Chapter 1 · Mechanics

1-12 Kepler’s Laws and Universal Gravitation

❓ Lesson question

Why does one satellite appear fixed above Earth, and what determines the time a satellite takes to complete an orbit?

You will learn to

1Describe and apply Kepler’s laws of planetary motion.
2Explain how the orbital period depends on orbital radius for planets and satellites.
3Determine gravitational force and gravitational field strength.

Consider the night sky. Some satellites appear fixed relative to Earth's surface, while others move rapidly across the sky. Planets also follow regular orbits around the Sun, and their motion has been studied for centuries. These motions are not random. Gravity continuously accelerates each object toward the body it orbits, while its tangential velocity carries it forward. As a result, the object remains in continuous free fall, following a curved orbital path rather than falling directly toward the central body.

✍️ Predict before you continue

Write your answer before reading on. A prediction you got wrong and then examined teaches you more than an answer you were handed.

(a) What determines whether a satellite appears fixed relative to Earth’s surface or moves rapidly across the sky?

(b) Suppose a satellite’s instantaneous velocity is unchanged when the gravitational force changes. How would a stronger or weaker gravitational force affect its orbit?

Write a line for each before this opens.

🔬 Explore — Orbital motion investigation

What you’ll do: Analyse the orbital radii and periods of Jupiter’s four largest moons to identify a mathematical pattern.

⏱️ 12 min🧰 The provided table of orbital radius (r) and orbital period (T). · A calculator.

1Use the values of orbital radius, r, and orbital period, T, given for each moon.
2For each moon, calculate T², r³, and the ratio T²/r³.
3Record the calculated value of T²/r³ in the final column of the table.
4Compare the values of T²/r³ for all four moons.

Analyse what you collected. (a) Does the ratio remain approximately constant for the four moons, or does it vary significantly? (b) What property of the central planet determines this constant value, and which physical laws explain it?

🎛️ Simulation — orbits

Moon	r (×10⁸ m)	T (days)	T²	r³	T² / r³
Io	4.22	1.77	3.13	75.2	0.04169
Europa	6.71	3.55	12.60	302.1	0.04171
Ganymede	10.7	7.15	51.12	1225.0	0.04173
Callisto	18.8	16.7	278.89	6644.7	0.04197

The four values agree to within 0.68% of each other.

They are not identical, and they do not need to be — measurement is never exact. What matters is that four moons at radii differing by a factor of more than four share one number. The only thing those four moons have in common is Jupiter, so the constant must be a property of Jupiter and not of any moon. That is what T²/r³ = 4π²/GM says: the right-hand side contains no property of the orbiting body at all.

In Explore, you calculated $T^{2} / r^{3}$ for Jupiter's four largest moons. The similar values for the four moons reveal a common pattern. This nearly constant ratio indicates that the moons orbit the same central mass and is consistent with an inverse-square gravitational force.

Newton's law of universal gravitation

Every particle attracts every other along the line joining them; this attraction is the gravitational force. For two masses $m_{1}$ and $m_{2}$ a distance $r$ apart, it is proportional to the product of the masses and inversely proportional to the square of the distance:

F = \frac{G m _{1} m _{2}}{r ^{2}}

Here $G = 6.67 \times 1 0^{- 11}$ N·m²/kg² is the universal gravitational constant, the same for all bodies.

Building the model

A moon of mass $m$ orbits Jupiter (mass $M$ ) in a circle of radius $r$ at speed $v$ . The gravitational force between them, $GM m / r^{2}$ , provides the centripetal force that keeps the moon in its circular orbit:

\frac{GM m}{r ^{2}} = \frac{m v ^{2}}{r}

The moon's mass cancels, leaving its orbital speed:

v^{2} = \frac{GM}{r}

(1)

In one period the moon travels one circumference, so $v = 2 π r / T$ . Substituting into (1) and rearranging:

(\frac{2 π r}{T})^{2} = \frac{GM}{r} \Rightarrow \frac{T ^{2}}{r ^{3}} = \frac{4 π ^{2}}{GM}

(2)

Interpreting the result

Equation (2) is the pattern you measured. Its right-hand side contains no property of the moon — only $G$ and the central mass $M$ . That is why the four moons share one value: they all orbit Jupiter, so the constant is fixed by its mass.

A misconception worth killing here

"Satellites float because they are beyond gravity." They are not. At the International Space Station, about 400 km up, the field is still roughly 8.7 N/kg — nearly 90% of its value at the ground. Astronauts float because they are in continuous free fall, falling around the Earth rather than into it. Note that the derivation above only works because gravity is present and large: it is the centripetal force. A satellite beyond gravity would travel in a straight line and leave.

The relation $T^{2} \propto r^{3}$ is the third of Kepler's three laws, drawn from planetary data long before Newton explained them.

First law. Each planet moves in an ellipse with the Sun at one focus. The longest diameter is the major axis; half of it is the semi-major axis $a$ . For the Earth's orbit, $a$ is one astronomical unit ( $1 A U = 1.50 \times 1 0^{11}$ m).

Second law. The radius vector from the Sun to a planet sweeps out equal areas in equal times, so the areal velocity stays constant. With $θ$ the angle between the radius vector $r$ and the velocity vector $v$ :

\frac{1}{2} r v sin θ = \frac{1}{2} r_{1} v_{1} = \frac{1}{2} r_{2} v_{2} = constant

At perihelion (nearest, $r_{1}$ ) and aphelion (farthest, $r_{2}$ ) the velocity is perpendicular to the radius vector, so $sin θ = 1$ and $r_{1} v_{1} = r_{2} v_{2}$ : a planet moves faster when nearer the Sun.

Third law. The square of the period over the cube of the semi-major axis is the same constant for every body orbiting a given centre:

\frac{T ^{2}}{a ^{3}} = k = \frac{4 π ^{2}}{GM}

For a circular orbit the semi-major axis $a$ is the radius $r$ . Because $k$ is shared, any two bodies orbiting the same centre obey $T_{1}^{2} / a_{1}^{3} = T_{2}^{2} / a_{2}^{3}$ .

A force field is a region in which an object experiences a non-contact force. A gravitational field can be represented by vectors, which show the direction and relative strength of the field at each point, or by field lines: their direction shows the direction of the field, and the closer they are together, the stronger the field.

The gravitational field strength $g$ at a point is the gravitational force per unit mass acting on a small test mass placed there:

g = \frac{F}{m}

It is a vector, directed along the field, and is measured in newtons per kilogram (N/kg, equivalently m/s²). For a spherical mass $M$ , the field is radial and its magnitude at a distance $r$ from the centre is:

g = \frac{GM}{r ^{2}}

so it obeys an inverse-square law: if the distance from the mass doubles, $g$ falls to a quarter of its value. In a uniform field the field strength is the same at every point, so the field lines are parallel and equally spaced. In a radialfield the lines spread apart as $r$ increases and $g$ decreases. Earth's field is radial; however, over a small region close to the surface it is very nearly uniform, which is why $g \approx 9.8$ N/kg can be treated as constant there.

Point

• First law: planets move in elliptical orbits with the Sun at one focus. The semi-major axis of the Earth's orbit is 1 astronomical unit.
• Second law: the area swept per unit time by the radius vector is constant — $\frac{1}{2} r v sin θ$ . At perihelion and aphelion $sin θ = 1$ , so $r_{1} v_{1} = r_{2} v_{2}$ .
• Third law: $T^{2} / a^{3} = k$ is constant for all bodies orbiting the same centre, and $k = 4 π^{2} / GM$ depends only on the central mass.
• The attractive force between any two objects is the gravitational force, $F = G m_{1} m_{2} / r^{2}$ , with $G$ the universal gravitational constant.

📐 Equation box

Universal gravitation

F = G \frac{m _{1} m _{2}}{r ^{2}}

Orbital speed

v^{2} = \frac{GM}{r}

Kepler's third law

\frac{T ^{2}}{a ^{3}} = k = \frac{4 π ^{2}}{GM}

Areal velocity (second law)

\frac{1}{2} r_{1} v_{1} = \frac{1}{2} r_{2} v_{2}

Gravitational field strength

g = \frac{F}{m} = \frac{GM}{r ^{2}}

🛠️ Worked example (book p.73)

Situation

Three independent problems on gravitation and orbital motion.

(a) A comet moves in an elliptical orbit in which its distance from the Sun at aphelion is 5.0 times its distance at perihelion.

(b) An asteroid orbits the Sun with a semi-major axis of 4.0 astronomical units.

(c) Two objects, each of mass 60 kg, are placed 3.0 m apart.

Step 1 of 6 — try it yourself before you open the next one.

Now you try (p.73)

Calculate the mass of the Sun, given that the period of the Earth's orbit around the Sun is

3.156 \times 1 0^{7}

s and the semi-major axis of Earth's orbit is

1.496 \times 1 0^{11}

m.
Answer. Rearrange Kepler's third law for the central mass:

\frac{T ^{2}}{a ^{3}} = \frac{4 π ^{2}}{GM} \Rightarrow M = \frac{4 π ^{2} a ^{3}}{G T ^{2}}

M = \frac{4 π ^{2} ( 1.496 \times 1 0 ^{11} ) ^{3}}{( 6.67 \times 1 0 ^{- 11} ) ( 3.156 \times 1 0 ^{7} ) ^{2}} = 1.99 \times 1 0^{30} kg

Notice what has just happened: by timing one year and measuring one distance, you have weighed the Sun. Nothing about the Earth's own mass entered the calculation — it cancelled when the model was built, which is exactly why the method works.

When the distance of a comet from the Sun at aphelion is 2.5 astronomical units, and the distance from the Sun at perihelion is 1.5 astronomical units, determine the ratio of its speed at aphelion to its speed at perihelion.

The semi-major axis of a certain asteroid's orbit is 16 astronomical units. Using the fact that the Earth's orbit has a semi-major axis of 1.0 astronomical unit and an orbital period of 1.0 year, determine the orbital period of this asteroid.

Two objects, each with a mass of 60 kg, are 2.0 m apart. Determine the magnitude of the gravitational force acting between these two objects. Let the gravitational constant be

6.67 \times 1 0^{- 11}

N·m²/kg².

In a New Context

The geostationary orbit. A communications satellite is most useful when it stays above the same point on the equator, so that a dish on the ground can point in one fixed direction in the sky. For this, the satellite must orbit the Earth in the same time the Earth takes to rotate once on its axis, $T = 8.62 \times 1 0^{4}$ s (about one day). Rearranging the relation derived earlier:

r = 3 \frac{GM T ^{2}}{4 π ^{2}} \approx 4.2 \times 1 0^{7} m

which is about $3.6 \times 1 0^{7}$ m above the surface. At this radius the orbital speed is $v = 2 π r / T \approx 3.1 \times 1 0^{3}$ m/s. Only this one radiuslets the satellite orbit at the same angular speed as the Earth, so it stays above the same point.

Explain why a rocket launched eastward from near the equator requires less additional speed to reach an eastward orbit than a rocket launched westward from a high-latitude site.

Answer. Because the launch site is already moving. The Earth's surface turns eastward, and at the equator it does so fastest — about 465 m/s, since that is where the circle of latitude is largest. A rocket launched eastward from there starts with that 465 m/s already in the right direction, so the engines need to supply that much less of the roughly 7.8 km/s an orbit requires. At high latitude the circle of latitude is smaller, so the surface speed is lower and less is donated. And launching westward is worse than neutral: the rocket must first cancel the eastward motion it started with and then build the same speed the other way, so the site's rotation counts against it twice. This is why launch sites are built near the equator and why almost all orbits are prograde.

🧪 Nature of Science — Patterns and trends — the laws that waited seventy years

Kepler did not derive his laws from a theory of force. He inferred them from patterns in Tycho Brahe's precise naked-eye observations of planetary motion. The first two laws were published in Astronomia Nova in 1609, and the third in Harmonices Mundi in 1619. For nearly seventy years, these laws remained reliable empirical descriptions without a generally accepted dynamical explanation. In Principia (1687), Newton showed how they could be explained using his laws of motion and universal gravitation. Kepler's second law follows from the central direction of the gravitational force, while the first and third depend on its inverse-square variation with distance.

An answer to "your turn". Given only the planet's distance from the Sun, an astronomer could predict how long its year would be — before knowing anything about its mass, its composition, or why it moved at all. Measure the semi-major axis in AU, cube it, take the square root, and the answer is in Earth years. That is a genuine prediction: it says what will be observed, it could have been wrong, and it was checked against every planet found afterwards.

It is worth being clear about what the seventy years cost and what they did not. Kepler's laws worked perfectly well as descriptions throughout — nobody's prediction failed for want of Newton. What was missing was the reason, and with it any way to know where else the rules applied. Newton's explanation did not make the laws more accurate; it made them part of something larger, so that the same mathematics could be turned on a falling apple, a cannonball, and a moon nobody had yet seen.

Your turn: State one prediction that Kepler’s third law allowed astronomers to make about a newly discovered planet’s motion before Newton provided a gravitational explanation for the relationship.

✅ Lesson question, answered

Why does one satellite appear fixed above Earth, and what determines the time a satellite takes to complete an orbit?

Earth's gravitational force provides the centripetal force that keeps a satellite in orbit. The orbital period depends on the semi-major axis and Earth's mass. According to Kepler's third law:

T^{2} = \frac{4 π ^{2}}{GM} a^{3}

where $T$ is the orbital period and $a$ the semi-major axis (equal to the orbital radius $r$ for a circular orbit).

A geostationary satellite moves in a circular orbit above Earth's equator, in the same direction as Earth's rotation, and has an orbital period equal to Earth's rotational period. As a result, it remains above the same location and appears stationary in the sky. A satellite in a lower orbit, such as the International Space Station, has a smaller orbital radius and so, by $T^{2} \propto r^{3}$ , a shorter period — which is why it crosses the sky in minutes.

Key terms

gravitational force: — The mutual attraction between any two masses, F = Gm₁m₂/r².
universal gravitational constant: — G = 6.67 × 10⁻¹¹ N·m²/kg², the same everywhere.
perihelion / aphelion: — The nearest and farthest points of an orbit around the Sun.
semi-major axis: — Half the longest diameter of an elliptical orbit; the radius for a circle.
astronomical unit: — The semi-major axis of the Earth's orbit, 1.50 × 10¹¹ m.
gravitational field strength: — The force per unit mass, g = F/m = GM/r², in N/kg.
geostationary orbit: — The one orbit whose period equals a day, so the satellite stays above one point.

How the real paper is built

The ministry sets 50% multiple choice and 50% written answer. They test different things: the MCQ half rewards eliminating fast and spotting a distractor built from a plausible wrong method; the written half asks you to show that you understand why. Practise both halves — being fluent at one and weak at the other costs you exactly half the paper.

Two moons orbit the same planet. Compared with the nearer moon, the farther moon has:

A planet has an orbit with semi-major axis 4.0 AU. Its orbital period is:

The gravitational field strength at the surface of a planet is g. At a distance of two planet-radii from its centre, the field strength is:

Astronauts on the International Space Station appear weightless because:

A comet is 4.0 times farther from the Sun at aphelion than at perihelion. The ratio of its speed at aphelion to its speed at perihelion is:

Level 1 — direct substitution

One step. You are given the formula; put the numbers in.

1.1

(Exercise a) When the distance of a comet from the Sun at aphelion is 4.0 times its distance from the Sun at perihelion, determine the ratio of its speed at aphelion to its speed at perihelion.

1.2

(Exercise b) The semi-major axis of a certain asteroid's orbit is 9.0 astronomical units. Using the fact that the Earth's orbit has a semi-major axis of 1.0 astronomical unit and an orbital period of 1.0 year, determine the orbital period of this asteroid.

Level 2 — application in a new context

Two steps, or you must choose the right component before you can start.

2.1

(Exercise c) Two objects, each with a mass of 40 kg, are 3.0 m apart. Determine the magnitude of the gravitational force acting between these two objects. Let

G = 6.67 \times 1 0^{- 11}

N·m²/kg².

2.2

(Exercise d) Two objects, each with a mass of 50 kg, are 1.0 m apart. Determine the magnitude of the gravitational force acting between these two objects. Let

G = 6.67 \times 1 0^{- 11}

N·m²/kg².

Level 3 — exam challenge

Multi-step, or a case the textbook does not set and examiners do.

3.1

Derive an expression for the radius of a circular orbit in terms of the central mass

M

and the orbital period

T

. Use it to find the radius of a geostationary orbit, taking

M = 5.97 \times 1 0^{24}

kg and

T = 8.62 \times 1 0^{4}

s, and find the height above the surface and the orbital speed. Explain why only one radius gives a geostationary orbit.

3.2

Calculate Earth's gravitational field strength at its surface and at the altitude of the International Space Station, about 400 km up. Take

M = 5.97 \times 1 0^{24}

kg and

R = 6.37 \times 1 0^{6}

m. Use your answers to explain carefully why astronauts aboard the station appear weightless, and check your model by finding the station's orbital period.

📝 Exam-style question

Moon A orbits a planet at radius

r

with a period of 2.0 days. Moon B orbits the same planet at radius

3 r

. Determine the orbital period of moon B.

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