Gravity and Gravitation

7.1 Newton's Law of Gravitation

7.1.1 Statement of the Law

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 them. This force acts along the line joining the two particles.

If two point masses m₁ and m₂ are separated by a distance r, the magnitude of the mutual gravitational force F is:

where G is the universal gravitational constant, whose value in SI units is:

G = 6.674 × 10⁻¹¹ N m² kg⁻²

The value of G is the same everywhere in the universe and does not depend on the nature of the intervening medium, the masses involved, or their separation — it is a fundamental constant of nature, first measured experimentally by Henry Cavendish in 1798 using a sensitive torsion balance.

7.1.2 Vector Form of the Law

Since gravitational force is attractive and directed along the line joining the two masses, it can be written in vector form. If r⃗ is the position vector of m₂ relative to m₁, and r̂ is the unit vector along this direction, the force on m₂ due to m₁ is:

The negative sign indicates that the force is attractive, i.e. directed opposite to r̂ (opposite to the direction of increasing separation), pulling m₂ toward m₁. By Newton's third law, the force on m₁ due to m₂ is equal in magnitude and opposite in direction: F⃗₁₂ = −F⃗₂₁.

7.1.3 Characteristics of Gravitational Force

• It is always attractive, never repulsive.
• It obeys the inverse square law — the force falls off rapidly with distance.
• It is a central force, acting along the line joining the two masses.
• It is conservative — the work done in moving a mass between two points does not depend on the path taken.
• It is independent of the intervening medium — unlike electrostatic force in a dielectric.
• It is the weakest of the four fundamental forces of nature (gravitational, electromagnetic, strong nuclear, weak nuclear), yet it dominates on astronomical scales because it is always attractive and has infinite range.
• It obeys the principle of superposition: the net gravitational force on a particle due to several other masses is the vector sum of the forces due to each mass individually.

7.1.4 Principle of Superposition

If a mass m experiences gravitational forces F⃗₁, F⃗₂, F⃗₃, … due to masses M₁, M₂, M₃, … respectively, the resultant force on m is the vector sum:

This principle is essential when calculating the gravitational effect of extended or multiple bodies, such as the net pull on a satellite from the Earth, Moon, and Sun simultaneously.

Worked Example: Gravitational Force Between Two Masses

Problem: Calculate the gravitational force of attraction between two point masses of 50 kg and 80 kg separated by a distance of 2 m. (G = 6.674 × 10⁻¹¹ N m² kg⁻²)

Solution:

F = G m₁m₂ / r² = (6.674 × 10⁻¹¹ × 50 × 80) / (2)²

F = (6.674 × 10⁻¹¹ × 4000) / 4 = (2.6696 × 10⁻⁷) / 4

F = 6.674 × 10⁻⁸ N

Physical interpretation: This force is extraordinarily small compared to everyday forces, which explains why the gravitational attraction between ordinary objects is never noticed — gravitational effects only become significant when at least one of the masses is astronomically large.

7.2 Gravitational Field Strength

7.2.1 Concept of a Gravitational Field

A gravitational field is the region of space surrounding a mass within which another mass experiences a force of gravitational attraction. Rather than thinking of gravity as instantaneous "action at a distance," the field concept treats the mass as modifying the space around it, and any other mass placed in that space responds to the local field.

7.2.2 Definition of Gravitational Field Strength

The gravitational field strength (also called gravitational field intensity) at a point is defined as the gravitational force experienced per unit mass placed at that point.

It is a vector quantity, directed toward the source mass, with SI unit N kg⁻¹, which is dimensionally equivalent to m s⁻² (the unit of acceleration). This is not a coincidence: gravitational field strength at a point is numerically equal to the acceleration a test mass would experience if released there.

7.2.3 Gravitational Field Strength Due to a Point Mass

Consider a point mass M. By Newton's law of gravitation, the force it exerts on a small test mass m placed at a distance r is F = GMm/r². Using the definition above:

E = F/m = GMm/r² ÷ m

This is directed radially toward the mass M. Near the surface of the Earth (mass Mₑ, radius Rₑ), this field strength is exactly the familiar acceleration due to gravity:

This equation is one of the most important relations in the chapter, since it connects the macroscopic, measurable quantity g (≈ 9.8 m s⁻²) to the fundamental constant G and the Earth's mass and radius.

7.2.4 Field Due to a System of Masses

By the superposition principle, the net gravitational field at a point due to several masses is the vector sum of the fields due to each mass considered separately:

7.3 Gravitational Potential and Potential Energy

7.3.1 Gravitational Potential Energy: Derivation

The gravitational potential energy of a system of two masses is defined as the work done in bringing one mass from infinity to a given point in the gravitational field of the other, without any change in kinetic energy, with potential energy conventionally taken as zero at infinite separation.

Consider a mass M fixed at the origin, and a test mass m to be brought from infinity to a distance r from M along a radial line, moved quasi-statically (so that its kinetic energy remains unchanged throughout).

The gravitational force on m at a distance x from M is directed toward M (attractive): F(x) = −GMm/x²

Since force and potential energy are related by F = −dU/dx, the potential energy is obtained by integration:

U(r) = GMm · [−1/x]∞ʳ = GMm · (−1/r − 0)

Taking U(∞) = 0, the gravitational potential energy at distance r is:

Physical interpretation: The negative sign shows that the gravitational potential energy is always negative for a bound (attractive) system, with the least (most negative) value at r = 0 and increasing (becoming less negative) as r increases, approaching zero as r → ∞. This reflects the fact that work must be done against gravity to separate two masses to infinity.

7.3.2 Gravitational Potential: Definition and Derivation

The gravitational potential V at a point in a gravitational field is defined as the gravitational potential energy per unit mass at that point, or equivalently, the work done per unit mass in bringing a small test mass from infinity to that point without acceleration.

Gravitational potential is a scalar quantity with SI unit J kg⁻¹. It is always negative for a real (attractive) gravitational field, with V = 0 conventionally set at infinity, which represents the point of maximum potential.

7.3.3 Relation Between Gravitational Field Strength and Potential

Since potential energy is related to force by F = −dU/dr, dividing throughout by m gives the corresponding relation between field strength and potential:

This states that gravitational field strength at a point equals the negative gradient (rate of change) of gravitational potential with respect to distance. The negative sign indicates that the field points in the direction of decreasing potential.

Worked Example: Gravitational Potential Energy of a Satellite

Problem: Find the gravitational potential energy of a 2000 kg satellite located at a distance of 7000 km from the centre of the Earth. (Mass of Earth M = 6.0 × 10²⁴ kg, G = 6.674 × 10⁻¹¹ N m² kg⁻²)

Solution:

U = −GMm/r = −(6.674 × 10⁻¹¹ × 6.0 × 10²⁴ × 2000) / (7 × 10⁶)

U = −(6.674 × 6.0 × 2000 × 10¹³) / (7 × 10⁶) = −(80088 × 10¹³) / (7 × 10⁶)

U ≈ −1.14 × 10¹¹ J

7.4 Variation in Value of g Due to Altitude and Depth

7.4.1 Variation of g with Altitude (Height Above the Surface)

Assume the Earth is a uniform sphere of mass M and radius R. Consider a point at height h above the Earth's surface, so its distance from the Earth's centre is (R + h).

The acceleration due to gravity at the surface: g = GM/R²

The acceleration due to gravity at height h: gₕ = GM/(R+h)²

Dividing the two expressions:

gₕ/g = R²/(R+h)² = 1/(1 + h/R)²

For heights much smaller than the Earth's radius (h ≪ R), using the binomial approximation (1+x)ⁿ ≈ 1 + nx for small x:

Physical interpretation: The value of g decreases as altitude increases, since the point of measurement moves farther from the centre of mass of the Earth. This is why astronauts in low Earth orbit still experience substantial gravitational pull — their weightlessness is due to being in continuous free fall, not due to zero gravity.

7.4.2 Variation of g with Depth (Below the Surface)

Assume the Earth is treated as a uniform sphere of density ρ. By the shell theorem, only the mass enclosed within a sphere of radius (R − d) contributes to the gravitational field at a depth d below the surface; the mass in the spherical shell outside this radius exerts zero net gravitational field at interior points.

Mass of the Earth (uniform density): M = (4/3)πR³ρ

Mass enclosed within radius (R − d): M' = (4/3)π(R−d)³ρ

The acceleration due to gravity at depth d:

gd = GM'/(R−d)² = G · (4/3)π(R−d)³ρ / (R−d)² = (4/3)πGρ(R−d)

Since g = GM/R² = (4/3)πGρR, we can write (4/3)πGρ = g/R. Substituting:

Physical interpretation: The value of g decreases linearly with depth (assuming uniform density) and becomes zero at the centre of the Earth (d = R), since at the exact centre the gravitational pull from all directions cancels out completely.

Worked Example: Variation of g with Altitude

Problem: At what height above the Earth's surface does the acceleration due to gravity reduce to 4% less than its value at the surface? (Take R = 6400 km, and use the approximation for h ≪ R.)

Solution:

Using gₕ ≈ g(1 − 2h/R), the fractional decrease is 2h/R = 0.04

h = (0.04 × R)/2 = (0.04 × 6400)/2 = 256/2

h = 128 km

7.5 Centre of Mass and Centre of Gravity

7.5.1 Centre of Mass

The centre of mass (COM) of a system of particles is the point at which the entire mass of the system may be considered to be concentrated for the purpose of describing its translational motion. For a system of n particles of masses m₁, m₂, …, mₙ located at position vectors r⃗₁, r⃗₂, …, r⃗ₙ, the position vector of the centre of mass is defined as:

Along a single coordinate axis, this reduces to:

Xcm = Σmᵢxᵢ / Σmᵢ

The centre of mass is purely a consequence of how mass is distributed within a system; it does not depend on any external field. For a uniform, symmetric body (a sphere, cube, or rod of uniform density), the centre of mass coincides with the geometric centre.

7.5.2 Centre of Gravity

The centre of gravity (CG) of a body is the point at which the entire weight of the body may be considered to act, such that the net gravitational torque about this point due to all the particles of the body is zero.

For a body placed in a gravitational field that is uniform in magnitude and direction over the entire extent of the body (as is an excellent approximation for ordinary objects near the Earth's surface), the centre of gravity coincides exactly with the centre of mass.

7.5.3 Distinction Between Centre of Mass and Centre of Gravity

• Centre of mass depends only on the distribution of mass within the body; centre of gravity depends on both the mass distribution and the gravitational field in which the body is placed.
• For a body of ordinary size on Earth, g is essentially uniform across the body, so COM and CG coincide.
• For an extremely large or extended body (such as a very tall structure, or a body positioned so that one part is significantly closer to a gravitating mass than another part, e.g. a large satellite or a mountain-sized object), g varies measurably from one part of the body to another, and the centre of gravity shifts slightly toward the region of stronger field, no longer coinciding exactly with the centre of mass.
• Centre of mass exists for a system even in the complete absence of gravity (e.g. in deep space); centre of gravity has no meaning without a gravitational field.

Real-Life Application

The distinction between COM and CG becomes physically significant in analysing the stability of tall structures and in orbital mechanics, where gravity-gradient torque (arising from the slight separation of COM and CG in extended satellites) is used to passively stabilise the orientation of some spacecraft.

7.6 Motion of a Satellite: Orbital Velocity and Time Period

7.6.1 Orbital Velocity: Derivation

Consider a satellite of mass m moving in a stable circular orbit of radius r around a planet of mass M, with the necessary centripetal force being provided entirely by gravitational attraction.

Gravitational force provides the centripetal force required for circular motion:

GMm/r² = mv₀²/r

Cancelling m from both sides and one factor of r:

GM/r = v₀²

This is the orbital velocity — the speed a satellite must maintain in a circular orbit of radius r to avoid either falling toward the planet or drifting away. If the orbit is close to the planet's surface (r ≈ R), and using GM = gR², this simplifies to:

Physical interpretation: Orbital velocity decreases as the orbital radius increases — satellites in higher orbits move more slowly than those in lower orbits, since the gravitational pull (and hence the required centripetal acceleration) is weaker farther from the planet.

7.6.2 Time Period of a Satellite: Derivation

The time period T is the time taken for the satellite to complete one full revolution, equal to the circumference of the orbit divided by the orbital speed:

T = 2πr / v₀ = 2πr / √(GM/r) = 2πr · √(r/GM) = 2π √(r³/GM)

Squaring both sides gives T² = (4π²/GM) r³, i.e. T² ∝ r³, which is precisely Kepler's third law of planetary motion, here derived directly from Newton's law of gravitation combined with the dynamics of uniform circular motion. This connection — showing that Kepler's empirically discovered law follows necessarily from Newtonian gravitation — is one of the most historically significant results in the whole of classical physics.

Worked Example: Orbital Velocity and Period of a Satellite

Problem: A satellite revolves around the Earth in a circular orbit at a height of 1600 km above the Earth's surface. Calculate its orbital velocity and time period. (R = 6400 km, g = 9.8 m s⁻²)

Solution:

Orbital radius: r = R + h = 6400 + 1600 = 8000 km = 8 × 10⁶ m

Using GM = gR²: GM = 9.8 × (6.4 × 10⁶)² = 9.8 × 4.096 × 10¹³ = 4.014 × 10¹⁴ m³s⁻²

Orbital velocity: v₀ = √(GM/r) = √(4.014 × 10¹⁴ / 8 × 10⁶) = √(5.02 × 10⁷)

v₀ ≈ 7085 m s⁻¹ ≈ 7.09 km s⁻¹

Time period: T = 2πr/v₀ = (2 × 3.1416 × 8 × 10⁶) / 7085 = (5.027 × 10⁷) / 7085

T ≈ 7095 s ≈ 1.97 hours

7.7 Escape Velocity

7.7.1 Definition

Escape velocity is the minimum velocity with which an object must be projected from the surface of a planet so that it can permanently overcome the planet's gravitational pull and escape to infinity, arriving there with (at minimum) zero kinetic energy remaining.

7.7.2 Derivation Using Energy Conservation

Consider a body of mass m projected from the surface of a planet of mass M and radius R with velocity vₑ, and air resistance is neglected. The condition for "just escaping" is that the body reaches infinity with zero kinetic energy.

Total mechanical energy at the surface = Kinetic energy + Gravitational potential energy

Esurface = ½mvₑ² + (−GMm/R) = ½mvₑ² − GMm/R

Total mechanical energy at infinity (with zero kinetic energy and zero potential energy, by convention):

E∞ = 0 + 0 = 0

By the law of conservation of energy, Esurface = E∞:

½mvₑ² − GMm/R = 0

½vₑ² = GM/R

Using GM = gR², this can also be written as:

For the Earth (g = 9.8 m s⁻², R = 6.4 × 10⁶ m), this gives vₑ ≈ 11.2 km s⁻¹.

7.7.3 Relation Between Escape Velocity and Orbital Velocity

Comparing vₑ = √(2GM/R) with the orbital velocity at the same radius, v₀ = √(GM/R):

Physical interpretation: Escape velocity is always √2 (≈ 1.414) times the orbital velocity at the same distance from the centre of the planet. This is why a satellite cannot simply "speed up slightly" to escape orbit — it must increase its speed by a factor of √2, roughly a 41% increase in speed (a much larger increase in kinetic energy).

7.8 Potential and Kinetic Energy of the Satellite

7.8.1 Kinetic Energy of an Orbiting Satellite

For a satellite of mass m in a circular orbit of radius r with orbital velocity v₀ = √(GM/r), the kinetic energy is:

KE = ½mv₀² = ½m · (GM/r)

7.8.2 Potential Energy of an Orbiting Satellite

From the derivation in Section 7.3, the gravitational potential energy of the satellite-planet system at orbital radius r is:

7.8.3 Total Mechanical Energy of the Satellite

The total mechanical energy is the sum of kinetic and potential energy:

E = KE + PE = GMm/2r + (−GMm/r) = GMm/2r − GMm/r = GMm/2r − 2GMm/2r

Physical interpretation: The total energy of an orbiting satellite is negative, which is the mathematical signature of a bound system — the satellite does not have enough energy to escape to infinity and remains gravitationally trapped in orbit. Note also the important relation E = −KE, and PE = 2E, which follow directly from combining the expressions above; this factor-of-two relationship between kinetic and total energy is a general feature of any inverse-square-law circular orbit (a result related to the virial theorem).

7.8.4 Binding Energy of a Satellite

The binding energy of a satellite is the minimum energy that must be supplied to it to remove it from orbit to infinity (where its total energy becomes zero). Since the satellite's energy in orbit is E = −GMm/2r, the binding energy is simply the magnitude of this quantity:

Worked Example: Energy of an Orbiting Satellite

Problem: A satellite of mass 400 kg orbits the Earth at a height where its orbital radius is 7200 km. Calculate its kinetic energy, potential energy, and total energy. (M = 6.0 × 10²⁴ kg, G = 6.674 × 10⁻¹¹ N m² kg⁻²)

Solution:

GMm = 6.674 × 10⁻¹¹ × 6.0 × 10²⁴ × 400 = 1.602 × 10¹⁶

KE = GMm/2r = 1.602 × 10¹⁶ / (2 × 7.2 × 10⁶) = 1.602 × 10¹⁶ / 1.44 × 10⁷

KE ≈ 1.113 × 10⁹ J

PE = −GMm/r = −1.602 × 10¹⁶ / 7.2 × 10⁶

PE ≈ −2.225 × 10⁹ J

Total energy E = KE + PE = 1.113 × 10⁹ − 2.225 × 10⁹

E ≈ −1.113 × 10⁹ J (note that E = −KE, confirming the general relation)

7.9 Geostationary Satellite

7.9.1 Definition

A geostationary satellite is an artificial satellite that orbits the Earth in such a way that it appears stationary when viewed from a fixed point on the Earth's surface. This requires three specific conditions to be satisfied simultaneously:

• The orbit must be circular.
• The orbit must lie in the plane of the Earth's equator.
• The satellite's direction of revolution must be west to east, the same as the Earth's rotation, and its time period must exactly equal the Earth's rotational period (one sidereal day ≈ 23 hours 56 minutes, closely approximated as 24 hours for classroom calculations).

7.9.2 Derivation of the Height of a Geostationary Orbit

Using the time period formula derived in Section 7.6, T = 2π√(r³/GM), and setting T equal to the Earth's rotational period:

T² = 4π²r³/GM

r³ = GMT² / 4π²

Substituting G = 6.674 × 10⁻¹¹ N m² kg⁻², M = 6.0 × 10²⁴ kg, and T = 24 × 3600 s = 86400 s, this calculation yields:

Since the Earth's radius R ≈ 6400 km, the height of the geostationary orbit above the Earth's surface is:

7.9.3 Applications of Geostationary Satellites

• Telecommunication relay (television broadcast, long-distance telephone links).
• Weather monitoring satellites that continuously observe the same region of the Earth.
• Fixed-point satellite dish communication, since ground antennas do not need to track a moving satellite.

7.10 GPS (Global Positioning System)

7.10.1 Overview

The Global Positioning System (GPS) is a satellite-based navigation system that allows a receiver anywhere on or near the Earth's surface to determine its precise position (latitude, longitude, and altitude) and time. It consists of a constellation of about 24–32 satellites distributed in orbits at an altitude of approximately 20,200 km, arranged so that at least four satellites are visible from any point on Earth at any time.

Unlike geostationary satellites, GPS satellites are not geostationary; each completes roughly two orbits per sidereal day (time period ≈ 12 hours), and their orbital planes are inclined so as to provide global coverage rather than remaining fixed above one location.

7.10.2 Principle of Operation: Trilateration

Each GPS satellite continuously transmits a signal containing its precise orbital position and the exact time of transmission, generated using an onboard atomic clock. A GPS receiver on the ground measures the time delay between transmission and reception for signals from at least four different satellites, and multiplies this delay by the speed of light to determine its distance from each satellite. Using the known positions of the satellites and these four measured distances, the receiver's three-dimensional position (and a correction to its own clock) is calculated by a geometric method known as trilateration.

7.10.3 Role of Orbital Mechanics in GPS

The concepts developed earlier in this chapter — orbital velocity, time period, and the relation T² ∝ r³ — directly determine the design of the GPS satellite constellation: the orbital radius of about 26,600 km (measured from the Earth's centre) is chosen precisely because it produces the required 12-hour orbital period, ensuring predictable, repeating ground-track geometry for reliable global coverage.

At the level of precision GPS requires, both special and general relativistic effects on the onboard atomic clocks (arising from the satellite's orbital speed and its position in the Earth's gravitational field) become significant and must be corrected for; without such corrections, GPS position errors would accumulate at a rate of several kilometres per day.

7.10.4 Applications of GPS

• Navigation for aviation, maritime, and road transport.
• Precision agriculture and land surveying.
• Disaster management, emergency response, and search-and-rescue location tracking.
• Scientific applications including geodesy, tectonic plate motion monitoring, and precise timing synchronisation for communication networks.

Exam Tips and Key Takeaways

Important Formulas to Memorize

• F = Gm₁m₂/r²
• E = GM/r²
• V = −GM/r
• U = −GMm/r
• gₕ ≈ g(1−2h/R)
• gd = g(1−d/R)
• v₀ = √(GM/r)
• T = 2π√(r³/GM)
• vₑ = √(2GM/R) = √2·v₀
• KE = GMm/2r
• PE = −GMm/r
• Etotal = −GMm/2r

Common Board Questions

1. State and derive Newton's law of gravitation in vector form.
2. Derive the expression for variation of g with altitude and with depth.
3. Derive the orbital velocity and time period of a satellite, and show T² ∝ r³.
4. Derive the escape velocity and establish its relation with orbital velocity.
5. Distinguish between centre of mass and centre of gravity.
6. Explain the conditions required for a geostationary orbit and derive its height.
7. Explain the working principle of GPS.

Typical Numerical Pattern

• Given height/depth, find g
• Given orbital radius or height, find orbital velocity and time period
• Given planetary data, find escape velocity
• Given satellite mass and orbital radius, find KE, PE, and total energy

Important Reminders

• Always distinguish between quantities measured "from the Earth's centre" (r = R + h) and "above the surface" (h) — this is the single most common source of numerical errors in this chapter.
• Remember the sign convention: gravitational potential energy and potential are always negative for a bound system, with zero taken at infinity.
• When asked to compare escape velocity and orbital velocity, always quote the exact relation vₑ = √2 · v₀, since this is a frequently tested conceptual question.