What this chapter is about
Gravitation is the fundamental force of attraction that exists between any two masses in the universe. This chapter builds on Newton's laws of motion to explain how the same force that makes an apple fall to the ground also keeps the Moon in orbit around the Earth and the planets in orbit around the Sun. You will learn Newton's law of universal gravitation, which provides a single mathematical description for all gravitational phenomena.
The chapter connects to what you studied in earlier classes about the motion of celestial bodies and extends your understanding of forces. You will explore how gravitational force varies with distance, understand the concept of gravitational potential energy, and learn why satellites stay in orbit without falling down. The idea of escape velocity explains what speed a rocket needs to leave Earth permanently.
After studying this chapter, you should be able to calculate gravitational forces between objects, explain orbital motion of satellites and planets, understand Kepler's laws of planetary motion, and work with concepts like gravitational potential, acceleration due to gravity, and geostationary satellites. These ideas form the foundation for understanding space missions and the mechanics of our solar system.
Key ideas
- Universal gravitation: Every mass attracts every other mass with a force proportional to the product of their masses and inversely proportional to the square of the distance between their centres. This force acts along the line joining them.
- Acceleration due to gravity (g): Near Earth's surface, all objects experience the same gravitational acceleration of approximately 9.8 m/s², regardless of their mass. This value varies slightly with altitude, depth, and latitude.
- Gravitational potential energy: The work done against gravity to bring a mass from infinity to a point in a gravitational field. It is always negative, with zero reference at infinity.
- Kepler's laws: Planets move in elliptical orbits with the Sun at one focus; the line joining a planet to the Sun sweeps equal areas in equal times; the square of the orbital period is proportional to the cube of the semi-major axis.
- Orbital velocity: The horizontal speed needed for an object to remain in a stable circular orbit around a planet without falling or escaping.
- Escape velocity: The minimum speed an object must have at Earth's surface to escape Earth's gravitational pull completely, without any further propulsion.
- Geostationary satellites: Satellites that orbit Earth with a period of 24 hours, appearing stationary relative to a point on Earth's surface, used for communication purposes.
Formulas and facts to remember
Newton's law of universal gravitation: F = G × m₁ × m₂ / r² G = 6.67 × 10⁻¹¹ N m² kg⁻² (universal gravitational constant)
Acceleration due to gravity at Earth's surface: g = G × M / R² where M is Earth's mass and R is Earth's radius; g ≈ 9.8 m/s²
Variation of g with altitude h (when h << R): g' = g × (1 − 2h/R)
Variation of g with depth d: g' = g × (1 − d/R)
Gravitational potential energy: U = −G × M × m / r Negative sign indicates bound system; zero at infinity.
Orbital velocity for circular orbit: v₀ = √(G × M / r) = √(g × R) for orbit near surface
Escape velocity from Earth's surface: vₑ = √(2 × G × M / R) = √(2 × g × R) ≈ 11.2 km/s
Kepler's third law: T² ∝ a³, or T² = (4π² / G × M) × a³ where T is orbital period and a is semi-major axis.
Time period of satellite in circular orbit: T = 2π × √(r³ / G × M)
Worked examples
Example 1: Gravitational force between two masses
Two iron spheres, each of mass 50 kg, are placed with their centres 2 m apart in a physics laboratory. Calculate the gravitational force between them.
Solution: Using F = G × m₁ × m₂ / r² F = (6.67 × 10⁻¹¹) × 50 × 50 / (2)² F = (6.67 × 10⁻¹¹) × 2500 / 4 F = (6.67 × 10⁻¹¹) × 625 F = 4.17 × 10⁻⁸ N
This force is extremely small, which is why we do not notice gravitational attraction between everyday objects. Only when at least one mass is astronomical (like Earth) does gravity become significant.
Example 2: Value of g at a height
A weather balloon rises to a height of 32 km above Earth's surface. Find the acceleration due to gravity at this height. Take Earth's radius as 6400 km and g at surface as 9.8 m/s².
Solution: Since h = 32 km is much smaller than R = 6400 km, we use the approximation: g' = g × (1 − 2h/R) g' = 9.8 × (1 − 2 × 32/6400) g' = 9.8 × (1 − 64/6400) g' = 9.8 × (1 − 0.01) g' = 9.8 × 0.99 g' = 9.7 m/s²
The decrease is small because 32 km is tiny compared to Earth's radius.
Example 3: Escape velocity comparison
The Moon has mass approximately 1/81 of Earth's mass and radius approximately 1/3.7 of Earth's radius. If escape velocity from Earth is 11.2 km/s, find the escape velocity from the Moon's surface.
Solution: Escape velocity: vₑ = √(2GM/R) For Moon relative to Earth: vₑ(Moon)/vₑ(Earth) = √[(Mₘ/Mₑ) × (Rₑ/Rₘ)] vₑ(Moon)/11.2 = √[(1/81) × 3.7] vₑ(Moon)/11.2 = √(3.7/81) = √0.0457 = 0.214 vₑ(Moon) = 11.2 × 0.214 = 2.4 km/s
The Moon's lower escape velocity explains why it has no atmosphere; gas molecules easily exceed this speed and escape.
Common mistakes
Confusing mass and weight → Mass is the amount of matter (constant everywhere), while weight is the gravitational force on that mass (varies with location and g).
Using g = 9.8 m/s² for all situations → Remember that g varies with altitude, depth, and latitude; use the appropriate formula when the situation demands.
Forgetting the negative sign in gravitational potential energy → The negative sign is essential; it shows that energy must be supplied to separate the masses to infinity.
Thinking satellites fall because gravity is zero in orbit → Gravity acts on satellites; they are in continuous free fall but their horizontal velocity keeps them in orbit.
Applying Kepler's third law with wrong units → When using T² ∝ a³, ensure consistent units; mixing years with metres or days with kilometres gives incorrect results.
Quick revision
Newton's law: F = Gm₁m₂/r², where G = 6.67 × 10⁻¹¹ N m² kg⁻².
Acceleration due to gravity: g = GM/R² ≈ 9.8 m/s² at Earth's surface.
Escape velocity from Earth: vₑ = √(2gR) ≈ 11.2 km/s.
Orbital velocity for satellite near surface: v₀ = √(gR) ≈ 7.9 km/s.
Gravitational potential energy is negative and increases (becomes less negative) as distance increases.
Kepler's third law: T² ∝ a³ applies to all bodies orbiting the same central mass.