Physics · Lesson 11

Gravitational field

JAMBWAECNECO

What the syllabus asks

JAMB: gravitational field, Newton’s law of gravitation, g = GM/R², escape velocity, satellites

WAEC: gravitational field; Newton’s law of universal gravitation; gravitational constant G; gravitational field strength g and its relation to G; variation of g with altitude, depth and latitude; gravitational potential and potential energy; escape velocity; satellites and orbits (orbital speed, period, geostationary satellites, weightlessness); Kepler’s laws (qualitative)

NECO: gravitational field; Newton’s law of universal gravitation; gravitational constant G; gravitational field strength g and its relation to G; variation of g with altitude, depth and latitude; gravitational potential and potential energy; escape velocity; satellites and orbits (orbital speed, period, geostationary satellites, weightlessness); Kepler’s laws (qualitative)

What you will learn

State Newton’s law of universal gravitation and use F = Gm₁m₂/r².
Define gravitational field strength and derive g = GM/R².
Explain how g varies with height, depth and latitude, and calculate g at a height.
Define gravitational potential and calculate potential energy changes.
Derive and use the escape velocity formula.
Calculate orbital speed and period of satellites and describe geostationary satellites and their uses.
Explain weightlessness and state Kepler’s laws.

Key points

F = Gm₁m₂/r²; G = 6.67 × 10⁻¹¹ N m² kg⁻²; always attractive; equal and opposite.
Field strength g = F/m (N/kg = m/s²); g = GM/R² at the surface.
Above the surface g ∝ 1/r²; below, g decreases linearly to zero at the centre.
g is larger at the poles than at the equator (shape and rotation).
Potential V = −GM/r (zero at infinity); PE = −GMm/r; use mgh only for small heights.
Escape velocity √(2GM/R) = √(2gR) ≈ 11.2 km/s, independent of the body’s mass.
Orbital speed v = √(GM/r) ≈ 7.9 km/s near Earth; T = 2πr/v; T² ∝ r³.
Geostationary: above the equator, 24 h, west to east, about 36,000 km up; used for TV and communications.
Weightlessness is free fall (zero reaction), not zero gravity.
Kepler: elliptical orbits, equal areas in equal times, T² ∝ r³.

Start here: Newton’s law of universal gravitation

An orange falls from a tree in your compound; the Moon circles the Earth; the Earth circles the Sun. Isaac Newton realised that the same force explains all three: gravity. Every particle of matter in the universe attracts every other particle.

Newton’s law of universal gravitation: the force of attraction between two point masses is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centres.
F = G m₁m₂ / r², where G = 6.67 × 10⁻¹¹ N m² kg⁻² is the universal gravitational constant.

Key features:

The force is always attractive.
The forces on the two bodies are equal and opposite (Newton’s third law): the Earth pulls an apple with the same size force as the apple pulls the Earth. The apple moves visibly because its mass is tiny.
It is an inverse square law: double the distance → force becomes ¼; triple → 1/9.
For spheres such as planets, r is measured between their centres.
G is extremely small, so gravity is noticeable only when at least one mass is huge.

Worked example 11.1

Two students, each of mass 50 kg, sit 1.0 m apart. Find the gravitational force between them.

F = 6.67 × 10⁻¹¹ × 50 × 50/1.0² = 6.67 × 10⁻¹¹ × 2500.
F ≈ 1.67 × 10⁻⁷ N, far too small to notice.

≈ 1.7 × 10⁻⁷ N

Worked example 11.2

Find the force between the Earth (6.0 × 10²⁴ kg) and the Moon (7.4 × 10²² kg) if their centres are 3.84 × 10⁸ m apart.

F = 6.67 × 10⁻¹¹ × 6.0 × 10²⁴ × 7.4 × 10²²/(3.84 × 10⁸)².
Numerator ≈ 2.96 × 10³⁷; denominator ≈ 1.47 × 10¹⁷.
F ≈ 2.0 × 10²⁰ N. This huge force keeps the Moon in orbit and raises the ocean tides.

≈ 2.0 × 10²⁰ N

How was the law discovered? Newton compared the falling apple with the Moon. The Moon is about 60 Earth-radii from the Earth’s centre. If gravity weakens as 1/r², the Moon’s acceleration towards the Earth should be g/60² ≈ 9.8/3600 ≈ 0.0027 m/s². Using the Moon’s known orbit (radius 3.84 × 10⁸ m, period 27.3 days), its centripetal acceleration v²/r works out to almost exactly this value. The same force that drops an orange in Ibadan holds the Moon in its path: gravity is universal.

⚠️ Common mistake: forgetting to square r, or using distance from the surface instead of from the centre.

Try it yourself. If the distance between two masses is halved, what happens to the force?

Show solution

F ∝ 1/r²: (1/(½)²) = 4 times larger.

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Practice questions

1. Find the force between two 1000 kg masses 2 m apart.

Show worked answer

6.67 × 10⁻¹¹ × 10⁶/4 ≈ 1.7 × 10⁻⁵ N.

2. A planet has mass 3.0 × 10²³ kg and radius 2.0 × 10⁶ m. Find g at its surface.

Show worked answer

6.67 × 10⁻¹¹ × 3.0 × 10²³/4.0 × 10¹² ≈ 5.0 N/kg.

3. Find g at a height of 3R above the Earth’s surface (g₀ = 9.8).

Show worked answer

r = 4R: 9.8/16 ≈ 0.61 N/kg.

4. Find the escape velocity from a planet with g = 4 m/s² and R = 3.6 × 10⁶ m.

Show worked answer

√(2 × 4 × 3.6 × 10⁶) = ≈ 5.4 km/s.

5. State three conditions for a geostationary orbit.

Show worked answer

Above the equator; period 24 h; same direction as Earth’s rotation.

Key words

Gravitational field: A region where a mass experiences a gravitational force.
Universal gravitational constant G: 6.67 × 10⁻¹¹ N m² kg⁻²; the same everywhere.
Field strength g: Gravitational force per unit mass; unit N/kg (equal to m/s²).
Inverse square law: A quantity that varies as 1/r²; doubling r gives a quarter.
Weight: The gravitational force on a body, W = mg.
Gravitational potential V: Work done per unit mass in bringing a small mass from infinity to the point; J/kg.
Gravitational potential energy: Energy of a mass due to its position in a field, −GMm/r.
Escape velocity: The minimum launch speed needed to escape a planet’s gravity completely.
Satellite: A body orbiting a larger body.
Orbital speed: The speed needed to stay in a circular orbit at a given radius.
Period: The time for one complete orbit.
Geostationary satellite: A satellite above the equator with a 24-hour period, appearing fixed in the sky.

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