IB DP Physics · Theme D · Fields

Gravitational fields.

D.1 · Kepler’s laws, Newton’s law of gravitation, field strength, potential and orbits
Standard level and higher level

SL + HL · 5 hours Additional HL · 7 hours

The journey

Four parts.

Part 01 · SL + HL

Kepler’s laws

How the planets move round the Sun

Part 02 · SL + HL

Newton’s law of gravitation

The force between any two masses

Part 03 · SL + HL

Gravitational fields

Field strength and field lines

Part 04 · HL only

Energy and orbits

Potential, orbital speed and escape speed

Slides with a terracotta edge and an HL only flag are additional higher level. Every other slide is for SL and HL.

D.1 GRAVITATIONAL FIELDS · SL + HL

Part 01 · SL + HL

Kepler’s laws.

Three rules that describe every orbit, found before anyone knew why

D.1 GRAVITATIONAL FIELDS · SL + HL

The story · Prague, 1600–1619

Background story · not assessed

Tycho’s numbers, Kepler’s laws.

Colored engraving of Tycho Brahe seated inside his observatory, pointing upwards, beside a large brass quadrant fixed to the wall, with assistants reading the instrument and a clock — **Tycho Brahe** and his great mural quadrant, engraved in 1598

Printed Latin title page of Astronomia Nova — Kepler’s **Astronomia Nova**, 1609

Tycho measured planets by eye, with no telescope, to about one arcminute
Johannes Kepler joined him in Prague in 1600 and worked on the orbit of Mars
His best circular model missed Tycho’s data by eight arcminutes. He trusted the data and gave up circles
Two laws in 1609, the third in 1619

A small disagreement with good data overturned 2000 years of circles.

Quadrant: from Tycho’s Astronomiae instauratae mechanica, public domain. Title page: Johannes Kepler, public domain.

D.1 GRAVITATIONAL FIELDS · SL + HL

The shape of an orbit

Kepler’s first law.

Each planet moves in an ellipse
The Sun is at one focus of the ellipse
The distance from the Sun changes round the orbit
A circle is the special ellipse whose two foci coincide

AN ELLIPTICAL ORBITMUCH EXAGGERATED

D.1 GRAVITATIONAL FIELDS · SL + HL

What an ellipse is

Pin the two ends of a loose string to a board and pull it tight with a pencil. As the pencil moves, it draws an ellipse, and the two pins are the foci. For every point on the curve, the two distances to the foci add up to the same total: the length of the string.

How stretched are real orbits?

Much less than the drawing. The Earth is 147 million km from the Sun at its closest and 152 million km at its furthest, a difference of about 3%. Drawn to scale, its orbit could not be told from a circle by eye. That is why calculations in this topic treat orbits as circles.

Putting the Sun at the center of the ellipse. It is at a focus, to one side of the center, and nothing at all is at the other focus.

The same law holds for moons round planets and for satellites round the Earth: the body at the focus is whatever is being orbited.

The speed round an orbit

Kepler’s second law.

The line from the Sun to a planet sweeps out equal areas in equal times
Near the Sun the line is short, so the planet must move faster
Far from the Sun it moves slower

In a circular orbit the speed is constant.

ONE ORBIT, AT THE RIGHT SPEEDS

D.1 GRAVITATIONAL FIELDS · SL + HL

Why it speeds up: an energy argument

As the planet falls closer to the Sun, gravitational potential energy is transferred to kinetic energy, like a ball rolling down into a dip. As it climbs away again, it slows. The total energy stays the same all the way round.

Why it speeds up: a force argument

In a circular orbit the pull of the Sun is always at right angles to the velocity, so it changes only the direction. In an ellipse the pull has a component along the path: forwards while the planet approaches the Sun, backwards while it moves away.

The two shaded sectors look very different, one short and wide, the other long and thin, but they have the same area and take the same time.

Comparing orbits

Kepler’s third law.

\(T^{2} \propto r^{3}\)

\(T\): the orbital period
\(r\): the average radius of the orbit
So \(\dfrac{T^{2}}{r^{3}}\) is the same for every body orbiting the same center

Planet	\(r\) / AU	\(T\) / years	\(\frac{T^{2}}{r^{3}}\)
Mercury	0.387	0.241	1.00
Earth	1.00	1.00	1.00
Mars	1.52	1.88	1.00
Jupiter	5.20	11.9	1.00

D.1 GRAVITATIONAL FIELDS · SL + HL

What the law says in words

Planets further out take longer, and by more than you would guess from the longer path alone. Jupiter’s orbit is 5.2 times as long as the Earth’s, but its year is 11.9 times as long, so Jupiter must also be moving more slowly. A weaker pull, further from the Sun, needs a lower speed.

Why the units in the table work

With \(r\) in astronomical units and \(T\) in years, the Earth has \(r = 1\) and \(T = 1\), so the constant is exactly 1 for everything that orbits the Sun. In SI units the constant is a different number, but it is still the same for every planet.

Comparing a planet with a moon. The constant depends on the mass of the body at the center, so the law compares only bodies that orbit the same center.

Part 2 shows where the law comes from, and what the constant is.

Worked example

How long is the year of Ceres?.

Ceres orbits the Sun at \(r = 2.77 \,\mathrm{AU}\) Earth: \(r_\mathrm{E} = 1.00 \,\mathrm{AU}\), \(T_\mathrm{E} = 1.00\) year \(\dfrac{T^{2}}{r^{3}} = \dfrac{T_\mathrm{E}^{2}}{r_\mathrm{E}^{3}}\) \(T = T_\mathrm{E} \left(\dfrac{r}{r_\mathrm{E}}\right)^{\frac{3}{2}}\) \(T = 1.00 \times 2.77^{\frac{3}{2}} = \class{res}{4.61 \text{ years}}\)

Compare with a body whose orbit you know
Use any units, as long as they are the same for both bodies
The power \(\frac{3}{2}\) means “cube, then square root”

D.1 GRAVITATIONAL FIELDS · SL + HL

Quick check

Kepler’s laws.

01What shape is the orbit of a planet, and where is the Sun?

An ellipse, with the Sun at one focus

02Where in its orbit does a planet move fastest?

Where it is closest to the Sun

03What is the same for two equal time intervals on one orbit?

The area swept out by the line from the Sun

04Which quantity is the same for every planet round the Sun?

\(\frac{T^{2}}{r^{3}}\)

05A planet orbits at 4.0 AU. Its period?

\(4.0^{\frac{3}{2}} = 8.0\) years

06Can the third law compare the Moon with Mars?

No: they orbit different bodies

PRESS → TO REVEAL EACH ANSWER

Part 02 · SL + HL

Newton’s law of gravitation.

One force explains the falling apple, the Moon and all three of Kepler’s laws

D.1 GRAVITATIONAL FIELDS · SL + HL

The force between two masses

Newton’s universal law of gravitation.

\(F = G \dfrac{m_{1} m_{2}}{r^{2}}\)

Every mass attracts every other mass
\(r\) is the distance between their centers
\(G = 6.67 \times 10^{-11} \,\mathrm{N} \,\mathrm{m}^{2} \,\mathrm{kg}^{-2}\)

TWO MASSESEQUAL AND OPPOSITE FORCES

D.1 GRAVITATIONAL FIELDS · SL + HL

Why the two forces are equal

They are a Newton’s third law pair: the force of body 1 on body 2 and the force of body 2 on body 1. The equation shows it too, because swapping \(m_{1}\) and \(m_{2}\) changes nothing. The Earth pulls you with about 700 N and you pull the Earth with about 700 N. Only the accelerations differ, because the masses do.

Why you never notice it between everyday objects

\(G\) is tiny. Two 70 kg people standing 1.0 m apart attract each other with \(6.67 \times 10^{-11} \times \frac{70 \times 70}{1.0^{2}} = 3.3 \times 10^{-7} \,\mathrm{N}\), about the weight of a grain of fine sand. Gravity matters only when at least one of the masses is enormous.

Confusing \(G\) with \(g\). \(G\) is a universal constant, the same everywhere. \(g\) is the field strength at one place, 9.8 N kg⁻¹ only near the Earth’s surface.

How the force depends on distance

An inverse square law.

\(F \propto \dfrac{1}{r^{2}}\)
Double the separation: a quarter of the force
Three times the separation: one ninth
The force never quite reaches zero

FORCE AGAINST SEPARATION

D.1 GRAVITATIONAL FIELDS · SL + HL

Where the square comes from

Picture the influence of a mass spreading out evenly in all directions. At distance \(r\) it is spread over a sphere of area \(4 \pi r^{2}\). Twice as far away, the sphere has four times the area, so the influence at each point is four times weaker. Light from a lamp and the electric force of a charge fall off in the same way, for the same reason.

Using ratios

A satellite moves from 7000 km to 21 000 km from the center of the Earth. The distance has tripled, so the force is \(\frac{1}{3^{2}} = \frac{1}{9}\) of what it was. No masses and no \(G\) are needed.

Measuring \(r\) from the surface. A satellite 400 km above the ground is 6770 km from the center of the Earth, not 400 km: its weight is only about 11% less than on the ground.

When is a body a point?

Point masses and spheres.

The law is for point masses
A body acts as a point mass when it is small compared with the separation
A uniform sphere acts as if all its mass were at its center, for anything outside it

OUTSIDE THE SPHERE, THE SAME FIELD

D.1 GRAVITATIONAL FIELDS · SL + HL

Why the sphere result matters

An apple on a tree is a few meters from the nearest rock and 12 700 km from the rock on the far side of the Earth. Each piece of the Earth pulls it with a different force in a different direction. Newton proved that for a sphere all these pulls add up to exactly the force of one point mass at the center. That is why the distance in every calculation is measured to the center of the planet.

When it fails

The result needs a sphere whose density depends only on the distance from its center, and a point outside it. It does not work for an irregular asteroid seen from close by, and it does not work inside the planet.

The Earth and the Sun are 150 million km apart and the Sun is 1.4 million km across. Even without the sphere result, treating both as points is an excellent approximation.

The story · London, 1798

Background story · not assessed

Cavendish weighs the Earth.

Engraved cross-section of a room containing a long horizontal rod hung from a wire, with a small lead ball at each end and a large lead ball hung close to each of them — The **torsion balance**, from Cavendish’s paper of 1798

The force between laboratory masses is tiny: Newton never knew the value of \(G\)
John Michell designed the apparatus but died in 1793. Henry Cavendish rebuilt it
Two 158 kg lead balls pulled two 0.73 kg balls on a 1.8 m rod, twisting the wire that held it
His result: the Earth is 5.45 times as dense as water. Today’s value is 5.51

From his result comes \(G\), and from \(G\) the mass of the Earth.

Drawing: Henry Cavendish, Philosophical Transactions, 1798, public domain.

D.1 GRAVITATIONAL FIELDS · SL + HL

Worked example

The force that holds the Moon.

Earth \(5.97 \times 10^{24} \,\mathrm{kg}\) · Moon \(7.35 \times 10^{22} \,\mathrm{kg}\) Center to center: \(r = 3.84 \times 10^{8} \,\mathrm{m}\) \(F = G \dfrac{m_{1} m_{2}}{r^{2}}\) \(F = \dfrac{6.67 \times 10^{-11} \times 5.97 \times 10^{24} \times 7.35 \times 10^{22}}{(3.84 \times 10^{8})^{2}}\) \(F = \class{res}{1.98 \times 10^{20} \,\mathrm{N}}\)

Distances in meters, between centers
The same force acts on the Earth, towards the Moon

This force is the Moon’s centripetal force.

D.1 GRAVITATIONAL FIELDS · SL + HL

Checking it against the Moon’s motion

The Moon goes round once in 27.3 days, \(2.36 \times 10^{6} \,\mathrm{s}\). Its centripetal acceleration is \(\frac{4 \pi^{2} r}{T^{2}} = 2.72 \times 10^{-3} \,\mathrm{m} \,\mathrm{s}^{-2}\), so the force needed is \(7.35 \times 10^{22} \times 2.72 \times 10^{-3} = 2.0 \times 10^{20} \,\mathrm{N}\). The force that gravity supplies and the force that circular motion needs agree.

Newton’s own check

The Moon is 60 Earth radii away, so by the inverse square law gravity there should be \(\frac{1}{60^{2}} = \frac{1}{3600}\) of its value at the surface: \(\frac{9.8}{3600} = 2.7 \times 10^{-3} \,\mathrm{m} \,\mathrm{s}^{-2}\). The falling apple and the orbiting Moon obey the same law.

Work out the powers of ten separately: \(-11 + 24 + 22 - 16 = 19\). It catches most calculator slips.

Gravitation and circular motion

Where Kepler’s third law comes from.

A body of mass \(m\) in a circular orbit round mass \(M\) Gravitational force = centripetal force \(G \dfrac{M m}{r^{2}} = \dfrac{4 \pi^{2} m r}{T^{2}}\) \(T^{2} = \dfrac{4 \pi^{2}}{G M} r^{3}\)

Gravity is the only force: it provides all the centripetal force
The orbiting mass \(m\) cancels
The constant is \(\dfrac{4 \pi^{2}}{G M}\): it depends only on the central mass

D.1 GRAVITATIONAL FIELDS · SL + HL

The centripetal force used here

From A.2, the centripetal acceleration is \(\omega^{2} r\) with \(\omega = \frac{2 \pi}{T}\), which gives \(\frac{4 \pi^{2} r}{T^{2}}\). If a question gives a speed and not a period, use \(\frac{m v^{2}}{r}\) for the centripetal force.

Why the canceling matters

Because \(m\) cancels, an astronaut, a spanner and a space station at the same height all orbit together with the same period. This is why astronauts float inside the station: everything is falling round the Earth at the same rate. It is not because there is no gravity up there.

Learn the first line, not the last. “Gravitational force = centripetal force” is the starting point of every orbit calculation.

Kepler found the law from data in 1619. Newton explained it nearly seventy years later.

Worked example

Finding the mass of the Sun.

Earth’s orbit: \(r = 1.50 \times 10^{11} \,\mathrm{m}\) \(T = 1\) year \(= 3.16 \times 10^{7} \,\mathrm{s}\) \(T^{2} = \dfrac{4 \pi^{2}}{G M} r^{3}\) so \(M = \dfrac{4 \pi^{2} r^{3}}{G T^{2}}\) \(M = \dfrac{4 \pi^{2} \times (1.50 \times 10^{11})^{3}}{6.67 \times 10^{-11} \times (3.16 \times 10^{7})^{2}}\) \(M = \class{res}{2.00 \times 10^{30} \,\mathrm{kg}}\)

An orbit measures the mass at its center
Only \(r\) and \(T\) of the orbiting body are needed
The mass of the Earth does not appear

D.1 GRAVITATIONAL FIELDS · SL + HL

Why this method is so useful

Nobody can put a star or a planet on a balance. But anything with a satellite can be weighed: watch the satellite, measure its period and the size of its orbit, and the central mass follows. The mass of Jupiter comes from its moons, the mass of the Earth from the Moon, and the mass of the black hole at the center of our galaxy from the stars that orbit it.

What cannot be found

The orbit tells you nothing about the mass of the orbiting body, because that mass canceled in the derivation. To find the mass of a planet with no moon, such as Venus, astronomers had to wait for a spacecraft to fly past it.

Leaving the period in days or years. With \(G\) in SI units, \(T\) must be in seconds and \(r\) in meters.

Quick check

Newton’s law of gravitation.

01The separation of two masses is doubled. What happens to the force?

It falls to a quarter

02One of the masses is doubled. What happens to the force?

It doubles

03The Earth pulls the Moon. How big is the force of the Moon on the Earth?

The same size, in the opposite direction

04From where is \(r\) measured for a planet?

From its center

05What provides the centripetal force on a satellite?

The gravitational force

06Which mass sets the constant in \(T^{2} \propto r^{3}\)?

The central mass \(M\)

PRESS → TO REVEAL EACH ANSWER

Part 03 · SL + HL

Gravitational fields.

What a mass does to the space around it

D.1 GRAVITATIONAL FIELDS · SL + HL

Force per unit mass

Gravitational field strength.

\(g = \dfrac{F}{m}\)

A field is a region where a mass experiences a force
\(g\) is the force per unit mass on a small point mass placed at that point
A vector, in N kg⁻¹, pointing the way the force acts

Why “small”?

A test mass

It must not disturb the mass that produces the field

Same quantity, two names

N kg⁻¹ = m s⁻²

The field strength at a point equals the acceleration of free fall there

D.1 GRAVITATIONAL FIELDS · SL + HL

Field strength round a planet

The field of a point mass or sphere.

\(g = \dfrac{F}{m} = G \dfrac{M}{r^{2}}\)

\(M\): the mass that produces the field
An inverse square law again
Valid at the surface and outside it

THE EARTH’S FIELD STRENGTH

D.1 GRAVITATIONAL FIELDS · SL + HL

Where the equation comes from

Put a test mass \(m\) at distance \(r\) from \(M\). The force on it is \(G \frac{M m}{r^{2}}\). Divide by \(m\) and the test mass disappears, leaving \(G \frac{M}{r^{2}}\): the field depends only on the mass that makes it and on how far away you are.

Comparing planets

Surface gravity depends on both mass and radius. Mars has 0.107 of the Earth’s mass and 0.532 of its radius, so \(g_\mathrm{Mars} = 9.8 \times \frac{0.107}{0.532^{2}} = 3.7 \,\mathrm{N} \,\mathrm{kg}^{-1}\). A planet with less mass can still have strong surface gravity if it is small enough.

Thinking \(g\) is zero in orbit. At the height of the space station it is still about 8.7 N kg⁻¹, nearly 90% of its value on the ground.

Drawing a field

Gravitational field lines.

An arrow shows the direction of the force on a mass: always towards the mass that makes the field
Lines closer together: a stronger field
Lines never cross
Close to the surface the field is nearly uniform

TWO FIELDS TO KNOW

D.1 GRAVITATIONAL FIELDS · SL + HL

Why a field near the ground counts as uniform

The lines really do point to the center of the Earth, 6370 km below. Across a room, or even across a city, they are so nearly parallel, and the distance to the center changes so little, that \(g\) has the same size and direction everywhere. Climbing 1.0 km changes \(g\) by only 0.03%.

Why lines cannot cross

A field line shows the direction of the force on a mass at that point. If two lines crossed, a mass there would have two different force directions at once. Where the fields of two bodies overlap, they add as vectors to give one resultant direction.

When you sketch a radial field, use a ruler, space the lines evenly round the body, make them meet the surface at right angles and put an arrow on every one.

Worked example

The Earth’s field, on the ground and in orbit.

\(M = 5.97 \times 10^{24} \,\mathrm{kg}\) · radius \(R = 6.37 \times 10^{6} \,\mathrm{m}\) \(g = G \dfrac{M}{r^{2}}\) At the surface: \(r = R\) \(g = \dfrac{6.67 \times 10^{-11} \times 5.97 \times 10^{24}}{(6.37 \times 10^{6})^{2}} = \class{res}{9.81 \,\mathrm{N} \,\mathrm{kg}^{-1}}\)

400 km up: \(r = R + h = 6.77 \times 10^{6} \,\mathrm{m}\) \(g = \dfrac{6.67 \times 10^{-11} \times 5.97 \times 10^{24}}{(6.77 \times 10^{6})^{2}}\) \(g = \class{res}{8.69 \,\mathrm{N} \,\mathrm{kg}^{-1}}\)

D.1 GRAVITATIONAL FIELDS · SL + HL

Adding fields

The resultant field of two bodies.

Field strengths add as vectors
On the line between two bodies they point opposite ways: subtract

Zero resultant field at P \(G \dfrac{M_\mathrm{E}}{x^{2}} = G \dfrac{M_\mathrm{M}}{(d - x)^{2}}\) \(\dfrac{x}{d - x} = \sqrt{\dfrac{M_\mathrm{E}}{M_\mathrm{M}}} = \sqrt{81.2} = 9.01\) \(x = \dfrac{9.01}{10.01} d = \class{res}{0.90 d}\)

BETWEEN THE EARTH AND THE MOONNOT TO SCALE

D.1 GRAVITATIONAL FIELDS · SL + HL

Why the point is so close to the Moon

The Earth has 81 times the mass of the Moon. For the two fields to be equal, the Moon’s smaller mass must be made up for by a shorter distance, and because of the inverse square law the distances are in the ratio \(\sqrt{81} = 9\) to 1. The point is nine tenths of the way to the Moon, about 346 000 km from the Earth.

Other points on the line

Anywhere else between the two bodies, work out each field strength, then subtract the smaller from the larger. The resultant points towards the body whose field wins. Beyond either body, both fields point the same way, back towards the pair, so add them.

Take the square root of both sides before you rearrange. It turns a quadratic equation into a one-line ratio.

Only points on the line joining the two bodies are examined.

Quick check

Gravitational fields.

01Define gravitational field strength.

Force per unit mass on a small point mass

02A 3.0 kg mass feels 11.1 N on Mars. Field strength?

\(\frac{11.1}{3.0} = 3.7 \,\mathrm{N} \,\mathrm{kg}^{-1}\)

03\(g\) is 9.8 N kg⁻¹ at the surface. What is it at \(r = 2R\)?

\(\frac{9.8}{4} = 2.5 \,\mathrm{N} \,\mathrm{kg}^{-1}\)

04What do field lines that spread apart tell you?

The field is getting weaker

05Describe the field lines of a uniform field.

Parallel and evenly spaced

06Which way do gravitational field lines point?

Towards the mass that produces the field

PRESS → TO REVEAL EACH ANSWER

Part 04 · Additional higher level

Energy and orbits.

Potential energy and potential · equipotentials
Orbital speed · escape speed · atmospheric drag

D.1 GRAVITATIONAL FIELDS · ADDITIONAL HL

Energy of two masses

Gravitational potential energy.

\(E_\mathrm{p} = -G \dfrac{m_{1} m_{2}}{r}\)

The work done to assemble the system from infinite separation
Zero when the masses are infinitely far apart
Negative everywhere else, because gravity attracts

A MASS NEAR A PLANET

D.1 GRAVITATIONAL FIELDS · ADDITIONAL HL

Why it is negative

Gravity only attracts. As two masses come together from far apart, the field does work on them and they speed up: they gain kinetic energy, so they must have lost potential energy. Starting from zero and losing energy leaves a negative amount. To separate them again you would have to give that energy back. Negative potential energy alone does not make a system bound: a body moving fast enough has enough kinetic energy to escape. It is bound only if its total energy is negative, as in an orbit.

How this fits with \(mg \Delta h\)

Raise a mass from \(r\) to \(r + \Delta h\) near the surface. The change in \(-\frac{G M m}{r}\) is \(G M m \left(\frac{1}{r} - \frac{1}{r + \Delta h}\right) \approx \frac{G M m}{r^{2}} \Delta h = mg \Delta h\). The equation from A.3 is the short-distance version of this one, valid while \(g\) hardly changes.

Squaring \(r\). Force and field strength have \(r^{2}\) underneath. Energy and potential have \(r\).

“Higher” always means “less negative”. Moving away from a planet increases the potential energy towards zero.

Energy per unit mass

Gravitational potential.

\(V_\mathrm{g} = -G \dfrac{M}{r}\)

The work done per unit mass in bringing a mass from infinity to the point
A scalar, in J kg⁻¹. Zero at infinity
Potential energy of a mass \(m\) there: \(E_\mathrm{p} = m V_\mathrm{g}\)

THE EARTH’S POTENTIAL

D.1 GRAVITATIONAL FIELDS · ADDITIONAL HL

Potential is to energy what field strength is to force

Field strength is force per kilogram. Potential is potential energy per kilogram. Both describe the place, not the body you put there. The potential at the Earth’s surface, −62.5 MJ kg⁻¹, means that every kilogram resting on the ground has 62.5 MJ less potential energy than it would have far out in space.

Adding potentials is easy

Potential is a scalar, so the potentials of two bodies simply add as numbers, with no directions to think about. At a point between the Earth and the Moon the total potential is the sum of two negative numbers. This is one reason energy methods are often simpler than force methods.

Giving potential a direction. It has a sign but no direction: the minus sign means “lower than at infinity”, not “downwards”.

Worked example

Lifting a 500 kg satellite.

From the surface, \(R = 6.37 \times 10^{6} \,\mathrm{m}\), to \(r = 2R\) \(W = m \Delta V_\mathrm{g}\) with \(V_\mathrm{g} = -\dfrac{G M}{r}\) \(V_\mathrm{surface} = -\dfrac{3.98 \times 10^{14}}{6.37 \times 10^{6}} = -6.25 \times 10^{7} \,\mathrm{J} \,\mathrm{kg}^{-1}\) \(V_{2R} = -3.13 \times 10^{7} \,\mathrm{J} \,\mathrm{kg}^{-1}\) \(W = 500 \times (-3.13 + 6.25) \times 10^{7} = \class{res}{1.56 \times 10^{10} \,\mathrm{J}}\)

\(G M = 3.98 \times 10^{14} \,\mathrm{N} \,\mathrm{m}^{2} \,\mathrm{kg}^{-1}\) for the Earth
\(\Delta V_\mathrm{g}\) = final − initial: here it is positive
The work does not depend on the path taken

D.1 GRAVITATIONAL FIELDS · ADDITIONAL HL

Why \(mg \Delta h\) fails here

With \(g = 9.81 \,\mathrm{N} \,\mathrm{kg}^{-1}\) and \(\Delta h = 6.37 \times 10^{6} \,\mathrm{m}\) it gives \(3.1 \times 10^{10} \,\mathrm{J}\), twice the right answer. The field strength falls to a quarter on the way up, so the average force is much less than the weight at the ground. Potential takes this into account automatically.

What the answer does not include

This is only the energy needed to raise the satellite. To stay at that height it must also be moving at the orbital speed, which needs kinetic energy as well. Later slides add the two together.

The sign tells you who does the work. Moving outwards, \(\Delta V_\mathrm{g}\) is positive and an engine must supply the energy. Moving inwards, it is negative and the field supplies it.

Linking the two descriptions

Field strength is the potential gradient.

\(g = -\dfrac{\Delta V_\mathrm{g}}{\Delta r}\)

The gradient of a potential–distance graph gives the field strength
Steep graph: strong field
The minus sign: the field points towards lower potential

THE TANGENT AT 2R

D.1 GRAVITATIONAL FIELDS · ADDITIONAL HL

Why the gradient is the field

Move a 1 kg mass outwards by a small distance \(\Delta r\). You push against a force of \(g\) newtons, so you do work \(g \Delta r\), and that is the gain in potential. So the size of \(g\) is \(\frac{\Delta V_\mathrm{g}}{\Delta r}\): the potential gained per meter. A contour map works the same way: where the height changes quickly over a short distance, the hill is steep.

Checking with numbers

Between \(r = 6.37 \times 10^{6} \,\mathrm{m}\) and \(6.47 \times 10^{6} \,\mathrm{m}\), the potential rises from −62.5 to −61.5 MJ kg⁻¹. The gradient is \(\frac{1.0 \times 10^{6}}{1.0 \times 10^{5}} = 10 \,\mathrm{N} \,\mathrm{kg}^{-1}\), close to the 9.8 expected near the surface.

On a graph of \(g\) against \(r\), the area under the curve between two distances is the change in potential. Gradient one way, area the other.

Mapping a field with potential

Equipotential surfaces.

A surface on which the potential is the same everywhere
No work is done moving a mass along one
Field lines cross them at right angles
Equal steps of potential: surfaces closer together where the field is stronger

ROUND THE EARTHSTEPS OF 10 MJ kg⁻¹

D.1 GRAVITATIONAL FIELDS · ADDITIONAL HL

Why the right angle

If a field line met an equipotential at any other angle, the field would have a component along the surface. Moving a mass along the surface would then need work, and the potential would change, which contradicts the surface being an equipotential. So the field can have no component along it.

Reading the map

These are the contour lines of gravity. Round a spherical planet they are spheres centered on the planet. In the uniform field near the ground they are horizontal planes, evenly spaced: every meter you climb adds the same 9.8 J kg⁻¹. A satellite in a circular orbit stays on one equipotential, which is why gravity does no work on it and its speed stays constant.

Given equipotentials, you can draw the field lines: at right angles to every surface, pointing from higher potential to lower. Given field lines, you can sketch the equipotentials the same way.

The story · published 1728

Background story · not assessed

Newton’s cannon on a mountain.

Engraving of the Earth with a mountain at the top. Curved paths leave the mountain top: short ones fall back to the ground, longer ones reach further round, and the longest go all the way round the Earth — From **A Treatise of the System of the World**, 1728

Fire a cannonball horizontally from a very high mountain
Faster shots land further away, as the ground curves away beneath them
Fast enough, and the ball falls all the way round without ever landing
Printed the year after Newton died. The first artificial satellite flew 229 years later

An orbit is free fall that keeps missing the ground.

Engraving: Isaac Newton, 1728, public domain.

D.1 GRAVITATIONAL FIELDS · ADDITIONAL HL

Staying in a circular orbit

Orbital speed.

\(G \dfrac{M m}{r^{2}} = \dfrac{m v^{2}}{r}\) \(v_\mathrm{orbital} = \sqrt{\dfrac{G M}{r}}\)

One speed for each radius. The mass of the satellite cancels
A larger orbit needs a lower speed

TWO CIRCULAR ORBITS

D.1 GRAVITATIONAL FIELDS · ADDITIONAL HL

A worked case

For the space station, \(r = 6.77 \times 10^{6} \,\mathrm{m}\): \(v = \sqrt{\frac{3.98 \times 10^{14}}{6.77 \times 10^{6}}} = 7.67 \times 10^{3} \,\mathrm{m} \,\mathrm{s}^{-1}\). One orbit is \(2 \pi r = 4.25 \times 10^{7} \,\mathrm{m}\), so the period is \(\frac{4.25 \times 10^{7}}{7.67 \times 10^{3}} = 5540 \,\mathrm{s}\), about 92 minutes. The crew see sixteen sunrises a day.

Why there is no choice of speed

At a given radius gravity supplies a fixed centripetal acceleration, \(\frac{G M}{r^{2}}\). Circular motion at that radius needs \(\frac{v^{2}}{r}\). Only one speed makes the two match. Go faster and gravity cannot bend the path tightly enough: the orbit becomes an ellipse reaching further out. Go slower and the path drops inside the circle.

Thinking a higher orbit needs a higher speed. Further out the pull is weaker, so less speed is needed. But reaching a higher orbit still takes more energy.

Kinetic, potential and total

The energy of a satellite in orbit.

Kinetic: \(E_\mathrm{k} = \dfrac{1}{2} m v^{2} = \dfrac{G M m}{2r}\)
Potential: \(E_\mathrm{p} = -\dfrac{G M m}{r}\)
Total: \(E = E_\mathrm{k} + E_\mathrm{p} = -\dfrac{G M m}{2r}\)

A larger orbit has more total energy but less kinetic energy.

ENERGY AGAINST ORBITAL RADIUS

D.1 GRAVITATIONAL FIELDS · ADDITIONAL HL

Where the kinetic energy comes from

Put \(v^{2} = \frac{G M}{r}\) from the last slide into \(\frac{1}{2} m v^{2}\). The kinetic energy is always exactly half the size of the potential energy, so the total is negative and equal to minus the kinetic energy.

What a negative total means

The satellite is bound. It does not have enough kinetic energy to climb out to infinity, where its potential energy would be zero. To free it you would have to supply \(\frac{G M m}{2r}\), the size of its total energy.

The apparent paradox

To move a satellite to a higher orbit you must give it energy, yet it ends up slower. The potential energy rises by twice as much as the kinetic energy falls, so the total still goes up.

These three results are not in the data booklet. Derive them from \(v_\mathrm{orbital}\) and \(E_\mathrm{p}\), which are.

Worked example

Raising a 500 kg satellite to a higher orbit.

From \(r_{1} = 6.77 \times 10^{6} \,\mathrm{m}\) to \(r_{2} = 4.22 \times 10^{7} \,\mathrm{m}\) \(E = -\dfrac{G M m}{2r}\) \(\Delta E = \dfrac{G M m}{2} \left(\dfrac{1}{r_{1}} - \dfrac{1}{r_{2}}\right)\) \(\Delta E = \dfrac{3.98 \times 10^{14} \times 500}{2} \left(\dfrac{1}{6.77 \times 10^{6}} - \dfrac{1}{4.22 \times 10^{7}}\right)\) \(\Delta E = \class{res}{1.23 \times 10^{10} \,\mathrm{J}}\)

	Low orbit	High orbit
\(E_\mathrm{k}\) / GJ	+14.7	+2.4
\(E_\mathrm{p}\) / GJ	−29.4	−4.7
Total / GJ	−14.7	−2.4

Kinetic energy falls by 12.3 GJ. Potential energy rises by 24.7 GJ.

D.1 GRAVITATIONAL FIELDS · ADDITIONAL HL

Reading the table

In each column the potential energy is minus twice the kinetic energy and the total is minus the kinetic energy, as the last slide showed. Check that pattern in your own answers: if it fails, a sign has gone wrong.

Where the energy comes from

The rocket engine does 12.3 GJ of work on the satellite. The speeds follow from \(v = \sqrt{\frac{G M}{r}}\): 7.67 km s⁻¹ in the low orbit and 3.07 km s⁻¹ in the high one, which is the orbit of a geostationary satellite. The engine fires forwards to speed the satellite up, and it still arrives moving more slowly, because it loses speed as it climbs.

Finding only the change in potential energy. A change of orbit changes the kinetic energy as well: use the total.

Leaving for good

Escape speed.

Just escapes: total energy zero \(\dfrac{1}{2} m v_\mathrm{esc}^{2} - \dfrac{G M m}{r} = 0\) \(v_\mathrm{esc} = \sqrt{\dfrac{2 G M}{r}}\) From the Earth’s surface \(\sqrt{\dfrac{2 \times 3.98 \times 10^{14}}{6.37 \times 10^{6}}} = \class{res}{11.2 \,\mathrm{km} \,\mathrm{s}^{-1}}\)

FOUR LAUNCH SPEEDS

D.1 GRAVITATIONAL FIELDS · ADDITIONAL HL

What escape speed is

It is the least speed a body needs at a given point so that, with no engine and no further push, it never falls back. It slows all the way out, but gravity weakens so quickly with distance that the body never quite stops. The direction of launch does not matter, as long as it misses the planet: energy is a scalar.

How it compares with orbital speed

At the same radius, \(v_\mathrm{esc} = \sqrt{2} \times v_\mathrm{orbital}\). A satellite already in a circular orbit has half the kinetic energy it would need to escape. Increase its speed by 41% and it leaves.

Thinking a rocket must reach 11.2 km s⁻¹ to leave the Earth. Escape speed is for a body that coasts. A rocket whose engine keeps firing could leave at walking pace.

The mass of the escaping body cancels: the escape speed is the same for a molecule and for a spacecraft.

Worked example

From the launch pad to a low orbit.

A 500 kg satellite, at rest on a planet that does not rotate \(R = 6.37 \times 10^{6} \,\mathrm{m}\) · orbit \(r = 6.77 \times 10^{6} \,\mathrm{m}\) \(E_\mathrm{surface} = -\dfrac{G M m}{R}\) · \(E_\mathrm{orbit} = -\dfrac{G M m}{2r}\) \(\Delta E = G M m \left(\dfrac{1}{R} - \dfrac{1}{2r}\right)\) \(\Delta E = 3.98 \times 10^{14} \times 500 \left(\dfrac{1}{6.37 \times 10^{6}} - \dfrac{1}{1.354 \times 10^{7}}\right)\) \(\Delta E = \class{res}{1.65 \times 10^{10} \,\mathrm{J}}\)

On the pad: potential energy only
In orbit: potential and kinetic
Most of the energy goes into speed, not height

To escape from that orbit: a further \(\dfrac{G M m}{2r} = 1.47 \times 10^{10} \,\mathrm{J}\).

D.1 GRAVITATIONAL FIELDS · ADDITIONAL HL

Speed, not height

Split the 16.5 GJ into its parts. Raising the satellite 400 km increases its potential energy by only 1.8 GJ. Giving it the orbital speed of 7.67 km s⁻¹ takes 14.7 GJ. Getting into orbit is mostly about going sideways very fast.

Why “a planet that does not rotate”

On the real Earth a satellite on the launch pad at the equator is already moving east at 465 m s⁻¹ with the ground. It starts with some kinetic energy, which is why rockets are launched eastwards from sites near the equator. The exam model leaves this out, so the starting kinetic energy is zero.

The energy to escape from an orbit is minus its total energy: the amount that brings the total up to zero.

A small resistive force

A satellite in the thin upper atmosphere.

A small viscous drag does negative work: the total energy falls
Total energy \(-\dfrac{G M m}{2r}\) falls, so \(r\) gets smaller
\(v = \sqrt{\dfrac{G M}{r}}\), so the speed rises

Drag makes the satellite go faster.

A SLOWLY DECAYING ORBITMUCH EXAGGERATED

D.1 GRAVITATIONAL FIELDS · ADDITIONAL HL

How a resistive force can speed something up

As the orbit shrinks, the satellite falls towards the Earth and loses potential energy. Half of the potential energy lost becomes kinetic energy, and the other half is transferred to the atmosphere and the satellite as thermal energy by the drag. The kinetic energy rises, the potential energy falls by twice as much, and the total falls.

Why it gets worse

Lower down, the air is denser and the satellite is faster, so the drag is larger and the orbit decays more quickly. Each orbit is very nearly a circle, but the height falls faster and faster until the satellite burns up. The space station loses about 2 km of height a month and has to be boosted by rocket engines.

A model answer has three steps: total energy decreases, so the radius decreases, so the speed increases. Quote the equation that justifies each step.

Only this qualitative argument is required.

Quick check · additional higher level

Energy and orbits.

01Where is gravitational potential zero?

At infinity

02The potential at a point is −40 MJ kg⁻¹. Potential energy of 2.0 kg there?

\(2.0 \times (-40) = -80 \,\mathrm{MJ}\)

03How much work is done moving a mass along an equipotential?

None

04What does the gradient of a potential–distance graph give?

The field strength (with a minus sign)

05The orbital speed at some radius is 5.0 km s⁻¹. Escape speed there?

\(\sqrt{2} \times 5.0 = 7.1 \,\mathrm{km} \,\mathrm{s}^{-1}\)

06Drag acts on a satellite. What happens to its speed?

It increases, as the orbit shrinks

PRESS → TO REVEAL EACH ANSWER

Summary

Gravitational fields on one slide.

Kepler’s laws

How orbits behave

Ellipses with the Sun at a focus. Equal areas in equal times
\(T^{2} \propto r^{3}\)

Newton’s law

An inverse square force

\(F = G \frac{m_{1} m_{2}}{r^{2}}\), between centers
Gravity is the centripetal force of an orbit

Field strength

Force per unit mass

\(g = \frac{F}{m} = G \frac{M}{r^{2}}\), a vector
Radial round a planet, uniform near its surface

Potential · HL

Energy per unit mass

\(V_\mathrm{g} = -\frac{G M}{r}\), zero at infinity
\(W = m \Delta V_\mathrm{g}\) and \(g = -\frac{\Delta V_\mathrm{g}}{\Delta r}\)

Equipotentials · HL

At right angles to the field

No work is done along one
Closer together where the field is stronger

Orbits · HL

Two speeds

\(v_\mathrm{orbital} = \sqrt{\frac{G M}{r}}\) and \(v_\mathrm{esc} = \sqrt{\frac{2 G M}{r}}\)
Drag: smaller orbit, higher speed

D.1 GRAVITATIONAL FIELDS · SL + HL

Summary

Key words.

EllipseThe shape of an orbit

Orbital periodThe time for one complete orbit

Point massSmall compared with the separation

Inverse square lawForce \(\propto \dfrac{1}{r^{2}}\)

Gravitational field strengthForce per unit mass. A vector

Field lineDirection of the force on a mass

Uniform fieldSame strength and direction everywhere

Gravitational potentialWork done per unit mass from infinity

Potential gradientPotential change per meter

Equipotential surfaceJoins points at the same potential

Orbital speedSpeed for a circular orbit

Escape speedThe least speed to reach infinity

D.1 GRAVITATIONAL FIELDS · SL + HL