitation, published in 1687. This chapter studies that law, the acceleration due to gravity, and its variation with height, depth, and rotation of the Earth.

We then study gravitational potential energy and the escape velocity - the minimum speed needed to leave a gravitational field forever. We examine the motion of satellites in circular orbits and derive their orbital velocity and time period. Kepler's laws of planetary motion describe the shapes and speeds of planetary orbits, and their derivation from Newton's law shows the deep unity of the physics of the heavens and the Earth.

Gravitation is remarkable because it is the weakest of the four fundamental forces, yet it dominates at astronomical scales because it is always attractive and has infinite range. It is also the first force ever described by a quantitative law, and its study laid the foundation of modern science.

2. The Universal Law of Gravitation

Newton's universal law of gravitation states that every particle 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 them. If two bodies of masses m1 and m2 are separated by a distance r, the gravitational force between them is:

F = G * m1 * m2 / r^2

Here G is the universal gravitational constant, with the value G = 6.674 x 10^-11 N m^2/kg^2. The force acts along the line joining the two bodies and is always attractive. For extended bodies like the Earth, the force acts as if the entire mass were concentrated at the centre.

The gravitational force is mutual: each body attracts the other with an equal and opposite force, consistent with Newton's third law. Because the force decreases as the square of the distance, it is called an inverse-square force. The force becomes weaker with distance but never becomes zero, so gravitation has infinite range.

3. Acceleration Due to Gravity

The acceleration produced in a freely falling body by the Earth's gravity is called the acceleration due to gravity, denoted g. From Newton's law and Newton's second law, if M is the mass of the Earth and R its radius, then:

g = G * M / R^2

The value of g at the Earth's surface is approximately 9.8 m/s^2. Using G = 6.674 x 10^-11, M = 6 x 10^24 kg, and R = 6.4 x 10^6 m, we obtain this value. The acceleration due to gravity is independent of the mass of the falling body, which is why all bodies fall with the same acceleration in the absence of air resistance.

The value of g varies with height, depth, and the rotation of the Earth. At a height h above the surface, the acceleration becomes:

g' = g * (R/(R + h))^2

For small heights, this is approximately g' = g(1 - 2h/R). At a depth d below the surface, the acceleration decreases linearly:

g' = g(1 - d/R)

At the centre of the Earth, g is zero. The value of g is also slightly less at the equator than at the poles because of the Earth's rotation and its equatorial bulge.

4. Gravitational Potential Energy and Escape Velocity

The gravitational potential energy of a body of mass m at a distance r from the centre of the Earth, with the zero of potential energy at infinity, is:

U = -G * M * m / r

The negative sign indicates that the potential energy is zero at infinity and decreases (becomes more negative) as the body approaches the Earth. The gravitational potential at a point is the potential energy per unit mass:

V = -G * M / r

The escape velocity is the minimum velocity required for a body to escape from the gravitational field of the Earth permanently. It is found by equating the kinetic energy of the body to the magnitude of its gravitational potential energy at the surface:

(1/2) m v_esc^2 = G * M * m / R

v_esc = sqrt(2 G M / R) = sqrt(2 g R)

Using g = 9.8 m/s^2 and R = 6.4 x 10^6 m, the escape velocity from the Earth is about 11.2 km/s. The escape velocity is independent of the mass of the escaping body; a cricket ball and a spacecraft both need the same speed to escape from Earth.

5. Satellites and Orbital Velocity

A satellite is a body that revolves around a planet in a closed orbit. The Moon is a natural satellite of the Earth, and artificial satellites such as communication and weather satellites are placed into orbit by rockets. For a satellite to move in a circular orbit of radius r around the Earth, the gravitational force must provide the required centripetal force:

G * M * m / r^2 = m * v^2 / r

The orbital velocity is therefore:

v_o = sqrt(G M / r)

For a satellite close to the Earth's surface, r is approximately equal to the Earth's radius R, and the orbital velocity is about 7.92 km/s. The time period of the satellite is:

T = 2 pi r / v_o = 2 pi sqrt(r^3 / (G M))

A satellite with a time period equal to the Earth's rotational period and moving in the equatorial plane appears stationary to an observer on Earth; such satellites are called geostationary satellites. They orbit at a height of about 36,000 km above the Earth's surface.

6. Kepler's Laws of Planetary Motion

Kepler formulated three laws of planetary motion based on the careful observations of Tycho Brahe. The first law states that the planets move in elliptical orbits with the Sun at one focus. The second law, the law of equal areas, states that the line joining a planet to the Sun sweeps out equal areas in equal intervals of time. This law implies that a planet moves faster when it is closer to the Sun and slower when it is farther away.

The third law, the law of periods, states that the square of the period of revolution of a planet is proportional to the cube of the semi-major axis of its orbit:

T^2 proportional to a^3

Kepler's laws are empirical - they describe the observed motion of planets. Newton showed that these laws follow from his universal law of gravitation combined with his laws of motion, providing a theoretical basis for them. The derivation of Kepler's third law for a circular orbit gives:

T^2 = (4 pi^2 / G M) r^3

so that the ratio T^2/r^3 is a constant for all planets of the Sun.

7. Variation of g and Weightlessness

The apparent weight of a body is the normal reaction that a surface exerts on it. Inside a satellite orbiting the Earth, both the satellite and the bodies inside it are in free fall toward the Earth, so the normal reaction is zero. The bodies therefore appear weightless, although their mass and the gravitational force on them are unchanged. This is the phenomenon of weightlessness experienced by astronauts in orbit.

Weightlessness is not the absence of gravity; it is the absence of the normal reaction. In fact, at the height of a typical orbit, the value of g is only about 10 percent less than its value on the surface. The astronauts appear weightless because they and their spacecraft accelerate toward the Earth at the same rate.

Quick Revision Tables

Quantity Formula Value on Earth
Gravitational force F = G m1 m2 / r^2 G = 6.674 x 10^-11 N m^2/kg^2
Acceleration due to gravity g = G M / R^2 9.8 m/s^2
Escape velocity v_esc = sqrt(2 g R) 11.2 km/s
Orbital velocity v_o = sqrt(G M / r) 7.92 km/s (near surface)
Gravitational PE U = -G M m / r Negative
Kepler's Law Statement
First law Planets move in elliptical orbits, Sun at one focus
Second law Equal areas are swept in equal times
Third law T^2 proportional to a^3

Mind Map

graph TD A["GRAVITATION"] --> B["Universal Law"] A --> C["Acceleration due to Gravity"] A --> D["Escape Velocity"] A --> E["Satellites"] A --> F["Kepler's Laws"] A --> G["Weightlessness"] B --> B1["F = G m1 m2 / r^2"] B --> B2["G = 6.674 x 10^-11"] C --> C1["g = GM/R^2 = 9.8 m/s^2"] C --> C2["g' = g(1 - 2h/R), g' = g(1 - d/R)"] D --> D1["v_esc = sqrt(2GM/R) = sqrt(2gR)"] D --> D2["11.2 km/s"] E --> E1["v_o = sqrt(GM/r)"] E --> E2["Geostationary at 36000 km"] F --> F1["Elliptical orbits"] F --> F2["Equal areas in equal times"] F --> F3["T^2 proportional to a^3"] G --> G1["Free fall, zero normal reaction"]

Important Diagrams (SVG)

Diagram 1: Force Between Two Masses - Universal Law

UNIVERSAL LAW OF GRAVITATION m1 m2 F F F = G m1 m2 / r^2 Always attractive, along the joining line Inverse-square force, infinite range GOLDEN RULE Doubling the distance reduces the gravitational force to one-fourth - it falls as the square of the distance!

Diagram 2: Escape Velocity and Satellite Orbits

ESCAPE AND ORBITAL VELOCITY EARTH SATELLITE ESCAPE PATH Orbital: v_o = sqrt(GM/r) ~ 7.92 km/s Escape: v_esc = sqrt(2GM/R) ~ 11.2 km/s GOLDEN RULE Escape velocity is sqrt(2) times the orbital velocity, and it is independent of the mass of the escaping body!

Common Mistakes

  1. Confusing g (acceleration due to gravity, 9.8 m/s^2) with G (universal gravitational constant, 6.674 x 10^-11).
  2. Believing the escape velocity depends on the mass of the body; it depends only on the mass and radius of the planet.
  3. Using the formula g' = g(1 - 2h/R) for large heights; it is valid only when h is much smaller than R.
  4. Thinking that weightlessness in a satellite means zero gravity; gravity is still present, but the normal reaction is zero.
  5. Applying Kepler's third law with T^2 proportional to r^3 without specifying the semi-major axis for elliptical orbits.
  6. Forgetting the negative sign in gravitational potential energy U = -GMm/r.
  7. Confusing orbital velocity with escape velocity; escape velocity is sqrt(2) times the orbital velocity.

Exam Tips

  1. State Newton's universal law of gravitation with the formula and the value of G.
  2. Derive g = GM/R^2 from Newton's law and second law, and give its value as 9.8 m/s^2.
  3. Write the variations of g with height and depth: g' = g(1 - 2h/R) and g' = g(1 - d/R).
  4. Define escape velocity and derive v_esc = sqrt(2gR), giving the value 11.2 km/s.
  5. Derive orbital velocity v_o = sqrt(GM/r) and time period T = 2 pi sqrt(r^3/GM).
  6. State Kepler's three laws and give the numerical relation T^2 = (4 pi^2/GM) r^3.
  7. Explain weightlessness in a satellite using the concept of free fall and zero normal reaction.

Conclusion

In this chapter we studied the universal law of gravitation, F = Gm1m2/r^2, and derived the acceleration due to gravity g = GM/R^2 = 9.8 m/s^2. We saw how g varies with height, depth, and rotation, and we derived the escape velocity v_esc = sqrt(2gR) = 11.2 km/s and the orbital velocity v_o = sqrt(GM/r). The study of satellites led us to geostationary orbits, and Kepler's three laws of planetary motion described the elliptical orbits, equal areas, and the T^2 proportional to a^3 law that Newton derived from gravitation. Finally, we understood weightlessness as the condition of zero normal reaction during free fall. Gravitation reveals a single force governing the fall of an apple and the motion of planets - the deepest unification in classical physics.