Unit 7: Gravitation

Newton's Law of Gravitation

Newton's law of universal gravitation states that every point mass attracts every other point mass in the universe 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.

F = G \frac{m_1 m_2}{r^2}
where:

F – gravitational force (N)

G – universal gravitational constant = 6.67 × 10^{-11} N·m^2/kg^2

m₁, m₂ – masses of the two bodies (kg)

r – distance between the centres of the two masses (m)

The force is always attractive and follows an inverse‑square law.

Numerical Example 1

Calculate the gravitational force between the Earth (mass M_E = 5.97 × 10^{24} kg) and a 1 kg object placed on the Earth's surface (radius R_E = 6.37 × 10^6 m).

Substitute values into the formula:

F = (6.67×10^{-11}) × (5.97×10^{24}) × (1) / (6.37×10^6)^2

Carrying out the calculation:

F ≈ 9.80 N

This value is essentially the weight of the 1 kg mass near Earth's surface (g ≈ 9.8 m/s^2).

Gravitational Field Strength

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

g = F/m = GM/r^2

It is a vector quantity directed towards the centre of the mass creating the field. Its SI unit is N/kg, which is equivalent to m/s^2.

Numerical Example 2

Find the gravitational field strength at a height of 2000 km above Earth's surface.

Distance from Earth's centre: r = R_E + h = 6.37×10^6 + 2.0×10^6 = 8.37×10^6 m.

Apply g = GM/r^2:

g = (6.67×10^{-11})(5.97×10^{24}) / (8.37×10^6)^2 ≈ 5.68 m/s^2

Thus, at 2000 km altitude the field strength drops to about 58 % of its surface value.

Gravitational Potential and Potential Energy

Gravitational potential (V) at a point is the work done per unit mass to bring a small test mass from infinity to that point without acceleration.

V = -GM/r

The negative sign indicates that work must be done against the gravitational field to move the mass away from the source.

Gravitational potential energy (U) of a mass m at distance r is:

U = mV = -GMm/r

At the Earth's surface (r = R):

U_{surface} = -GMm/R

For small heights (h ≪ R) the change in potential energy approximates to the familiar expression:

ΔU ≈ mgh

Numerical Example 3

Compute the gravitational potential energy of a 500 kg satellite orbiting at a radius of 7.0 × 10^6 m from Earth's centre.

Use U = -GMm/r:

U = -(6.67×10^{-11})(5.97×10^{24})(500) / (7.0×10^6) ≈ -2.84×10^{10} J

The negative sign shows the satellite is bound to Earth.

Variation in the Value of g

The acceleration due to gravity is not constant over Earth's surface; it varies with altitude, depth, latitude, and Earth's shape.

Altitude Effect

For heights h much smaller than Earth's radius (h ≪ R):

g' = g (1 - 2h/R)

Depth Effect

At a depth d below the surface (assuming uniform density):

g' = g (1 - d/R)

Latitude Effect

Due to Earth's rotation and its oblate spheroid shape, g is maximum at the poles and minimum at the equator.

Numerical Example 4

Estimate g at a height of 10 km above sea level.

Using g' = g (1 - 2h/R) with g = 9.8 m/s^2, h = 1.0×10^4 m, R = 6.37×10^6 m:

g' = 9.8 (1 - 2×1.0×10^4 / 6.37×10^6) ≈ 9.8 (1 - 0.00314) ≈ 9.77 m/s^2

The reduction is only about 0.03 m/s^2, illustrating the weak altitude dependence near the surface.

Centre of Mass and Centre of Gravity

Centre of mass (CM) is the point where the total mass of a system can be considered to be concentrated for the purpose of analysing translational motion.

For a system of point masses:

x_{cm} = (∑ m_i x_i) / (∑ m_i)

(Similar expressions hold for y and z coordinates.)

Centre of gravity (CG) is the point where the total weight of the body appears to act. In a uniform gravitational field, the CG coincides with the CM.

Numerical Example 5

Find the centre of mass of two masses: 3 kg at x = 0 m and 5 kg at x = 4 m.

x_{cm} = (3×0 + 5×4) / (3+5) = 20 / 8 = 2.5 m

The CM lies 2.5 m from the origin, closer to the larger mass.

Motion of a Satellite

A satellite in a circular orbit of radius r experiences a centripetal force provided by gravity.

Orbital Velocity

v_{orb} = √(GM/r) = √(gR^2/r)

Time Period

T = 2π √(r^3/GM)

This is Kepler’s third law for circular orbits.

Surface Orbit Approximation

If the satellite orbits just above Earth's surface (r ≈ R):

T ≈ 2π √(R/g) ≈ 84 minutes

Numerical Example 6

Determine the orbital speed and period of a satellite at an altitude of 400 km.

Orbital radius: r = R + h = 6.37×10^6 + 0.4×10^6 = 6.77×10^6 m.

Orbital velocity:

v = √(GM/r) = √[(6.67×10^{-11})(5.97×10^{24}) / 6.77×10^6] ≈ 7.66×10^3 m/s = 7.66 km/s

Time period:

T = 2π √(r^3/GM) = 2π √[(6.77×10^6)^3 / ((6.67×10^{-11})(5.97×10^{24}))] ≈ 5.55×10^3 s ≈ 92.5 min

These values agree with typical low‑Earth‑orbit satellites.

Escape Velocity

Escape velocity is the minimum speed needed for an object to break free from the gravitational attraction of a massive body without further propulsion.

v_{esc} = √(2GM/R) = √(2gR)

Numerical Example 7

Calculate Earth's escape velocity.

v_{esc} = √[2 × (6.67×10^{-11}) × (5.97×10^{24}) / (6.37×10^6)] ≈ 1.12×10^4 m/s = 11.2 km/s

Thus, a projectile must be launched at roughly 11.2 km/s to leave Earth's gravitational influence.

Potential and Kinetic Energy of a Satellite

For a satellite of mass m in a circular orbit of radius r:

Kinetic Energy:

KE = ½ m v_{orb}^2 = GMm/(2r)

Gravitational Potential Energy:

PE = -GMm/r

Total Mechanical Energy:

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

The total energy is negative, indicating a bound orbit; the magnitude equals the kinetic energy.

Numerical Example 8

Find KE, PE, and total energy for the 500 kg satellite from Example 3 (r = 7.0×10^6 m).

KE = GMm/(2r) = (6.67×10^{-11})(5.97×10^{24})(500) / (2×7.0×10^6) ≈ 1.42×10^{10} J

PE = -GMm/r = -2.84×10^{10} J (from Example 3)

E = KE + PE = -1.42×10^{10} J

Notice that |KE| = |E| and PE = 2E, as expected for a circular orbit.

Geostationary Satellite

A geostationary satellite appears stationary relative to an observer on Earth because its orbital period matches Earth's rotation period (24 h) and it orbits above the equator.

Orbital Radius

Setting T = 24 h = 86400 s in the period formula:

r = [GMT^2/(4π^2)]^{1/3}

Substituting Earth’s values gives r ≈ 4.22×10^7 m.

Height Above Surface

h = r - R ≈ 4.22×10^7 - 6.37×10^6 ≈ 3.58×10^7 m ≈ 35 800 km

Often quoted as ≈ 36 000 km.

Applications

Communication (TV, telephone, internet)

Weather monitoring

Navigation aids

Numerical Example 9

Verify the orbital speed of a geostationary satellite.

Use v = √(GM/r) with r = 4.22×10^7 m:

v = √[(6.67×10^{-11})(5.97×10^{24}) / 4.22×10^7] ≈ 3.07×10^3 m/s ≈ 3.07 km/s

Despite the high altitude, the speed is much lower than low‑Earth‑orbit satellites because the orbital radius is large.

Global Positioning System (GPS)

The GPS consists of a constellation of at least 24 satellites orbiting in medium Earth orbit (MEO) at an altitude of about 20 200 km (orbital radius ≈ 26 600 km). Each satellite transmits precise timing signals; a receiver determines its position by trilateration from signals of at least four satellites.

Working Principle

Each satellite broadcasts its precise orbital parameters (ephemeris) and a timestamp.

The receiver measures the signal travel time, converting it to a pseudo‑range.

With four satellites, the receiver solves for three spatial coordinates and clock bias.

Applications

Navigation (automotive, aviation, marine)

Surveying and geodesy

Tracking (assets, wildlife, personal devices)

Timing synchronization for networks

Numerical Example 10

Approximate the orbital period of a GPS satellite.

Orbital radius: r ≈ 2.66×10^7 m.

Period: T = 2π √(r^3/GM):

T = 2π √[(2.66×10^7)^3 / ((6.67×10^{-11})(5.97×10^{24}))] ≈ 4.32×10^4 s ≈ 12.0 h

GPS satellites thus complete two orbits per sidereal day, ensuring consistent ground coverage.

Summary

This chapter has covered the fundamental concepts of gravitation: Newton's law, field strength, potential, variation of g, centre of mass/gravity, satellite dynamics, escape velocity, energy considerations, geostationary and GPS satellites. Each topic was supported by definitions, mathematical expressions, and worked numerical problems to reinforce understanding and prepare learners for quantitative applications in physics and engineering.