Gravitation and Circular Motion (SL)

Every planet, moon, star and satellite is held on its path by the same force: gravity, acting between any two masses in the universe. This workbook builds up the physics of that force from scratch — starting with the force itself and the field it creates, then turning to the mathematics of circular motion that explains why a satellite orbits rather than falls straight down. Space turns out to be the cleanest possible laboratory for circular motion, since gravity is often the only force acting.

By the end of this workbook you should be able to:

state and apply Newton's law of gravitation for point masses, and describe the conditions under which an extended body can be treated as a point mass
define gravitational field strength as the force per unit mass on a small point mass, and sketch gravitational field lines
describe circular motion in terms of angular velocity, centripetal acceleration and centripetal force, and explain why a centripetal force changes an object's direction even when its speed stays constant
state Kepler's three laws of orbital motion, and derive Kepler's third law, T² ∝ r³, from Newton's law of gravitation and circular motion
derive and use the equation for orbital speed, v_orb = √(GM / r)

Name

1. Newton's law of gravitation

Every object with mass attracts every other object with mass. This might seem surprising — you are not aware of being pulled towards your desk, your chair, or the person sitting next to you — but the attraction is there. It is simply too weak to notice unless at least one of the masses involved is enormous, such as a planet.

Isaac Newton's insight, in the seventeenth century, was that the force pulling an apple to the ground and the force keeping the Moon in orbit around the Earth are exactly the same force. He proposed that every pair of masses in the universe attracts every other pair — this is why the law is called universal gravitation.

Key idea. The gravitational force between two point masses is proportional to the product of the masses, and inversely proportional to the square of the distance between them. It is always attractive — gravity only ever pulls, never pushes.

gravitational force, F = Gm₁m₂ / r²

G, the gravitational constant: the constant of proportionality in Newton's law of gravitation, G = 6.67 × 10⁻¹¹ N m² kg⁻². It is often called "Big G" to distinguish it from g, the gravitational field strength. G is believed to have the same value everywhere in the universe and at all times.

Because G is so small, the gravitational force between two ordinary-sized objects — two people, or a person and a building — is far too small to detect. It only becomes significant when at least one mass is planet-sized or larger.

Fig. 1.1 The gravitational force between two point masses falls away with the square of their separation: doubling the separation cuts the force to a quarter, and quadrupling it cuts the force to a sixteenth (vectors not drawn to scale). The same size force acts on both masses, in opposite directions — an example of Newton's third law.

Treating extended bodies as point masses

Newton's law is stated for point masses — masses concentrated at a single location. Real objects, such as planets, are not points; they are extended spheres of matter. Fortunately, Newton's shell theorem shows that a spherically symmetric object (uniform density, or made of uniform concentric shells) attracts other masses exactly as if all of its mass were concentrated at its centre — provided the other mass is entirely outside it.

Conditions for the point-mass approximation: an extended body can be treated as a point mass, with all of its mass acting at its centre, when it is spherically symmetric (or approximately so, like a planet or star) and the distance to the other mass is measured from its centre, to a location outside the body itself.

Worked example 1.1

Two asteroids, of mass 2.4 × 10¹² kg and 6.0 × 10¹² kg, have their centres 850 m apart. Calculate the gravitational force of attraction between them.

Working:
F = Gm₁m₂ / r²
F = (6.67 × 10⁻¹¹ × 2.4 × 10¹² × 6.0 × 10¹²) ÷ 850² ... (complete the calculation, in N)

Check your understanding

1The Earth has a mass of 6.0 × 10²⁴ kg and the Sun has a mass of 2.0 × 10³⁰ kg. Their centres are 1.5 × 10¹¹ m apart. Calculate the gravitational force of attraction between them.

F = Gm₁m₂/r² = (6.67 × 10⁻¹¹ × 6.0 × 10²⁴ × 2.0 × 10³⁰) ÷ (1.5 × 10¹¹)² ≈ 3.6 × 10²² N.

2Estimate the gravitational force of attraction between two students of mass 60 kg, sitting 1.0 m apart, then use your result to explain why we never notice the gravitational pull of nearby objects.

F = Gm₁m₂/r² = (6.67 × 10⁻¹¹ × 60 × 60) ÷ 1.0² = 2.4 × 10⁻⁷ N. This is roughly ten million times smaller than the weight of a single grain of sand, so it is completely undetectable — G is so small that gravitational forces only become significant when at least one mass is planet-sized or larger.

3The Moon is not a perfect sphere of uniform density. State the condition under which it is still reasonable to use Newton's law of gravitation with the Moon treated as a point mass, and identify what distance should be used for r in that case.

Provided the Moon is approximately spherically symmetric (its density varies mainly with depth rather than direction) and the other mass lies outside the Moon's surface, the shell theorem lets us treat it as a point mass with all its mass acting at its centre. The distance r used in the equation must then be measured from the Moon's centre, not from its surface.

2. Gravitational field strength and field lines

Newton's law tells us the force between two specific masses. But it is often more useful to describe the space around a mass in general — a region where any other mass would feel a force is called a gravitational field. Rather than asking "what force would a 5 kg rock feel here?" and then a different question for a 50 kg rock, we describe the field once, per unit mass, and can then find the force on any mass we like.

Key idea. Gravitational field strength, g, at a point is defined as the gravitational force per unit mass experienced by a small point mass placed at that point.

gravitational field strength, g = F / m

Test mass: a small point mass, small enough that its own gravitational field has a negligible effect on the field being measured. Gravitational field strength has SI unit N kg⁻¹, which is equivalent to m s⁻² — numerically, g is the same as the acceleration due to gravity, since a = F/m by Newton's second law.

Combining g = F/m with Newton's law of gravitation, F = GMm/r², gives an equation for the field strength around any point mass (or spherically symmetric mass) M:

g = GM / r²

Like gravitational force, gravitational field strength follows an inverse square law: double the distance from the centre of a planet and the field strength falls to a quarter; treble it and the field strength falls to a ninth.

Field lines

A gravitational field can be drawn as a pattern of field lines. Each line shows the direction of the force that a test mass would feel if placed on it — and because gravity is always attractive, field lines always point towards the mass creating the field. Field lines are closer together where the field is stronger, and they never cross (a test mass cannot feel a force in two different directions at the same point).

Fig. 2.1 Field lines point towards the mass creating the field, since gravity is attractive. a A radial field, such as around a planet — lines get closer together nearer the surface, showing the field is stronger there. b A uniform field, such as in a small region close to a planet's surface, where the field lines are evenly spaced and parallel.

Live simulation: the inverse-square field

Drag the slider to move a test mass away from a planet's surface, and watch how both the field lines and the g–r graph respond. The planet shown has a surface field strength g₀ = 9.8 N kg⁻¹.

Field strength explorer

Distance from centre, r = 1.0 R

g = 9.80 N kg−1

Fig. 2.2 As the test mass moves further from the planet's centre, the field lines it crosses become more widely spaced and its field-strength arrow shrinks — both showing the same inverse square relationship traced out on the graph.

Worked example 2.1

Calculate the gravitational field strength at the surface of Mars, given its mass is 6.42 × 10²³ kg and its radius is 3.39 × 10⁶ m.

Working:
g = GM / r²
g = (6.67 × 10⁻¹¹ × 6.42 × 10²³) ÷ (3.39 × 10⁶)² ... (complete the calculation, in N kg⁻¹)

Check your understanding

4A planet has a mass of 4.9 × 10²⁴ kg. Calculate the gravitational field strength at a point 8.2 × 10⁶ m from its centre.

g = GM/r² = (6.67 × 10⁻¹¹ × 4.9 × 10²⁴) ÷ (8.2 × 10⁶)² ≈ 4.9 N kg⁻¹.

5State two rules that gravitational field lines must always obey, and explain, in terms of field lines, what it means for a field to be "uniform".

Field lines never cross, and they are always closer together where the field is stronger. A field is uniform where its field lines are evenly spaced and parallel, so that the field strength has the same magnitude and direction at every point in that region.

6A gravitational field strength of 5.0 N kg⁻¹ is measured at the surface of a small moon. Determine the gravitational field strength at a distance of three moon-radii from its centre.

Field strength follows an inverse square law with distance from the centre, so at three times the radius the field strength falls to 1/3² = 1/9 of its surface value: g = 5.0 ÷ 9 = 0.56 N kg⁻¹.

3. Circular motion

Space is the perfect place to study circular motion. A satellite coasting around the Earth has no engine firing, no air resistance, no surface to push against — gravity is the only force acting on it, and yet it moves in a curved path rather than a straight line. To understand why, we first need to describe circular motion itself, before returning to gravity specifically in the sections that follow.

Angular velocity

For an object moving in a circle of radius r, the period, T, is the time taken for one complete revolution, and the frequency, f, is the number of revolutions per second (f = 1/T). Rather than tracking the object's position with distances, it is often more convenient to track the angle it has swept through, measured in radians. The rate of change of this angle is the angular velocity, ω:

angular velocity, ω = Δθ / Δt = 2π / T = 2πf

Angular velocity is measured in rad s⁻¹. Because the object travels a distance of one circumference, 2πr, in one period T, its (tangential) speed is v = 2πr / T, which combines with the equation above to give a direct link between the two descriptions of speed:

tangential velocity, v = ωr

Centripetal acceleration and force

An object moving at constant speed around a circle is still accelerating, because its velocity — a vector — is continuously changing direction, even though its magnitude stays fixed. This acceleration points towards the centre of the circle at every instant, which is why it is called centripetal ("centre-seeking") acceleration.

Fig. 3.1 At any point on a circular path, the velocity v is tangential to the circle, while the centripetal acceleration a points radially inward, towards the centre — the two are always perpendicular.

Combining v = ωr with the standard relationship between changing velocity direction and acceleration gives two equivalent forms:

centripetal acceleration, a = v² / r = ω²r

By Newton's second law, a resultant force must cause this acceleration. This resultant is called the centripetal force — not a new, separate kind of force, but simply whatever combination of real forces (tension, gravity, friction, the normal force, and so on) happens to supply the net inward force needed. Because it is always directed perpendicular to the object's velocity, a centripetal force can never speed the object up or slow it down — it only ever changes the direction of the velocity, which is exactly why the motion stays circular rather than straight.

centripetal force, F = mv² / r = mω²r

Key idea. "Centripetal force" is not a new physical force acting alongside gravity, tension, friction, etc. — it is the name given to the resultant of whatever real forces are acting, when that resultant points towards the centre of a circular path. Because this resultant is always perpendicular to the velocity, it changes the object's direction, not the magnitude of its speed. Always identify the actual, physical force(s) providing it before using F = mv²/r.

Live simulation: circular motion vectors

Adjust the radius and the angular velocity of the orbiting mass and watch how the tangential velocity vector and the centripetal force vector respond.

Circular motion explorer

Radius, r = 1.5 m

Angular velocity, ω = 2.0 rad s−1

v = 3.00 m s−1
a = 6.00 m s−2

Fig. 3.2 A mass (2.0 kg) moves at constant angular velocity around a circle of radius r; the blue arrow shows its tangential velocity, the amber arrow its centripetal acceleration.

Worked example 3.1

A stone of mass 60 g is tied to a string and whirled in a horizontal circle of radius 50 cm. The string snaps when the tension exceeds 14 N. Calculate the maximum speed at which the stone can be whirled without the string snapping.

Working:

The tension provides the centripetal force, so at the maximum speed: F = mv²/r ⇒ 14 = (0.060)v² / (0.50)

v² = 14 × 0.50 / 0.060 ... (complete the rearrangement and take the square root to find v, in m s⁻¹)

Check your understanding

7A car of mass 1200 kg goes around a bend of radius 25 m at a constant 11.1 m s⁻¹. Calculate the magnitude of the centripetal force acting on it, and state which real force provides it.

F = mv²/r = 1200 × (11.1)² / 25 ... (complete the calculation, in N). This force is provided by friction between the tyres and the road.

8A satellite completes one orbit of the Earth every 92 minutes at an angular velocity ω. Calculate ω in rad s⁻¹.

ω = 2π/T = 2π / (92 × 60) ... (complete the calculation, in rad s⁻¹)

9Explain why a passenger in a car going around a bend at constant speed still feels a sideways force from their seatbelt, even though their speed is not changing.

Speed being constant does not mean velocity is constant, because velocity is a vector and its direction is continuously changing as the car turns. This changing direction is itself an acceleration (centripetal acceleration, directed towards the centre of the turn). By Newton's second law this requires a resultant force towards the centre; the seatbelt (along with friction from the seat) provides part of this force on the passenger, which is felt as a sideways push. Because this force is always perpendicular to the passenger's velocity, it changes their direction but not their speed.

4. Kepler's laws of orbital motion

Long before Newton explained why planets move as they do, the astronomer Johannes Kepler worked out, from decades of careful naked-eye observations by Tycho Brahe, exactly how they move. His three laws, published in the early 1600s, describe the motion of every planet, moon, and satellite — and, as you will see, they follow directly from Newton's law of gravitation and circular motion, applied to an orbit.

Kepler's three laws.
1. The orbit of a planet is an ellipse, with the Sun at one focus.
2. A line joining a planet to the Sun sweeps out equal areas in equal times.
3. The square of a planet's orbital period is proportional to the cube of its orbital radius (semi-major axis): T² ∝ r³.

Live simulation: exploring Kepler's laws

This simulation (inspired by the interactive Kepler's laws tool by Dr Jones Physics) has three tabs, one for each law. Use the controls in each tab to see the law in action.

Orbital eccentricity

Eccentricity, e = 0.00

Circular orbit (e = 0): the Sun sits at the centre.

Fig. 4.1 Tab 1: dragging the eccentricity slider morphs the orbit from a circle towards a stretched ellipse, with the Sun fixed at one focus. Tab 2: equal-area sweeps in equal time intervals, shown shaded. Tab 3: orbital period versus radius for the simulated system, compared with the T² ∝ r³ prediction.

Simulation design inspired by the Kepler's Laws interactive by Dr Jones Physics (drjonesphysics.com).

Deriving Kepler's third law

For a planet of mass m in a (near-)circular orbit of radius r around a star of mass M, gravity provides the centripetal force. This derivation connects two ideas you already know — Newton's law of gravitation and circular motion — into a brand new result, so it is worth working through one step at a time.

Must learn This is a must learn derivation. Make sure you can reproduce every step below from Newton's law of gravitation and the centripetal force equation, rather than just quoting the final result.

Start with Newton's law of gravitation set equal to the centripetal force required for a circular orbit:

GMm / r² = mv² / r

Cancel the mass of the orbiting planet, m, from both sides:

GM / r² = v² / r

Replace v with the orbital speed, v = 2πr / T:

GM / r² = (2πr / T)² / r

Simplify and rearrange to find the final relationship:

T² = (4π² / GM) r³

Since 4π²/GM is constant for a given central mass M, this confirms T² ∝ r³ — and shows the constant of proportionality depends only on the mass being orbited, not on the orbiting mass or its speed.

Worked example 4.1

Io, one of Jupiter's moons, orbits at a mean radius of 4.22 × 10⁸ m with a period of 1.53 × 10⁵ s. Use this to estimate the mass of Jupiter.

Working:

T² = (4π² / GM) r³ ⇒ M = 4π² r³ / (G T²)

M = 4π² × (4.22 × 10⁸) ³ / [(6.67 × 10⁻¹¹) × (1.53 × 10⁵) ²] ... (complete the calculation, in kg)

Check your understanding

10State, in your own words, what Kepler's second law tells us about a planet's speed at different points in its elliptical orbit.

Because the planet sweeps out equal areas in equal times, it must move faster when it is closer to the Sun (where the swept triangle is short and wide) and slower when it is further away (where the swept triangle is long and narrow), so that both triangles have the same area in the same time interval.

11Two moons orbit the same planet. Moon A has an orbital radius 4 times greater than moon B. Calculate the ratio of their orbital periods, T_A / T_B.

T² ∝ r³, so T_A/T_B = (r_A/r_B)^3/2 = 4^3/2 = 8. Moon A takes 8 times longer to complete an orbit than moon B.

12Explain why Kepler's third law derivation above (using circular-orbit centripetal force) still gives the correct relationship between T and r for an elliptical orbit, provided r is taken as the semi-major axis.

A full derivation for an ellipse requires more advanced mathematics than the circular-orbit case, but it produces exactly the same result, T² = (4π²/GM)a³, with the orbital radius r replaced by the semi-major axis a. The circular-orbit derivation shown is a special case (e = 0) of this more general result, which is why it correctly predicts the constant of proportionality 4π²/GM.

5. Orbital speed

A satellite orbiting a planet is undergoing circular motion, and the only force providing the centripetal force is gravity. This lets us find exactly how fast a satellite must travel to stay in a given circular orbit — using nothing more than Newton's law of gravitation and the centripetal force equation from Section 3.

For a satellite of mass m in a stable circular orbit of radius r around a planet of mass M, gravity alone supplies the centripetal force:

GMm / r² = mv² / r

The satellite's mass m cancels — orbital speed does not depend on the mass of the orbiting object, only on the mass being orbited and the orbital radius:

orbital speed, v_orb = √(GM / r)

Key idea. A satellite in a smaller (lower) orbit must travel faster to stay in that orbit, since v_orb = √(GM/r) increases as r decreases. This is why low-orbiting satellites, such as the International Space Station, complete an orbit in around 90 minutes, while satellites in much higher orbits can take 24 hours or longer.

Live simulation: orbital speed and orbital radius

Drag the slider to place a satellite into a wider or narrower circular orbit, and watch how the orbital speed needed to stay in that orbit changes. The planet shown has a surface orbital speed v₀ = 7.9 km s⁻¹ (roughly Earth's value, ignoring the atmosphere).

Orbital speed explorer

Orbital radius, r = 1.0 R

v_orb = 7.90 km s−1

Fig. 5.1 As the satellite's orbital radius increases, both the length of its velocity arrow and the v_orb–r graph show the same inverse-square-root relationship, v_orb ∝ 1/√r.

Worked example 5.1

Calculate the orbital speed of a satellite orbiting the Earth (mass 5.97 × 10²⁴ kg) at an altitude of 400 km, so that its distance from the Earth's centre is 6.77 × 10⁶ m.

Working:

v_orb = √(GM/r) = √[(6.67 × 10⁻¹¹ × 5.97 × 10²⁴) / (6.77 × 10⁶)] ... (complete the calculation, in km s⁻¹)

Check your understanding

13A satellite orbits a planet of mass 4.9 × 10²⁴ kg at a radius of 8.0 × 10⁶ m. Calculate its orbital speed.

v_orb = √(GM/r) = √[(6.67 × 10⁻¹¹ × 4.9 × 10²⁴) / (8.0 × 10⁶)] ... (complete the calculation, in m s⁻¹)

14Two satellites, A and B, orbit the same planet in circular orbits. Satellite A has a smaller orbital radius than satellite B. Use v_orb = √(GM/r) to state and explain which satellite has the greater orbital speed.

Since v_orb = √(GM/r), for the same central mass M the orbital speed decreases as the orbital radius r increases. Satellite A has the smaller radius, so it has the greater orbital speed. Physically, satellite A experiences a stronger gravitational field strength (g = GM/r² is larger at smaller r), so a larger centripetal force per unit mass is available, requiring a greater speed to keep it moving in its (smaller) circular path rather than spiralling inward.

Glossary

Gravitational field: A region of space in which a mass experiences a force due to the presence of another mass.
Gravitational field strength, g: The gravitational force per unit mass at a point in a field; g = F/m = GM/r², units N kg⁻¹.
Field line: A line showing the direction of the gravitational force on a small point mass at each point; for an attractive field, lines point towards the mass creating the field.
Point mass: An approximation treating an object's entire mass as concentrated at a single point, valid outside a spherically symmetric body (measuring from its centre).
Centripetal force: The resultant of the real force(s) acting on an object moving in a circle, directed towards the centre and perpendicular to its velocity; F = mv²/r = mω²r.
Centripetal acceleration: The acceleration of an object moving in a circle at constant speed, directed towards the centre; a = v²/r = ω²r.
Angular velocity, ω: The rate of change of angle swept, in radians per second; ω = 2π/T = 2πf, related to linear speed by v = ωr.
Kepler's laws: Three empirical laws describing orbital motion: orbits are ellipses with the Sun/planet at one focus; equal areas are swept in equal times; T² ∝ r³.
Orbital speed: The speed needed to maintain a stable circular orbit at a given radius; v_orb = √(GM/r).