Gravitation and Circular Motion

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, moving through the energy ideas it creates, and finishing with the circular motion that explains why the Moon stays up. 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, and describe the conditions under which an extended body can be treated as a point mass
define gravitational field strength, sketch gravitational field lines, and combine fields from more than one mass
derive and use the equation for gravitational potential energy, including the convention that it is zero at infinite separation and negative everywhere else
define gravitational potential, relate it to field strength through the potential gradient, and describe equipotential surfaces
describe circular motion in terms of angular velocity, centripetal acceleration and centripetal force
state Kepler's three laws of orbital motion and derive Kepler's third law from Newton's law of gravitation
calculate orbital and escape speeds, and describe qualitatively how a small drag force changes a satellite's height and speed

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.

Answer:
F = Gm₁m₂ / r²
F = (6.67 × 10⁻¹¹ × 2.4 × 10¹² × 6.0 × 10¹²) ÷ 850²

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

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 test mass placed at that point.

gravitational field strength, g = F / m

Test mass: a 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.

Combining gravitational fields

Gravitational field strength is a vector quantity. If a location is within the fields of two or more masses, the resultant field is found by adding the individual field vectors — remembering that each points towards its own mass.

Worked example 2.1

At a point P between the Earth and the Moon, on the line joining their centres, the gravitational field strength due to the Earth is 4.20 × 10⁻³ N kg⁻¹ (towards the Earth) and due to the Moon is 1.10 × 10⁻³ N kg⁻¹ (towards the Moon). Calculate the resultant gravitational field strength at P.

Answer:
Taking the direction towards the Earth as positive:
resultant g = (+4.20 × 10⁻³) + (−1.10 × 10⁻³)

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

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.

7Two stars, A and B, are far apart. At a certain point Q, the field due to star A is 6.0 × 10⁻⁶ N kg⁻¹ towards A, and the field due to star B is 2.0 × 10⁻⁶ N kg⁻¹, also towards A (Q lies beyond B, so both fields point the same way). Calculate the resultant field strength at Q.

Since both fields point in the same direction, they add directly: resultant g = 6.0 × 10⁻⁶ + 2.0 × 10⁻⁶

3. Gravitational potential energy

You have previously used ΔE_p = mgΔh to calculate changes in gravitational potential energy near the Earth's surface. That equation only works because, over the small heights involved, g barely changes — it can be treated as a constant. Once we consider distances comparable to a planet's radius (an orbit, a journey to the Moon, an escaping rocket), g is no longer constant, and we need a more general definition.

Key idea. The gravitational potential energy of a system is defined as the work done to assemble the system — that is, to bring its components together — from an infinite separation.

Why choose infinity? Two masses only stop attracting each other completely when they are infinitely far apart. Choosing this as our reference point means every real, finite separation has a well-defined gravitational potential energy relative to the same, universally agreed zero.

From work done to an equation

We know that work done = force × distance moved (in the direction of the force). If the gravitational force were constant, this would be simple. But as a mass m is brought in from infinity towards a mass M, the force grows continuously, following the inverse square law — so we cannot just multiply one force value by the whole distance. Instead, the total work done is the sum of many small contributions, each with the force appropriate to that separation. Graphically, this total is the area under a force–distance graph.

Live simulation: gravitational potential energy as an area

The graph shows how the force on a 1 kg test mass varies as it is brought in from a large distance towards a planet's surface (R). Drag the slider to bring the test mass closer, and watch the shaded area — the work done, and so the magnitude of E_p — grow towards the value predicted by the equation.

Building E_p from the area under F–r

Current distance, r = 8.0 R

work done so far ≈ 0 J
|E_p| from formula = 12.5 J

Fig. 3.1 The shaded area under the force–distance graph, from the current position out towards infinity, is the work that would be needed to remove the mass to infinity — the magnitude of its gravitational potential energy at that point.

The calculus version

Summing infinitely many, infinitesimally small contributions of force × distance is exactly what integration does. Although you are not required to reproduce this derivation, it is worth seeing briefly, because it shows precisely where the equation for E_p comes from.

The work done by the gravitational force as a mass m moves from a separation R out to infinity is:

W = ∫_R^∞ (GMm / r²) dr = [ GMm / r ]_R^∞ = 0 − (GMm / R) = −GMm / R

Gravity is attractive, so it does negative work as the mass moves away — it is pulling the mass back in, not pushing it out. The work done by a force such as this is always equal to minus the change in potential energy, and since E_p is defined to be zero at infinity, this negative work done by gravity turns out to be exactly equal to the gravitational potential energy stored at R:

gravitational potential energy, E_p = −Gm₁m₂ / r

Key idea. E_p is always negative (or zero, only at r = infinity). As r increases, E_p increases too — becoming a smaller negative number — approaching, but never quite reaching, zero. As r decreases, E_p becomes an increasingly large negative number.

Worked example 3.1

Calculate the gravitational potential energy of a 1200 kg spacecraft at a distance of 8.0 × 10⁶ m from the centre of a planet of mass 5.5 × 10²⁴ kg.

Answer:
E_p = −Gm₁m₂/r
E_p = −(6.67 × 10⁻¹¹ × 5.5 × 10²⁴ × 1200) ÷ (8.0 × 10⁶)

Check your understanding

8Calculate the gravitational potential energy of the Earth–Moon system. (Mass of Earth = 6.0 × 10²⁴ kg, mass of Moon = 7.3 × 10²² kg, separation = 3.8 × 10⁸ m.)

E_p = −Gm₁m₂/r = −(6.67 × 10⁻¹¹ × 6.0 × 10²⁴ × 7.3 × 10²²) ÷ (3.8 × 10⁸)

9Explain, in terms of the definition of gravitational potential energy, why E_p can never be positive for two masses at a finite separation.

E_p is defined as the work done to assemble the system from infinite separation, where E_p = 0. Because gravity is attractive, bringing the masses together from infinity requires no energy input at all — if anything, energy would be released. So E_p can only stay the same (at infinity) or decrease below zero as the masses are brought closer, meaning it is always negative or zero, never positive.

10A satellite moves from a distance of 2R to a distance of 4R from a planet's centre, where R is the planet's radius. State, with a reason, whether its gravitational potential energy increases or decreases, and whether it becomes more or less negative.

Since E_p = −GMm/r, increasing r makes the magnitude of E_p smaller, so E_p becomes less negative — that is, E_p increases (moves closer to zero) as the satellite moves further away.

11Explain why ΔE_p = mgΔh would give a very inaccurate answer for the change in gravitational potential energy of a rocket travelling from the Earth's surface into a high orbit, but works well for a book lifted from the floor onto a table.

ΔE_p = mgΔh assumes g is constant over the height change involved. For a book on a table, the height change is tiny compared with the Earth's radius, so g barely changes and the approximation is excellent. For a rocket reaching orbit, the height change is comparable to (or larger than) the Earth's radius, over which g falls significantly — so a constant-g formula badly overestimates or underestimates the true change, and the full E_p = −GMm/r equation must be used instead.

4. Gravitational potential

Gravitational potential energy, E_p, depends on which two masses you are considering — it belongs to a specific pair. Just as we generalised force (which depends on a specific test mass) into field strength (which doesn't), we can generalise E_p into a quantity that describes the field itself, independent of whatever test mass happens to be placed in it.

Key idea. Gravitational potential, V_g, at a point is the gravitational potential energy per unit mass at that point: V_g = E_p / m. Like E_p, it is defined to be zero at infinity, so V_g is always negative (or zero) and approaches zero as r → ∞.

Substituting E_p = −GMm/r for a test mass m in the field of a mass M, and dividing by m, gives:

gravitational potential, V_g = −GM / r

E_p vs V_g. Think of E_p as the total price of a shopping trolley, and V_g as the price per kilogram on the shelf label. The price-per-kilogram (V_g) is a property of the shop shelf itself — it doesn't change depending on how much you buy. The total price (E_p) depends on both the price per kilogram and how much mass you actually put in the trolley: E_p = mV_g. Units: E_p is in joules (J); V_g is in joules per kilogram (J kg⁻¹).

Potential difference and work done

If a mass m moves between two points with different gravitational potentials, work is done on or against the gravitational field. The gravitational potential difference, ΔV_g, between two points is the work done per unit mass in moving between them, so:

work done, W = mΔV_g

Moving a mass to a point of higher potential (further from M, since V_g is negative and increases towards zero) requires positive work done on the mass, against the attractive pull of gravity — exactly as raising a book against Earth's gravity requires work.

Field strength as a potential gradient

Field strength and potential are two ways of describing the same field, so they must be linked. The field strength at a point equals the negative of the rate at which potential changes with distance:

g = −ΔV_g / Δr

The minus sign matters: V_g becomes less negative (increases) as r increases, but g points inward, towards decreasing r. A steep potential gradient (V_g changing rapidly with distance) corresponds to a strong field; where V_g is nearly flat, the field is weak.

Worked example 4.1

Near a certain moon, the gravitational potential is −1.62 × 10⁵ J kg⁻¹ at a distance of 1.20 × 10⁶ m from its centre, and −1.44 × 10⁵ J kg⁻¹ at 1.35 × 10⁶ m. Estimate the field strength in this region.

Working:

g ≈ −ΔV_g / Δr = −[(−1.44 × 10⁵) − (−1.62 × 10⁵)] / [(1.35 × 10⁶) − (1.20 × 10⁶)]

g ≈ −(1.8 × 10⁴) / (1.5 × 10⁵) ... (complete the division to find g, in N kg⁻¹)

Equipotential surfaces

An equipotential surface joins all the points around a mass that share the same gravitational potential. Around a single point (or spherical) mass, these surfaces are concentric spheres — shown here in cross-section as concentric circles.

Fig. 4.1 Equipotential surfaces (dashed circles, each at a constant V_g) are always perpendicular to field lines (solid arrows), shown here by the small right-angle marker where a field line crosses one of the equipotentials.

Key idea. Equipotential surfaces are always perpendicular to field lines, and no work is done moving a mass along an equipotential surface (since V_g — and so E_p — does not change). Field lines point from regions of higher potential towards regions of lower (more negative) potential.

12Calculate the gravitational potential at the surface of Mars, given its mass is 6.42 × 10²³ kg and its radius is 3.39 × 10⁶ m.

V_g = −GM/r = −(6.67 × 10⁻¹¹ × 6.42 × 10²³) / (3.39 × 10⁶) ... (complete the calculation, in J kg⁻¹)

13A 500 kg probe moves from a point where V_g = −2.0 × 10⁶ J kg⁻¹ to a point where V_g = −1.4 × 10⁶ J kg⁻¹. Calculate the work done on the probe by an external engine, assuming it moves at constant speed.

W = mΔV_g = 500 × [(−1.4 × 10⁶) − (−2.0 × 10⁶)] ... (complete the calculation, in J)

14Explain why no work is done in moving a satellite along a single equipotential surface, even though a gravitational force acts on it throughout the motion.

Work done depends on ΔV_g (W = mΔV_g). Every point on an equipotential surface shares the same V_g by definition, so ΔV_g = 0 between any two points on it, and therefore W = 0 — even though a force acts on the satellite throughout, that force is always perpendicular to the direction of motion along the surface, so it does no work.

5. 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. 5.1 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:

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. 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. 5.2 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 5.1

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)

15A 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.

16A 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)

17Explain 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.

6. 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. 6.1 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 6.1

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)

18State, 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.

19Two 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 ... (complete the calculation)

20Explain 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.

7. Orbital speed, escape speed and drag

We now bring circular motion and gravitational fields back together. 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 it must travel to stay in a given orbit — and, taking the idea further, how fast an object must travel to escape a planet's gravity altogether.

Orbital speed

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)

Escape speed

To escape a planet's gravity entirely means reaching r = ∞ with (at minimum) zero speed remaining — that is, total mechanical energy of exactly zero. Setting kinetic energy at launch equal in magnitude to the (negative) gravitational potential energy holding the object back:

½mv_esc² = GMm / r

escape speed, v_esc = √(2GM / r)

Key idea. Comparing the two results shows v_esc = √2 × v_orb at the same radius — escape speed is always √2 (≈ 1.41) times the speed needed for a circular orbit at that same distance, regardless of the planet.

The energy of a satellite in orbit

A satellite in a stable circular orbit has kinetic energy E_k = ½mv_orb² = GMm/2r, and gravitational potential energy E_p = −GMm/r. Its total mechanical energy is therefore:

E_T = E_k + E_p = GMm/2r − GMm/r = −GMm / 2r

Notice that E_T = −E_k = ½E_p: the total energy is always negative for a bound orbit (confirming the satellite cannot escape without extra energy being added), and exactly half the magnitude of the potential energy.

Fig. 7.1 Kinetic energy (teal), gravitational potential energy (navy) and total mechanical energy (amber, dashed) of a satellite, plotted against orbital radius r (schematic, not to scale). Note E_T always sits exactly halfway between E_p and zero.

Live simulation: launch speed and orbit type

Fire a projectile horizontally from a tall tower on an airless planet (Newton's classic "cannonball" thought experiment) and see how its path depends on launch speed. Then switch on atmospheric drag and see what happens to a satellite already in orbit.

Cannonball & drag explorer

Launch speed, v = 7.0 km s−1

Enable atmospheric drag

v_orb ≈ 7.9 km s−1 | v_esc ≈ 11.2 km s−1

Fig. 7.2 Below orbital speed the path falls back to the surface; at orbital speed it closes into a circle; between orbital and escape speed it becomes an ellipse; at or above escape speed it never returns. With drag enabled, a satellite spirals slowly inward, paradoxically speeding up as it loses height and total energy.

Key idea — the effect of drag. A satellite experiencing a small amount of atmospheric drag loses total mechanical energy over time, so its orbital radius gradually decreases. Because E_T = −GMm/2r, a smaller r actually corresponds to a more negative (lower) total energy, but a larger kinetic energy and orbital speed (v_orb = √(GM/r) increases as r decreases) — so drag causes a satellite to spiral inward while speeding up, even as it continuously loses energy to friction. This continues until the satellite reaches the denser lower atmosphere, where drag increases sharply and it re-enters.

Worked example 7.1

Calculate the orbital speed and escape speed for a satellite at the Earth's surface (radius 6.37 × 10⁶ m, mass 5.97 × 10²⁴ kg), ignoring atmospheric drag.

Working:

v_orb = √(GM/r) = √[(6.67 × 10⁻¹¹ × 5.97 × 10²⁴) / (6.37 × 10⁶)] ... (complete the calculation, in m s⁻¹)

v_esc = √2 × v_orb ... (use your answer above to complete this, in m s⁻¹)

21A 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)

22Calculate the escape speed from the Moon's surface (mass 7.35 × 10²² kg, radius 1.74 × 10⁶ m), and suggest why the Moon has essentially no atmosphere.

v_esc = √(2GM/r) = √[(2 × 6.67 × 10⁻¹¹ × 7.35 × 10²²) / (1.74 × 10⁶)] ... (complete the calculation, in m s⁻¹). This escape speed is low enough that fast-moving gas molecules in a would-be atmosphere regularly exceed it and are lost to space over geological time, which is why the Moon has essentially no atmosphere today.

23A satellite in low orbit experiences a small amount of drag. Explain why its orbital speed increases over time, even though drag is removing energy from the system.

Drag removes total mechanical energy, making E_T = −GMm/2r more negative, which corresponds to a smaller orbital radius r. Since v_orb = √(GM/r), a smaller r means a larger v_orb — so the satellite's orbital speed actually increases, even as its total energy decreases, because the drop in (increasingly negative) potential energy outweighs the modest energy lost to drag.

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 test 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).
Gravitational potential energy, E_p: The work done to assemble a system of masses from an infinite separation; E_p = −Gm₁m₂/r, always negative, zero at infinity.
Gravitational potential, V_g: The gravitational potential energy per unit mass at a point; V_g = −GM/r, units J kg⁻¹.
Equipotential surface: A surface joining points of equal gravitational potential; always perpendicular to field lines, with zero work done moving along it.
Centripetal force: The resultant of the real force(s) acting on an object moving in a circle, directed towards the centre; 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.
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).
Escape speed: The minimum speed needed to escape a gravitational field completely (reach infinity with zero remaining speed); v_esc = √(2GM/r) = √2 × v_orb.

Gravitation and Circular Motion

1. Newton's law of gravitation

Treating extended bodies as point masses

Check your understanding

2. Gravitational field strength and field lines

Field lines

Live simulation: the inverse-square field

Field strength explorer

Combining gravitational fields

Check your understanding

3. Gravitational potential energy

From work done to an equation

Live simulation: gravitational potential energy as an area

Building Ep from the area under F–r

The calculus version

Check your understanding

4. Gravitational potential

Potential difference and work done

Field strength as a potential gradient

Equipotential surfaces

5. Circular motion

Angular velocity

Centripetal acceleration and force

Live simulation: circular motion vectors

Circular motion explorer

6. Kepler's laws of orbital motion

Live simulation: exploring Kepler's laws

Orbital eccentricity

Deriving Kepler's third law

7. Orbital speed, escape speed and drag

Orbital speed

Escape speed

The energy of a satellite in orbit

Live simulation: launch speed and orbit type

Cannonball & drag explorer

Glossary

Building E_p from the area under F–r