DP Physics HL D.1 Gravitational Fields

DP Physics HL D.1 Gravitational Fields - Full notes

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DP Physics HL | D.1 Gravitational Fields

DP PHYSICS · THEME D: FIELDS

D.1

Gravitational Fields

Full notes for HL · IB-style

Contents

1	Newton's law of gravitation
2	Gravitational field strength
3	Orbits and Kepler's laws
4	Gravitational potential and energy
5	Gravitation in context
6	Exam technique and further practice
7	Common misconceptions and links
8	Exam-style question with markscheme
9	Check your understanding
10	Summary and topic map

These are full teaching notes. Read each section in order, try the Think about it question before you read on, and do every Quick check before moving to the next section.

Before you start

Rate yourself on each statement now, and again after you've worked through the notes. Shade R (not yet), A (getting there) or G (confident).

I can…	R	A	G
use Newton's law of gravitation and the inverse-square law
define and calculate gravitational field strength and draw field patterns
state and use Kepler's three laws
derive and use orbital speed and T² ∝ r³
describe geostationary orbits and determine masses from orbits
explain weightlessness in orbit
calculate gravitational potential energy and potential
calculate escape speed and orbital energies; describe orbital decay

Sections marked (HL) are Additional Higher Level content: potential, potential energy, escape speed, orbital energy. Every other section is common to SL and HL.

Guiding questions

How are the properties of a gravitational field quantified? How are gravitational fields used to describe the motion of planets and satellites? How does energy change as objects move in gravitational fields?

How this topic is examined

D.1 appears in Papers 1 and 2. Expect Newton's law, field strength, orbital calculations and Kepler's laws; HL adds potential, escape speed and orbital energy.

Always measure r from the centre of the planet, not the surface.

Command term	What the examiner wants
State	a brief answer with no explanation
Outline	a brief account or summary
Describe	a detailed account without reasons
Explain	a detailed account including causes, reasons or mechanisms
Calculate	a numerical answer with working and correct units
Determine	obtain the only possible answer, often from a graph or data
Show that	obtain the given result, showing every step (give one more significant figure)
Estimate	an approximate value using reasonable assumptions
Sketch	a graph or diagram showing the general shape and key features
Deduce / Suggest	reach a conclusion from given information; propose a reasonable explanation

How to use these notes

Each section teaches one part of the syllabus in full. You'll see these boxes throughout; each has one job:

Box	What it's for
Think about it	a question to get you thinking before a section
Key point	the one idea you must not forget
Common mistake	an error examiners see every year, and how to avoid it
In the real world	where the physics shows up outside the classroom
Try this	something quick you can do yourself to test the idea
Worked example	full solutions with IB-style marks (M1 method, A1 answer, R1 reasoning)
Quick check	short questions to test yourself at the end of each section

1 Newton's law of gravitation

Every mass attracts every other mass.

Think about it

Why don't you feel the gravitational pull of the person sitting next to you?

Gravity holds planets in orbit, keeps our atmosphere in place and shapes galaxies. Newton showed that the force making an apple fall and the force keeping the Moon in orbit are the same.

1.1 The law

Two point masses attract each other with a force proportional to the product of their masses and inversely proportional to the square of their separation: F = GMm/r², with G = 6.67 × 10⁻¹¹ N m² kg⁻². Spherical bodies behave as point masses located at their centres. The forces on the two bodies are equal and opposite (Newton's third law).

Figure 1. Two masses attract with equal and opposite forces; r is measured between centres.

WORKED EXAMPLE 1 Earth–Moon (standard)

Calculate the force between Earth (6.0 × 10²⁴ kg) and the Moon (7.3 × 10²² kg) at 3.8 × 10⁸ m.

Solution

F = 6.67 × 10⁻¹¹ × 6.0 × 10²⁴ × 7.3 × 10²²/(3.8 × 10⁸)²M1

= 2.0 × 10²⁰ NA1

WORKED EXAMPLE 2 Two people (standard)

Estimate the force between two 70 kg people 1.0 m apart.

Solution

F = 6.67 × 10⁻¹¹ × 70 × 70/1.0²M1

= 3.3 × 10⁻⁷ N (negligible)A1

Common mistake

Using the altitude above the surface instead of the distance between centres.

1.2 The inverse-square law

Because F ∝ 1/r², doubling the separation reduces the force to a quarter; tripling reduces it to a ninth. Ratios avoid repeated calculation.

Figure 2. The inverse-square law: force against separation.

WORKED EXAMPLE 3 Ratio (standard)

A satellite at 2R from Earth's centre has weight 1000 N. Find its weight at 4R.

Solution

F ∝ 1/r²: factor 1/4M1

= 250 NA1

Quick check

1. How does F change if the distance triples?

Answers: 1. falls to 1/9

1.3 Forces from several masses

When more than one mass acts on a body, find each force with Newton's law and add them as vectors. By symmetry, a mass equidistant from two equal masses feels a resultant along the line of symmetry.

Figure 3. Forces on a small mass from two equal masses: the horizontal components cancel.

WORKED EXAMPLE 4 Symmetric case (exam level)

Two 1.0 × 10²⁴ kg masses are 6.0 × 10⁷ m apart. A 1000 kg probe is 4.0 × 10⁷ m above the midpoint. Find the resultant force on it.

Solution

Distance to each = 5.0 × 10⁷ m; F = GMm/r² = 26.7 NM1

Vertical components: 2 × 26.7 × (4/5) = 42.7 N towards the midpointA1

1.4 Newton's cannon

Newton imagined a cannon on a high mountain. Fired faster and faster, the ball lands further away, until at orbital speed the curve of its fall matches the curve of the Earth and it never lands: it is in orbit.

Figure 4. Newton's thought experiment: at orbital speed the projectile falls around the Earth.

Quick check

1. Why does an orbiting satellite not hit the ground?

Answers: 1. it falls at the same rate as the Earth's surface curves away

1.5 Measuring G

In 1798 Cavendish measured the tiny attraction between lead spheres with a torsion balance. Knowing G then allowed Earth's mass to be calculated from g = GM/R², 'weighing the Earth' for the first time.

WORKED EXAMPLE 5 Mass of the Earth (exam level)

Use g = 9.81 N kg⁻¹ and R = 6.37 × 10⁶ m to find Earth's mass.

Solution

M = gR²/G = 9.81 × (6.37 × 10⁶)²/6.67 × 10⁻¹¹M1

= 5.97 × 10²⁴ kgA1

In the real world

G is the least precisely known fundamental constant: its value is known only to about 1 part in 50 000, because gravitational forces between laboratory masses are so small.

1.6 Tides

The Moon's gravitational pull is stronger on the side of Earth nearest to it than on the far side. This difference stretches Earth's oceans into two bulges, so most coasts have two high tides a day. When the Sun and Moon line up (new and full moon) their effects add, giving larger spring tides; at right angles they give smaller neap tides.

Figure 5. The difference in the Moon's pull across Earth produces two tidal bulges (not to scale).

WORKED EXAMPLE 6 Difference in pull (exam level)

Find the ratio of the Moon's gravitational field strength at Earth's near side to that at its far side (Earth–Moon distance 3.84 × 10⁸ m, R = 6.37 × 10⁶ m).

Solution

Ratio = ((d + R)/(d − R))²M1

= (3.904/3.776)² = 1.07A1

2 Gravitational field strength

Force per unit mass.

2.1 Defining g

A gravitational field is a region where a mass experiences a force. Gravitational field strength g = F/m is the force per unit mass on a small test mass (N kg⁻¹). Around a spherical mass, g = GM/r². At Earth's surface g ≈ 9.81 N kg⁻¹, equal to the free-fall acceleration.

WORKED EXAMPLE 7 Surface g (standard)

Calculate g at the surface of Mars (M = 6.4 × 10²³ kg, R = 3.4 × 10⁶ m).

Solution

g = 6.67 × 10⁻¹¹ × 6.4 × 10²³/(3.4 × 10⁶)²M1

= 3.7 N kg⁻¹A1

WORKED EXAMPLE 8 At altitude (standard)

Find g at a height of 400 km above Earth.

Solution

r = 6.77 × 10⁶ m; g = GM/r²M1

= 8.7 N kg⁻¹A1

2.2 Field lines

Gravitational field lines show the direction of the force on a test mass. They point towards masses; their density shows field strength. Around a planet the field is radial; close to the surface over small distances it is approximately uniform (parallel, equally spaced lines).

Figure 6. Radial field around a planet and the nearly uniform field near its surface.

Quick check

1. Why is the field near Earth's surface treated as uniform?

Answers: 1. over heights much smaller than R, g barely changes

2.3 Variation of g with distance

Outside a planet, g falls as 1/r². Inside a uniform planet, only the mass closer to the centre contributes, so g increases linearly with r from zero at the centre to its surface value.

Figure 7. Gravitational field strength inside and outside a uniform planet.

WORKED EXAMPLE 9 Twice the radius (standard)

Find g at 2R and at R/2 (uniform Earth).

Solution

2R: 9.81/4 = 2.45 N kg⁻¹A1

R/2: 9.81/2 = 4.9 N kg⁻¹A1

2.4 Weight on other worlds

Weight W = mg depends on the local field strength, while mass is the same everywhere.

Figure 8. Weight of a 70 kg person on different worlds (surface or cloud-top values).

WORKED EXAMPLE 10 Lunar jump (exam level)

An astronaut can jump 0.40 m high on Earth. Estimate the height on the Moon with the same take-off speed.

Solution

h = v²/2g, so h ∝ 1/gM1

h = 0.40 × 9.81/1.62 = 2.4 mA1

2.5 Comparing planets

Surface g ∝ M/R², so comparing a planet's mass and radius with Earth's gives its surface gravity directly.

Body	M / ME	R / RE	g / N kg⁻¹
Moon	0.012	0.27	1.62
Mars	0.11	0.53	3.71
Earth	1	1	9.81
Jupiter	318	11.2	24.8

WORKED EXAMPLE 11 Jupiter (exam level)

Show that g at Jupiter's cloud tops is about 25 N kg⁻¹.

Solution

g = 9.81 × 318/11.2²M1

= 24.9 N kg⁻¹A1

2.6 Combining fields and neutral points

Field strengths from several masses add as vectors. On the line between two masses there is a neutral point where the fields cancel; it is closer to the smaller mass.

Figure 9. Field strengths of Earth and Moon along the line joining them (Earth's mass = 81 × Moon's mass).

WORKED EXAMPLE 12 Neutral point (exam level)

Find the distance from Earth's centre where the fields of Earth and Moon cancel (separation 3.84 × 10⁸ m).

Solution

GME/x² = GMM/(d − x)²: x/(d − x) = 9M1

x = 0.9d = 3.46 × 10⁸ mA1

2.7 Variations of g at Earth's surface

g is slightly larger at the poles (9.83 N kg⁻¹) than at the equator (9.78 N kg⁻¹) because Earth is flattened and rotates. Local rock density also changes g slightly, which geophysicists use to find mineral deposits.

WORKED EXAMPLE 13 Mountain top (exam level)

Find the percentage decrease in g at the top of Everest (8.8 km altitude).

Solution

(6370/6378.8)² = 0.9972M1

≈ 0.28% smallerA1

2.8 Testing the inverse-square law with data

Satellite measurements of g at different distances from Earth's centre can test the inverse-square law: a graph of g against 1/r² should be a straight line through the origin with gradient GM.

r / 10⁶ m	6.37	7.0	10	20	42
g / N kg⁻¹	9.81	8.13	3.98	1.00	0.226

Figure 10. g against 1/r² (r in units of 10⁶ m).

WORKED EXAMPLE 14 Gradient (exam level)

The gradient is 3.98 × 10¹⁴ N m² kg⁻¹ when r is in metres. Find Earth's mass.

Solution

Gradient = GMM1

M = 3.98 × 10¹⁴/6.67 × 10⁻¹¹ = 5.97 × 10²⁴ kgA1

2.9 Launching a satellite (energy overview)

Getting a satellite into orbit needs energy to raise it against gravity and to give it orbital speed. Launching eastward near the equator takes advantage of Earth's rotation (about 0.46 km s⁻¹ at the equator), which is why many launch sites are at low latitudes.

WORKED EXAMPLE 15 Kinetic energy needed (standard)

Find the kinetic energy a 1000 kg satellite needs for low Earth orbit at 7.7 km s⁻¹.

Solution

Ek = ½(1000)(7700)²M1

= 3.0 × 10¹⁰ JA1

3 Orbits and Kepler's laws

Gravity provides the centripetal force.

Think about it

The International Space Station is falling towards Earth all the time. Why does it never hit the ground?

3.1 Kepler's three laws

1. Planets move in ellipses with the Sun at one focus. 2. A line from the Sun to a planet sweeps out equal areas in equal times, so planets move fastest when closest to the Sun. 3. The square of the period is proportional to the cube of the mean orbital radius: T² ∝ r³.

Figure 11. Kepler's second law: equal areas are swept out in equal times.

WORKED EXAMPLE 16 Third law (standard)

Mars orbits at 1.52 AU. Find its period in Earth years.

Solution

T² = 1.52³ = 3.51M1

T = 1.87 yearsA1

3.2 Circular orbits

For a circular orbit of radius r, gravity provides the centripetal force: GMm/r² = mv²/r. So v = √(GM/r), independent of the satellite's mass. The period T = 2πr/v gives T² = (4π²/GM)r³, which is Kepler's third law.

Figure 12. In a circular orbit, gravity is perpendicular to the velocity.

Figure 13. Orbital speed against orbital radius for Earth satellites.

WORKED EXAMPLE 17 ISS (exam level)

The ISS orbits 400 km above Earth. Find its speed and period.

Solution

r = 6.77 × 10⁶ m; v = √(GM/r) = 7.67 km s⁻¹M1 A1

T = 2πr/v = 5550 s = 92.5 minA1

Common mistake

Thinking heavier satellites must orbit faster. The satellite's mass cancels.

3.3 Testing Kepler's third law

Plotting T² against a³ for the planets gives a straight line through the origin. In years and astronomical units, the gradient is 1 for the Sun's planets.

Figure 14. T² against a³ for the planets from Mercury to Jupiter.

WORKED EXAMPLE 18 Asteroid (exam level)

An asteroid has a period of 4.6 years. Find its orbital radius in AU.

Solution

a³ = 4.6² = 21.2M1

a = 2.8 AUA1

3.4 Geostationary orbits

A geostationary satellite orbits above the equator, in the same direction as Earth rotates, with a period of 24 h (more precisely one sidereal day), so it stays above one point. All geostationary satellites share the same radius.

Figure 15. A geostationary satellite stays above one point on the equator.

WORKED EXAMPLE 19 Radius (exam level)

Calculate the radius of a geostationary orbit.

Solution

r³ = GMT²/4π² with T = 86 400 sM1

r = 4.2 × 10⁷ m (about 36 000 km altitude)A1

In the real world

About 550 satellites share the geostationary belt; satellite TV dishes point at them permanently.

3.5 Weighing planets and stars

Measuring a moon's orbital radius and period gives the central body's mass: M = 4π²r³/GT². This is how the masses of the Sun, planets and even black holes are found.

WORKED EXAMPLE 20 Jupiter (exam level)

Io orbits Jupiter at 4.22 × 10⁸ m with period 1.77 days. Find Jupiter's mass.

Solution

M = 4π²(4.22 × 10⁸)³/(6.67 × 10⁻¹¹ × (1.53 × 10⁵)²)M1

= 1.9 × 10²⁷ kgA1

WORKED EXAMPLE 21 The Sun (exam level)

Find the Sun's mass from Earth's orbit (r = 1.50 × 10¹¹ m, T = 3.16 × 10⁷ s).

Solution

M = 4π²r³/GT²M1

= 2.0 × 10³⁰ kgA1

3.6 Speed in an elliptical orbit

In an elliptical orbit the speed changes continuously: it is greatest at the closest point (perihelion) and least at the furthest point (aphelion), as Kepler's second law requires.

Figure 16. Speed around an elliptical orbit with eccentricity 0.5.

Quick check

1. Where in its orbit does Earth move fastest?

Answers: 1. at perihelion, in early January

3.7 Gravity assists

A spacecraft passing close to a moving planet can gain or lose speed relative to the Sun by exchanging momentum with the planet. Missions such as Voyager and Juno used gravity assists to reach the outer planets with far less fuel.

Figure 17. A gravity assist: the probe's path bends around a moving planet and it leaves faster relative to the Sun.

Quick check

1. Where does the extra kinetic energy of the probe come from?

Answers: 1. the planet's orbital kinetic energy (the planet slows by an immeasurably small amount)

3.8 Detecting exoplanets

A star and its planet orbit their common centre of mass, so the star wobbles. Doppler shifts in the star's light reveal the wobble period (the planet's orbital period); Kepler's third law then gives the orbital radius.

Figure 18. Radial velocity of a star wobbling due to an orbiting planet.

WORKED EXAMPLE 22 Orbital radius (exam level)

The star 51 Pegasi has mass 2.2 × 10³⁰ kg and wobbles with period 4.23 days. Find the planet's orbital radius.

Solution

r³ = GMT²/4π² with T = 3.65 × 10⁵ sM1

r = 7.9 × 10⁹ m (0.053 AU)A1

In the real world

The first exoplanet around a Sun-like star, 51 Pegasi b, was found this way in 1995, earning Mayor and Queloz the 2019 Nobel Prize.

3.9 Weightlessness

Astronauts in orbit feel weightless because they and their spacecraft are in free fall together around Earth, not because gravity is absent. At ISS altitude g is still about 8.7 N kg⁻¹, about 89% of its surface value.

WORKED EXAMPLE 23 Centripetal acceleration (standard)

Show that the centripetal acceleration of the ISS equals g at its altitude.

Solution

v²/r = 7670²/6.77 × 10⁶ = 8.7 m s⁻²M1

= GM/r² = 8.7 N kg⁻¹A1

Common mistake

Saying there is no gravity in space.

3.10 Binary stars

Two stars orbiting each other both move in circles about their common centre of mass, with the same period. The heavier star moves in the smaller circle. Measuring the period and separation gives the total mass of the pair.

Figure 19. A binary system: both stars orbit their centre of mass.

Quick check

1. Which star in a binary has the larger orbital speed?

Answers: 1. the lighter one (larger radius, same period)

3.11 Space debris and orbital lifetime

Low Earth orbit is crowded with thousands of satellites and millions of debris fragments travelling at about 7.5 km s⁻¹. Even small fragments carry enormous kinetic energy, so collisions can destroy satellites and create more debris.

WORKED EXAMPLE 24 Debris energy (standard)

Find the kinetic energy of a 10 g fragment at 7.5 km s⁻¹ relative to a satellite.

Solution

Ek = ½(0.010)(7500)²M1

= 2.8 × 10⁵ J (like a car at 70 km h⁻¹)A1

3.12 Types of satellite orbit

Orbit	Altitude	Period	Uses
low Earth orbit	200–2000 km	90–130 min	ISS, imaging, internet constellations
polar orbit	LEO, passing over poles	about 100 min	mapping, weather (scans whole Earth)
medium Earth orbit	about 20 000 km	about 12 h	GPS
geostationary	35 800 km	24 h	communications, weather

Quick check

1. Why are polar orbits useful for mapping?

Answers: 1. Earth rotates beneath, so the whole surface is covered

4 Gravitational potential and energy

Energy in gravitational fields.

4.1 Gravitational potential energy (HL)

The gravitational potential energy of two masses is Ep = −GMm/r, defined as zero at infinite separation. It is negative because work must be done to separate the masses to infinity. Near Earth's surface, mgΔh is an approximation for small height changes.

WORKED EXAMPLE 25 Satellite (exam level)

Calculate the gravitational potential energy of a 1000 kg satellite at r = 7.0 × 10⁶ m.

Solution

Ep = −6.67 × 10⁻¹¹ × 5.97 × 10²⁴ × 1000/7.0 × 10⁶M1

= −5.7 × 10¹⁰ JA1

4.2 Gravitational potential (HL)

Gravitational potential Vg = −GM/r is the work done per unit mass to bring a small mass from infinity to that point. Field strength is the negative potential gradient: g = −ΔVg/Δr. The work to move mass m between two points is W = mΔVg.

Figure 20. Gravitational potential against distance from Earth's centre.

WORKED EXAMPLE 26 Work to move (exam level)

Find the work needed to move a 500 kg probe from Earth's surface to r = 2R.

Solution

ΔV = GM/2R = 3.13 × 10⁷ J kg⁻¹M1

W = 500 × 3.13 × 10⁷ = 1.6 × 10¹⁰ JA1

4.3 Equipotential surfaces (HL)

Equipotentials join points of equal potential. They are perpendicular to field lines, and no work is done moving along one. Around a planet they are concentric spheres, spaced further apart with distance for equal steps of potential.

Figure 21. Field lines and equipotentials around a planet.

Quick check

1. Why do equipotentials get further apart with distance?

Answers: 1. the field (potential gradient) gets weaker

4.4 Escape speed (HL)

The minimum launch speed to escape a planet's field completely (ignoring air resistance) gives total energy zero: ½mv² = GMm/R, so vesc = √(2GM/R). It is independent of the projectile's mass.

WORKED EXAMPLE 27 Earth (exam level)

Calculate the escape speed from Earth.

Solution

v = √(2 × 6.67 × 10⁻¹¹ × 5.97 × 10²⁴/6.37 × 10⁶)M1

= 11.2 km s⁻¹A1

WORKED EXAMPLE 28 Moon (exam level)

Find the Moon's escape speed (M = 7.35 × 10²² kg, R = 1.74 × 10⁶ m) and explain why it has no atmosphere.

Solution

v = 2.4 km s⁻¹A1

Fast gas molecules exceed this speed and escape over timeR1

4.5 Orbital energy (HL)

For a circular orbit: Ek = GMm/2r, Ep = −GMm/r and the total ET = −GMm/2r. Moving to a higher orbit increases the total energy (less negative) but decreases the kinetic energy (slower speed).

Figure 22. Kinetic, potential and total energy per kg of a satellite against orbital radius.

WORKED EXAMPLE 29 Raising an orbit (exam level)

A 2000 kg satellite moves from r = 7.0 × 10⁶ m to 8.0 × 10⁶ m. Find the energy required.

Solution

ΔE = (GMm/2)(1/r₁ − 1/r₂)M1

= 7.1 × 10⁹ JA1

4.6 Orbital decay (HL)

Atmospheric drag on low satellites removes energy, so the total energy becomes more negative, the orbit radius decreases and, surprisingly, the speed increases. Orbits eventually decay unless boosted.

Quick check

1. Why does a satellite speed up as its orbit decays?

Answers: 1. at smaller r, Ek = GMm/2r is larger

5 Gravitation in context

History, applications and models.

5.1 From Kepler to Newton

Tycho Brahe's precise naked-eye observations allowed Kepler to discover his three empirical laws (1609–1619). Newton (1687) showed that all three follow from a single inverse-square law of gravitation combined with his laws of motion, uniting the physics of the heavens and the Earth.

Quick check

1. Why was Newton's law of gravitation a major unification?

Answers: 1. it explained falling objects on Earth and planetary orbits with one law

5.2 Period against radius

The period of an orbit rises steeply with radius (T ∝ r3/2). Low satellites circle Earth in about 90 minutes; the Moon takes about 27 days.

Figure 23. Orbital period against radius for Earth satellites.

WORKED EXAMPLE 30 Reading the graph (standard)

Use T ∝ r3/2 to find the period of a satellite at 2.0 × 10⁷ m.

Solution

T = 24 × (2.0/4.22)1.5M1

= 7.8 hA1

5.3 GPS and gravity

GPS satellites orbit at about 20 000 km altitude with a 12-hour period. Receivers find their position by timing signals from at least four satellites; the precision depends on knowing each satellite's orbit, calculated from gravitation, and on relativistic corrections to the clocks.

WORKED EXAMPLE 31 GPS orbit (exam level)

Show that a 12-hour orbit has an altitude of about 20 000 km.

Solution

r = (GMT²/4π²)1/3 = 2.66 × 10⁷ mM1

altitude = 2.66 × 10⁷ − 6.37 × 10⁶ = 2.0 × 10⁷ mA1

5.4 Comparing fields: gravity and electric

Gravitational and electric fields follow the same inverse-square mathematics, which is developed further in D.2.

Feature	Gravitational field	Electric field
source	mass	charge
force law	F = GMm/r²	F = kq₁q₂/r²
field strength	g = GM/r² (N kg⁻¹)	E = kQ/r² (N C⁻¹)
direction	always attractive	attractive or repulsive
relative strength	very weak	about 10³⁹ times stronger for an electron and proton

Quick check

1. Why does gravity dominate on astronomical scales even though it is much weaker?

Answers: 1. planets and stars are electrically neutral overall, but masses always attract and add up

5.5 Models and limitations

Newton's law treats planets as point masses and ignores other bodies. It predicts orbits extremely well, but small discrepancies (such as Mercury's orbit) required Einstein's general relativity, which describes gravity as curved spacetime.

Quick check

1. Name one observation Newton's law could not fully explain.

Answers: 1. the precession of Mercury's orbit

6 Exam technique and further practice

Applying gravitation.

6.1 Method checklist

•Measure r from the centre of the planet: r = R + altitude.
•Use ratios for inverse-square problems.
•For orbits, equate GMm/r² to mv²/r or mω²r.
•Convert days and hours to seconds.
•Remember orbital speed and period are independent of the satellite's mass.

6.2 Multi-step problem: satellite

WORKED EXAMPLE 32 Weather satellite (exam level)

A 450 kg satellite orbits 800 km above Earth. Find (a) g at the satellite; (b) its speed; (c) its period in minutes.

Solution

(a) r = 7.17 × 10⁶ m; g = 7.75 N kg⁻¹M1 A1

(b) v = √(GM/r) = 7.45 km s⁻¹A1

6.3 Multi-step problem: geostationary link

WORKED EXAMPLE 33 Signal delay (exam level)

A phone call is relayed by a geostationary satellite. Find (a) the satellite's altitude; (b) the minimum delay for a signal to go up and down.

Solution

(a) r = 4.22 × 10⁷ m; altitude = 3.58 × 10⁷ mA1

(b) t = 2 × 3.58 × 10⁷/3.0 × 10⁸ = 0.24 sM1 A1

6.4 Graded practice

WORKED EXAMPLE 34 Double mass and radius (standard)

A planet has twice Earth's mass and twice its radius. Find its surface g.

Solution

g ∝ M/R² = 2/4M1

= 4.9 N kg⁻¹A1

WORKED EXAMPLE 35 Period ratio (standard)

A satellite's orbital radius is quadrupled. By what factor does its period change?

Solution

T ∝ r3/2M1

41.5 = 8A1

WORKED EXAMPLE 36 12-hour orbit (exam level)

Find the radius of an Earth orbit with a 12-hour period.

Solution

r = (GMT²/4π²)1/3M1

= 2.66 × 10⁷ mA1

6.5 Data: Jupiter's moons

The four largest moons of Jupiter obey Kepler's third law.

Moon	r / 10⁸ m	T / days
Io	4.22	1.77
Europa	6.71	3.55
Ganymede	10.7	7.15
Callisto	18.8	16.7

Figure 24. T² against r³ for Jupiter's moons.

WORKED EXAMPLE 37 Mass from gradient (exam level)

The gradient is 3.13 × 10⁻¹⁶ s² m⁻³. Find Jupiter's mass.

Solution

Gradient = 4π²/GMM1

M = 4π²/(6.67 × 10⁻¹¹ × 3.13 × 10⁻¹⁶) = 1.9 × 10²⁷ kgA1

6.6 Graded practice: fields

WORKED EXAMPLE 38 Field ratio (standard)

At what height above Earth's surface is g half its surface value?

Solution

r = R√2 = 9.01 × 10⁶ mM1

h = 2.64 × 10⁶ mA1

WORKED EXAMPLE 39 Density of a planet (exam level)

A planet has radius 4.0 × 10⁶ m and surface g = 6.0 N kg⁻¹. Find its mean density.

Solution

M = gR²/G = 1.44 × 10²⁴ kg; V = 2.68 × 10²⁰ m³M1

ρ = 5.4 × 10³ kg m⁻³A1

WORKED EXAMPLE 40 Resultant field (exam level)

Find the resultant field strength midway between two 5.0 × 10²⁴ kg planets 1.0 × 10⁹ m apart, and at a point 2.0 × 10⁸ m from one along the line.

Solution

Midway: equal and opposite, 0A1

g = G(5.0 × 10²⁴)(1/(2 × 10⁸)² − 1/(8 × 10⁸)²) = 7.8 × 10⁻³ N kg⁻¹M1 A1

6.7 Graded practice: orbits

WORKED EXAMPLE 41 Moon's period (standard)

Use the Moon's orbital radius (3.84 × 10⁸ m) to find its period in days.

Solution

T = 2π√(r³/GM)M1

= 2.37 × 10⁶ s = 27.4 daysA1

WORKED EXAMPLE 42 Speed of Earth (standard)

Find Earth's orbital speed around the Sun (r = 1.50 × 10¹¹ m, T = 1 year).

Solution

v = 2πr/TM1

= 29.8 km s⁻¹A1

WORKED EXAMPLE 43 Low orbit around the Moon (exam level)

Find the period of a spacecraft in a low circular orbit just above the Moon's surface (M = 7.35 × 10²² kg, R = 1.74 × 10⁶ m).

Solution

T = 2π√(R³/GM)M1

= 6.5 × 10³ s = 108 minA1

6.8 Graded practice: Kepler and mass

WORKED EXAMPLE 44 Saturn's moon (exam level)

Titan orbits Saturn at 1.22 × 10⁹ m with period 15.9 days. Find Saturn's mass.

Solution

M = 4π²r³/GT² with T = 1.37 × 10⁶ sM1

= 5.7 × 10²⁶ kgA1

WORKED EXAMPLE 45 Equal areas (standard)

A comet is 0.6 AU from the Sun at perihelion and 30 AU at aphelion. Explain qualitatively how its speed compares at the two points.

Solution

Equal areas in equal timesR1

Much faster at perihelion (about 50 times, since v ∝ 1/r at these points)A1

WORKED EXAMPLE 46 Kepler's third law in AU (standard)

Neptune orbits at 30 AU. Find its period in years.

Solution

T = 301.5M1

= 164 yearsA1

6.9 Graded practice: fields and weight

WORKED EXAMPLE 47 Mass on scales (standard)

A 2.0 kg mass is weighed on Earth and on Mars. Find each weight.

Solution

Earth: 19.6 NA1

Mars: 2.0 × 3.71 = 7.4 NA1

WORKED EXAMPLE 48 Pendulum on Mars (exam level)

A pendulum has period 2.0 s on Earth. Find its period on Mars.

Solution

T ∝ 1/√gM1

T = 2.0√(9.81/3.71) = 3.3 sA1

WORKED EXAMPLE 49 Field at Moon's orbit (standard)

Find Earth's gravitational field strength at the Moon's distance (3.84 × 10⁸ m).

Solution

g = GM/r²M1

= 2.7 × 10⁻³ N kg⁻¹A1

6.10 Graded practice: mixed problems

WORKED EXAMPLE 50 Two satellites (exam level)

Satellite A orbits at radius r, satellite B at 2r around the same planet. Find the ratios vB/vA and TB/TA.

Solution

v ∝ r−1/2: vB/vA = 0.71A1

T ∝ r3/2: TB/TA = 2.8A1

WORKED EXAMPLE 51 Free fall from orbit height (exam level)

A ball is dropped from rest at a height where g = 8.7 N kg⁻¹ (ignoring air). What is its initial acceleration, and why does an astronaut there not fall to Earth?

Solution

a = 8.7 m s⁻² downwardsA1

The astronaut also has a large horizontal speed, so falls around Earth (orbits)R1

WORKED EXAMPLE 52 Planet from moon data (exam level)

A moon orbits a planet at 2.0 × 10⁸ m every 5.0 days. Find the planet's mass and its surface g if its radius is 3.0 × 10⁷ m.

Solution

M = 4π²r³/GT² = 2.5 × 10²⁵ kgM1 A1

g = GM/R² = 1.9 N kg⁻¹A1

6.11 Graded practice: explanations

WORKED EXAMPLE 53 Why the Moon does not fall (exam level)

Explain why the Moon stays in orbit even though Earth's gravity pulls it towards Earth.

Solution

Gravity provides the centripetal force needed for circular motionR1

The Moon's tangential velocity carries it sideways as it falls, so its distance stays constantR1

WORKED EXAMPLE 54 Lower orbit faster (exam level)

Explain, using an equation, why a satellite in a lower orbit travels faster.

Solution

GMm/r² = mv²/r gives v = √(GM/r)M1

Smaller r gives larger vR1

WORKED EXAMPLE 55 Field lines (standard)

State two properties of gravitational field lines around a planet.

Solution

Point towards the centre of the planetA1

Closer together near the planet, where the field is strongerA1

6.12 Key terms

Term	Meaning
gravitational field strength	force per unit mass, g = GM/r²
field line	direction of force on a test mass
Kepler's third law	T² ∝ r³
geostationary orbit	equatorial orbit with 24 h period
gravitational potential (HL)	work per unit mass from infinity, −GM/r
escape speed (HL)	√(2GM/R)

6.13 Self-test

Try these without notes, then check the answers.

Quick check

1. How does g change if you move from R to 3R?

2. Does orbital speed depend on satellite mass?

3. Find the period of a satellite at r = 4R if one at R has period 85 min.

4. Why do astronauts feel weightless?

5. Where is the neutral point between Earth and Moon?

6. What provides the centripetal force for a planet?

Answers: 1. falls to 1/9 2. no 3. 85 × 8 = 680 min 4. they are in free fall with the craft 5. about 0.9 of the way to the Moon 6. the Sun's gravity

6.14 Self-test 2

More quick questions on orbits and fields.

Quick check

1. What is the period of a geostationary satellite?

2. State Kepler's second law.

3. Find g at a distance 4R from Earth's centre.

4. What provides the centripetal force on the Moon?

5. How does the speed of a satellite change if its orbital radius is quadrupled?

6. State one reason tides occur twice a day.

Answers: 1. 24 hours 2. a line from the Sun to a planet sweeps out equal areas in equal times 3. 9.81/16 = 0.61 N kg⁻¹ 4. Earth's gravitational pull 5. it halves 6. the Moon's pull is stronger on the near side than the far side, producing two bulges

6.15 Common exam errors

•Using altitude instead of orbital radius.
•Forgetting to square r in g = GM/r².
•Using hours instead of seconds in Kepler's third law.
•Saying there is no gravity in orbit.
•(HL) Giving a positive value for gravitational potential.

7 Common misconceptions and links

Students often think…	Actually…
there is no gravity in orbit	gravity provides the centripetal force
r is the altitude	r is measured from the centre
heavier satellites orbit faster	orbital speed is independent of satellite mass
higher orbits are faster	higher orbits are slower
potential energy is positive	gravitational potential energy is negative (zero at infinity)

How D.1 connects to the rest of the course

Topic	Connection
A.2 Forces and momentum	circular motion
D.2 Electric and magnetic fields	inverse-square fields and potential
A.4 Rigid body mechanics (HL)	angular momentum and Kepler's second law
E.5 Fusion and stars	gravitational collapse of stars

8 Exam-style question with markscheme

Try this under exam conditions (about 18 minutes) before reading the markscheme.

A satellite of mass 450 kg orbits Earth (M = 5.97 × 10²⁴ kg, R = 6.37 × 10⁶ m) at an altitude of 800 km.

(a) Calculate the gravitational field strength at the satellite. [2]

(b) Calculate the orbital speed. [2]

(d) Explain why an astronaut on board feels weightless. [2]

(e) State how the period would change if the satellite's mass doubled. [1]

(f) Calculate the total energy of the satellite in orbit. [2]

Markscheme

Part	Answer	Marks
(a)	r = 7.17 × 10⁶ m; g = GM/r²	M1
	7.75 N kg⁻¹	A1
(b)	v = √(GM/r)	M1
	7.45 × 10³ m s⁻¹	A1
(c)	T = 2πr/v = 6050 s	M1
	101 min	A1
(d)	astronaut and craft in free fall together	R1
	no normal reaction force	R1
(e)	no change	A1
(f)	E = −GMm/2r	M1
	−1.25 × 10¹⁰ J	A1

Examiner's comments. Most errors came from using the altitude instead of the orbital radius. In (d), 'no gravity' gained no credit.

9 Check your understanding

Section A: multiple choice

1. G has units of

A. N m² kg⁻² B. N kg⁻¹ C. m s⁻² D. N m kg⁻¹

2. If the distance between two masses doubles, the force

A. halves B. doubles C. becomes one quarter D. quadruples

3. Orbital speed of a satellite depends on

A. the satellite's mass B. the planet's mass and the orbit radius C. the satellite's shape D. the satellite's density

4. Kepler's third law states

A. T ∝ r B. T³ ∝ r² C. T ∝ 1/r D. T² ∝ r³

5. A geostationary satellite has a period of

A. 90 min B. 24 h C. 12 h D. 1 month

6. g at Earth's surface is about

A. 9.81 N kg⁻¹ B. 6.67 × 10⁻¹¹ N kg⁻¹ C. 1.6 N kg⁻¹ D. 0

7. Astronauts in orbit feel weightless because

A. gravity is zero B. they are outside the field C. the craft spins D. they are in free fall

8. g inside a uniform planet

A. is constant B. falls as 1/r² C. increases linearly with r D. is zero everywhere

9. Escape speed from a planet is

A. √(GM/R) B. 2GM/R C. GM/R² D. √(2GM/R)

Section B: structured

10. Calculate the gravitational force between two 1.0 kg masses 0.10 m apart. [2]

11. Derive an expression for the orbital speed of a satellite in a circular orbit. [3]

12. Calculate the orbital radius of a satellite with period 12 h around Earth. [3]

13. Explain why satellites in lower orbits have shorter periods. [2]

Answers

1. A 2. C 3. B 4. D 5. B 6. A 7. D 8. C 9. D

10. 6.7 × 10⁻⁹ N.

11. GMm/r² = mv²/r: v = √(GM/r).

12. r = (GMT²/4π²)1/3 = 2.66 × 10⁷ m.

13. Smaller r: higher speed and shorter path; T ∝ r3/2.

10 Summary and topic map

Idea	Remember
Newton's law	F = GMm/r²
Field strength	g = GM/r²
Kepler	ellipses; equal areas; T² ∝ r³
Orbits	v = √(GM/r)
Geostationary	T = 24 h, r ≈ 4.2 × 10⁷ m
Weightlessness	free fall, not zero gravity

Before the exam, make sure you…

•measure r from the centre
•use ratios for inverse-square problems
•equate gravity to centripetal force
•remember orbital speed is independent of satellite mass

Gravitational Fields on one page

Cover this page with a sheet of paper and try to rebuild it from memory. Whatever you can't recall is what to revise next.

Figure 25. Topic map for Gravitational Fields.

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