Motion in Gravitational Fields

How one inverse square law sets the field around a planet, the speed and period of every orbit, and the energy needed to change orbit or escape altogether.

65 min · Module 5: Advanced Mechanics

Assumes you have read Circular Motion.

The full teaching. Start here.

Syllabus requirementsModule 5: Advanced Mechanics

apply qualitatively and quantitatively Newton’s Law of Universal Gravitation to:
determine the force of gravity between two objects $F = \frac{GM m}{r ^{2}}$
investigate the factors that affect the gravitational field strength $g = \frac{GM}{r ^{2}}$
predict the gravitational field strength at any point in a gravitational field, including at the surface of a planet (ACSPH094, ACSPH095, ACSPH097)ACSPH094 ACSPH095 ACSPH097
investigate the orbital motion of planets and artificial satellites when applying the relationships between the following quantities:
gravitational force
centripetal force
centripetal acceleration
mass
orbital radius
orbital velocity
orbital period
predict quantitatively the orbital properties of planets and satellites in a variety of situations, including near the Earth and geostationary orbits, and relate these to their uses (ACSPH101)ACSPH101
investigate the relationship of Kepler’s Laws of Planetary Motion to the forces acting on, and the total energy of, planets in circular and non-circular orbits using: (ACSPH101)ACSPH101
$v = \frac{2 π r}{T}$
$\frac{r ^{3}}{T ^{2}} = \frac{GM}{4 π ^{2}}$
derive quantitatively and apply the concepts of gravitational force and gravitational potential energy in radial gravitational fields to a variety of situations, including but not limited to:
the concept of escape velocity $v_{esc} = \frac{2 GM}{r}$
total potential energy of a planet or satellite in its orbit $U = - \frac{GM m}{r}$
total energy of a planet or satellite in its orbit $U + K = - \frac{GM m}{2 r}$
energy changes that occur when satellites move between orbits (ACSPH096)ACSPH096
Kepler’s Laws of Planetary Motion (ACSPH101)ACSPH101

What this actually asks for

This is the longest topic in Module 5, and almost all of it grows from one equation. Newton's law of universal gravitation gives the field. Setting that force equal to the centripetal force from the last lesson gives every orbital relationship. Adding energy gives escape velocity and orbit changes.

Read the verbs. Dot point 1 says apply qualitatively and quantitatively, so you need both numbers and explanations. Dot point 5 says derive quantitatively, so the results for escape velocity and orbital energy must be shown from first principles, not just quoted. Kepler's third law also appears in derivation questions often enough that the four line proof should be automatic.

The short version

Every pair of masses attracts with a force

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

where $r$ is measured between centres. Dividing by the small mass gives the field strength, the force on each kilogram:

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

For a circular orbit, gravity is the centripetal force. Setting $GM m / r^{2} = m v^{2} / r$ gives

$v = \frac{GM}{r} \frac{r ^{3}}{T ^{2}} = \frac{GM}{4 π ^{2}}$

The satellite's mass cancels, so it has no effect on the orbit. Higher orbits are slower and take longer.

Potential energy in a radial field is $U = - GM m / r$ , zero at infinity. In a circular orbit $K = GM m /2 r$ , so the total is

$E = - \frac{GM m}{2 r}$

Raising a satellite increases its total energy but lowers its speed. Escape needs a total energy of zero, which gives $v_{esc} = 2 GM / r$ .

The fastest way to lose marks: using altitude in place of distance from the centre.

Newton's law of universal gravitation

One force for apples and moons

AnswersWhy should the force that drops an apple also hold the Moon in orbit?

Newton's insight was that the Moon is falling too. Left alone it would travel in a straight line; instead it curves around the Earth, so it is always falling towards the Earth, just never getting closer, because it is also moving sideways fast enough to keep missing.

He then asked how strongly the Earth pulls at the Moon's distance. The Moon's centripetal acceleration, worked out from its orbit, is about $1/3600$ of $g$ . The Moon is about $60$ Earth radii away, and $6 0^{2} = 3600$ . The pull falls as the inverse square of the distance from the Earth's centre.

The same law holds between any two masses. The force is proportional to each mass, inversely proportional to the square of the separation, and always attractive. By Newton's third law, the Earth pulls the Moon exactly as hard as the Moon pulls the Earth. The Earth just responds far less, because it is eighty times more massive.

Newton's law of universal gravitationon the NESA formulae sheet

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

Symbols: $F$ size of the attractive force each mass exerts on the otherN
$G$ universal gravitational constant, 6.67 × 10⁻¹¹N m² kg⁻²
$M$ one of the two masses, usually the largerkg
$m$ the other masskg
$r$ distance between the centres of the two massesm
Valid when: The two bodies are point masses or spheres whose mass is spread symmetrically, and r is measured between their centres. Newton's third law applies: each body feels the same size of force.
Not valid when: r is taken as an altitude above a surface rather than a distance from the centre, or the bodies are irregular and close together compared with their size.
Rearranged: for r: $r = \frac{GM m}{F}$ for M: $M = \frac{F r ^{2}}{G m}$
Where it turns up: Finding the pull between a planet and its moon, or between the Sun and a planet
Predicting how the force changes when a mass or a separation is scaled
Setting gravity equal to the centripetal force to analyse an orbit
Where marks go missing: Using the altitude above the surface in place of the distance from the centre
Forgetting to square the distance
Believing the larger mass pulls harder on the smaller one than the smaller pulls on the larger

CheckpointAnswer before reading on.

The Moon has a mass of $7.35 \times 1 0^{22}$ kg and its centre is $3.84 \times 1 0^{8}$ m from the centre of the Earth. The Earth's mass is $6.0 \times 1 0^{24}$ kg.

Calculate the size of the gravitational force between the Earth and the Moon.

Your answerN

Give it to 3 significant figures.

Hint 1Newton's law of universal gravitation uses both masses and the distance between centres.

Hint 2The distance is squared.

Hint 3 $F = \frac{6.67 \times 1 0 ^{- 11} \times 6.0 \times 1 0 ^{24} \times 7.35 \times 1 0 ^{22}}{( 3.84 \times 1 0 ^{8} ) ^{2}}$

CheckpointAnswer before reading on.

Two masses attract each other with a gravitational force $F$ . One of the masses is doubled and the distance between their centres is tripled.

What is the new force?

A $\frac{2}{9} F$
B $\frac{2}{3} F$
C $18 F$
D $\frac{4}{9} F$

Hint 1Treat each change as a separate factor on the force.

Hint 2Doubling one mass doubles the force.

Hint 3Tripling the distance divides the force by three squared.

The gravitational field

A field describes what a mass would feel at each point, before any mass is placed there. The gravitational field strength is the force per kilogram, and dividing Newton's law by the small mass $m$ removes it:

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

The same number is the acceleration of free fall at that point, which is why its units can be written either as N kg⁻¹ or m s⁻².

Gravitational field strengthon the NESA formulae sheet

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

Symbols: $g$ gravitational field strength, the force per kilogram at that point, equal to the free fall accelerationN kg⁻¹
$G$ universal gravitational constantN m² kg⁻²
$M$ mass of the body producing the fieldkg
$r$ distance from the centre of that body to the pointm
Valid when: The point is outside a spherically symmetric body, or anywhere around a point mass. It follows from dividing the universal gravitation force by the test mass m.
Not valid when: The point is inside the body, where only the mass closer to the centre contributes, or r is taken from the surface instead of the centre.
Rearranged: for M: $M = \frac{g r ^{2}}{G}$ for r: $r = \frac{GM}{g}$
Where it turns up: Predicting the surface gravity of another planet from its mass and radius
Finding the field at a satellite's altitude
Comparing fields by ratio without calculating G M, since g is proportional to M over r squared
Where marks go missing: Adding the altitude to nothing, using it alone as r
Thinking g is zero in orbit because astronauts float
Using the mass of the object in the field rather than the mass producing it

Where this comes from

Three factors control it, which is what dot point 1 asks you to investigate. More mass means a stronger field, in direct proportion. Greater distance from the centre means a weaker field, as the inverse square. And for a surface value, the planet's radius matters as much as its mass: a small dense planet can have stronger surface gravity than a larger, lighter one.

Field strength

Altitude is measured from the surface, but the law uses distance from the centre. At 400 km up the field is still almost 90 per cent of its surface value.

Planet mass1.0 Earth masses

Planet radius1.00 Earth radii

Altitude marked400 km

Surface field strength: 9.86 N kg⁻¹
Field at the marked altitude: 8.73 N kg⁻¹
As a fraction of the surface value: 88.5 per cent
Distance from the centre there: 6.771 × 10⁶ m

Every slider is a normal range input, so the arrow keys move it one step and Home and End jump to the extremes.

CheckpointAnswer before reading on.

Mars has a mass of $6.42 \times 1 0^{23}$ kg and a radius of $3.39 \times 1 0^{6}$ m.

Calculate the gravitational field strength at the surface of Mars.

Your answerN kg⁻¹

Give it to 3 significant figures.

Hint 1Field strength is the force on each kilogram, so the mass of the object does not appear.

Hint 2At the surface, the distance from the centre is the planet's radius.

Hint 3 $g = \frac{6.67 \times 1 0 ^{- 11} \times 6.42 \times 1 0 ^{23}}{( 3.39 \times 1 0 ^{6} ) ^{2}}$

Worked example4 marks

The field and the force on a satellite

A $500$ kg satellite orbits $2000$ km above the Earth's surface. Using $M = 6.0 \times 1 0^{24}$ kg and $R = 6.371 \times 1 0^{6}$ m, find the gravitational field strength at the satellite and the gravitational force on it. Compare the field with its value at the surface.

Find the distance from the centre

The law of gravitation measures r between centres. Altitude is from the surface, so the radius must be added.

$r = 6.371 \times 1 0^{6} + 2.000 \times 1 0^{6} = 8.371 \times 1 0^{6} m$

The field strength

The field depends only on the Earth's mass and the distance, so it can be found before the satellite's mass is used.

$g = \frac{GM}{r ^{2}} = \frac{6.67 \times 1 0 ^{- 11} \times 6.0 \times 1 0 ^{24}}{( 8.371 \times 1 0 ^{6} ) ^{2}} = 5.71 N kg^{- 1}$

The force follows from the field

Force is field times mass. This is quicker and less error prone than substituting into the full law again.

$F = m g = 500 \times 5.71 = 2.86 \times 1 0^{3} N$

Compare by ratio

An inverse square comparison can be done without G or M at all, which makes a useful check.

$\frac{g _{orbit}}{g _{surface}} = (\frac{R}{r})^{2} = (\frac{6.371}{8.371})^{2} = 0.58$

Answer

The field is $5.7$ N kg⁻¹ and the force is $2.9 \times 1 0^{3}$ N, towards the Earth's centre. That is $58$ per cent of the surface value.

Is that answer sensible?

The satellite is only about a third of an Earth radius above the surface, so the field should be weaker but the same order as at the surface. With these data values, the surface field works out to $9.86$ N kg⁻¹ rather than exactly $9.8$ , because $M$ is rounded to two figures. $0.58 \times 9.86 = 5.71$ , which matches the direct calculation.

CheckpointAnswer before reading on.

Using the Earth's mass of $6.0 \times 1 0^{24}$ kg and radius of $6.371 \times 1 0^{6}$ m, calculate the gravitational field strength at an altitude of $420$ km.

Your answerN kg⁻¹

Give it to 3 significant figures.

Hint 1The law uses the distance from the Earth's centre, not the height above the surface.

Hint 2 $r = 6.371 \times 1 0^{6} + 4.20 \times 1 0^{5}$ m.

Hint 3 $r = 6.791 \times 1 0^{6}$ m. Now use $g = GM / r^{2}$ .

CheckpointAnswer before reading on.

Astronauts on the International Space Station, about $420$ km above the Earth's surface, float freely inside it.

Which statement best explains this?

AThe astronauts and the station are falling around the Earth together with the same acceleration, so nothing needs to support the astronauts.
BAt the station's altitude the Earth's gravitational field is too weak to have a noticeable effect.
CThe station is above the atmosphere, and gravity needs air to act through.
DThe centrifugal force from the orbit balances gravity, so the net force on the astronauts is zero.

Hint 1Compare 420 km with the Earth's radius of 6371 km.

Hint 2What does gravity do to the station and to the astronaut?

Hint 3If both are accelerating towards the Earth at the same rate, what does the floor push with?

Orbits: gravity as the centripetal force

A satellite in a circular orbit is in uniform circular motion, and the only force on it is gravity. So gravity is the centripetal force. Every orbital relationship in dot point 2 comes from writing that sentence as an equation.

Derivation

Orbital speed and Kepler's third law

Starts from

gravity supplying the centripetal force for a circular orbit

Ends at

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

Holds only if

The orbit is circular
Gravity from the central body is the only force
The central mass is much larger than the orbiting mass, so it stays still

Orbital speed and Kepler's third law

4 steps

1
Set gravity equal to the centripetal force
This is the whole physics of the orbit. Everything after it is algebra.
$\frac{GM m}{r ^{2}} = \frac{m v ^{2}}{r}$
2
Solve for the speed
The satellite's mass m appears on both sides and cancels. That cancellation is itself examinable.
$v^{2} = \frac{GM}{r} \Rightarrow v = \frac{GM}{r}$
3
Bring in the period
An orbit covers one circumference per period, which links the speed to the period.
$(\frac{2 π r}{T})^{2} = \frac{GM}{r} \Rightarrow \frac{4 π ^{2} r ^{2}}{T ^{2}} = \frac{GM}{r}$
4
Rearrange
Collecting r on one side and T on the other gives a ratio that is the same for every body orbiting the same central mass.
$\frac{r ^{3}}{T ^{2}} = \frac{GM}{4 π ^{2}}$

Speed of a circular orbitnot on the formulae sheet, learn it

v = \frac{GM}{r}

Symbols: $v$ speed of the orbiting bodym s⁻¹
$G$ universal gravitational constantN m² kg⁻²
$M$ mass of the central body being orbitedkg
$r$ orbital radius, measured from the centre of the central bodym
Valid when: The orbit is circular and gravity from the central body is the only force, so gravity supplies the whole centripetal force: GMm/r² = mv²/r.
Not valid when: The orbit is elliptical, where the speed changes around the path, or other forces such as atmospheric drag are significant.
Rearranged: for r: $r = \frac{GM}{v ^{2}}$ for M: $M = \frac{v ^{2} r}{G}$
Where it turns up: Finding the speed of a satellite at a given altitude
Showing that orbital speed does not depend on the satellite's own mass
Showing that higher orbits are slower
Where marks go missing: Including the satellite's mass, which cancels
Using the altitude as r
Concluding that a higher orbit is faster because more energy was needed to reach it

Where this comes from

Kepler's third lawon the NESA formulae sheet

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

Symbols: $r$ orbital radius, or for an ellipse the semi major axism
$T$ orbital periods
$G$ universal gravitational constantN m² kg⁻²
$M$ mass of the central bodykg
Valid when: Several bodies orbit the same central mass, which is much heavier than any of them. The ratio r³/T² is then the same for all of them, which is what lets the central mass be measured.
Not valid when: The bodies being compared orbit different central masses, or the orbiting body's mass is comparable with the central mass, as for a binary star.
Rearranged: for T: $T = 2 π \frac{r ^{3}}{GM}$ for r: $r = 3 \frac{GM T ^{2}}{4 π ^{2}}$ for M: $M = \frac{4 π ^{2} r ^{3}}{G T ^{2}}$
Where it turns up: Finding the radius of a geostationary orbit from its period of one sidereal day
Measuring the mass of a planet from the orbit of one of its moons
Predicting the period of a planet from its distance to the Sun
Where marks go missing: Leaving the period in hours or days while using G in SI units
Taking a cube root where a square root is needed, or the reverse
Using the altitude of a satellite as r

Where this comes from

Read the results as relationships between the seven quantities the syllabus lists. The gravitational force and the centripetal force are the same force. The centripetal acceleration equals the field strength at that radius. The satellite's mass has no effect on anything. A larger orbital radius means a lower orbital speed and a longer orbital period, and the period grows faster than the radius, as $r^{3/2}$ .

Circular orbit speed

A satellite in a higher orbit moves more slowly and has further to go, so its period grows much faster than its radius.

The numbers behind this graph

Altitude (km)	Speed (km s⁻¹)	Period (h)
400	7.69	1.54
2000	6.91	2.11
10000	4.94	5.78
20200	3.88	11.95
35786	3.08	23.88

Central mass1.0 Earth masses

Altitude marked420 km

Orbital radius: 6.791 × 10⁶ m
Orbital speed: 7.68 km s⁻¹
Period: 1.54 h, or 93 min
Centripetal acceleration: 8.68 m s⁻²

Every slider is a normal range input, so the arrow keys move it one step and Home and End jump to the extremes.

CheckpointAnswer before reading on.

Two satellites orbit the Earth on the same circular orbit. Satellite A has four times the mass of satellite B.

How do their orbital speeds compare?

AThey have the same speed.
BA is faster, because the gravitational force on it is four times larger.
CA is slower, because a heavier satellite is harder to keep moving.
DA is twice as fast, because speed depends on the square root of the mass.

Hint 1Write the equation that sets gravity equal to the centripetal force.

Hint 2The satellite's mass appears on both sides.

Hint 3After cancelling, what does the speed depend on?

CheckpointAnswer before reading on.

An Earth observation satellite moves in a circular orbit $800$ km above the Earth's surface. The Earth's mass is $6.0 \times 1 0^{24}$ kg and its radius is $6.371 \times 1 0^{6}$ m.

Calculate the satellite's orbital speed.

Your answerm s⁻¹

Give it to 3 significant figures.

Hint 1Gravity is the only force on the satellite, so it is the centripetal force.

Hint 2Set $\frac{GM m}{r ^{2}} = \frac{m v ^{2}}{r}$ and solve for $v$ .

Hint 3 $v = GM / r$ , with $r = 7.171 \times 1 0^{6}$ m.

Exam question

Harder · about 4 min

3 marks

A satellite of mass $m$ moves in a circular orbit of radius $r$ around a planet of mass $M$ with period $T$ .

Show that $\frac{r ^{3}}{T ^{2}} = \frac{GM}{4 π ^{2}}$ .

Write your response, then mark it against the criteria. Writing it out first is the part that works.

Hint 1Which force provides the centripetal force?

Hint 2Set $\frac{GM m}{r ^{2}} = \frac{m v ^{2}}{r}$ .

Hint 3Substitute $v = \frac{2 π r}{T}$ and rearrange.

Where students lose marks on this one

Starting from the result and rearranging it into something else.
Why it happens: A show that question looks finished once the target appears.
Start from physics you can justify, here the force equation, and end at the target. Each line should follow from the one before.

Written for this site.

Satellites near the Earth and geostationary orbits

The third dot point asks for predictions about real orbits and their uses. Two families matter most.

Low Earth orbits sit a few hundred to about two thousand kilometres up, with periods of roughly ninety minutes to two hours and speeds near $7.5$ km s⁻¹. Being close gives the sharpest images, and on an inclined or polar orbit the Earth turns beneath the satellite so it eventually passes over everywhere. The cost is that any one place is overhead only briefly, and the thin upper atmosphere drags on the satellite, so it slowly loses height.

Geostationary orbits have a period of one rotation of the Earth, $23$ h $56$ min, lie above the equator and travel in the same direction as the Earth spins. The satellite therefore stays above one point, so a dish on the ground can be fixed in place. That makes the orbit ideal for communications and for continuous weather watching of one hemisphere. A period of one day fixes the radius, so every geostationary satellite sits at the same height, about $36000$ km up.

Worked example4 marks

A GPS satellite

GPS satellites orbit at an altitude of $20200$ km. Find their orbital speed and period, and explain why they cannot be used as geostationary satellites.

Orbital radius

Every orbital relationship uses distance from the Earth's centre.

$r = 6.371 \times 1 0^{6} + 2.02 \times 1 0^{7} = 2.657 \times 1 0^{7} m$

Speed from gravity as the centripetal force

The satellite's mass is not given and does not need to be, because it cancels.

$v = \frac{GM}{r} = \frac{6.67 \times 1 0 ^{- 11} \times 6.0 \times 1 0 ^{24}}{2.657 \times 1 0 ^{7}} = 3.88 \times 1 0^{3} m s^{- 1}$

Period from the speed

With the speed known, v = 2πr/T is quicker than Kepler's law and gives the same answer.

$T = \frac{2 π r}{v} = \frac{2 π \times 2.657 \times 1 0 ^{7}}{3.88 \times 1 0 ^{3}} = 4.30 \times 1 0^{4} s = 11.9 h$

Answer

The satellites travel at $3.9$ km s⁻¹ and go round in about $12$ hours, twice per day. A geostationary satellite needs a period of one day, which requires a larger radius, so a GPS satellite drifts across the sky. That suits navigation, which needs several satellites visible from every point on the Earth, from different directions.

Is that answer sensible?

The speed lies between a low orbit's $7.7$ km s⁻¹ and a geostationary orbit's $3.1$ km s⁻¹, and so does the radius. Kepler's third law gives the same check: the GPS radius is $0.63$ of the geostationary radius, and $0.6 3^{1.5} = 0.50$ , half a day.

CheckpointAnswer before reading on.

A geostationary satellite has an orbital period equal to one rotation of the Earth, $8.6164 \times 1 0^{4}$ s. The Earth's mass is $6.0 \times 1 0^{24}$ kg.

Calculate the radius of a geostationary orbit.

Your answerm

Give it to 3 significant figures.

Hint 1Kepler's third law links the radius, the period and the central mass.

Hint 2Rearrange to $r^{3} = \frac{GM T ^{2}}{4 π ^{2}}$ .

Hint 3Take the cube root at the end, not the square root.

Exam question

Harder · about 8 min

5 marks

A weather agency is choosing between a satellite in low Earth orbit at an altitude of $400$ km and a geostationary satellite at an orbital radius of $4.22 \times 1 0^{7}$ m. The Earth's mass is $6.0 \times 1 0^{24}$ kg and its radius is $6.371 \times 1 0^{6}$ m.

Compare the orbital properties of the two satellites, including their periods, and relate these to their uses in observing weather.

Write your response, then mark it against the criteria. Writing it out first is the part that works.

Hint 1Find each period with Kepler's third law, $T = 2 π r^{3} / GM$ .

Hint 2A geostationary satellite's period matches the Earth's rotation. What does it therefore see?

Hint 3What does a low orbit gain in detail, and lose in coverage?

Where students lose marks on this one

Describing a geostationary satellite as not moving.
Why it happens: It appears fixed in the sky.
It moves at about $3$ km s⁻¹. It appears fixed because it goes round once per rotation of the Earth, in the same direction, above the equator.

Written for this site.

Kepler's laws

Kepler found his three laws from careful observation of the planets, decades before Newton. Newton's achievement was to show that all three follow from an inverse square force.

Law of orbits. Each planet moves on an ellipse with the Sun at one focus. A circle is the special case with both foci together.
Law of areas. The line from the Sun to a planet sweeps out equal areas in equal times.
Law of periods. $r^{3} / T^{2}$ is the same for every planet, and equals $GM /4 π^{2}$ for the Sun.

An elliptical orbit and the law of areas

1The Sun, at one focus
The Sun sits at a focus, not at the centre. The distance from the Sun to the planet therefore changes around the orbit. For most planets the ellipse is nearly a circle; the Earth's distance varies by only about three per cent.
2Closest point, perihelion
At perihelion the planet is closest to the Sun, has its lowest potential energy and so its highest kinetic energy. It moves fastest here.
3Furthest point, aphelion
At aphelion the planet is furthest from the Sun. Its potential energy is highest, so its kinetic energy and speed are lowest. It has not stopped, and it is still being pulled towards the Sun, which is why it turns back.
4Area swept near the Sun
In a given time near the Sun, the planet covers a long stretch of orbit, but the radius is short, so the swept area is a wide, short wedge.
5Equal area swept far away
In the same time far away, the planet covers only a short stretch, but the radius is long. The area comes out the same. Equal areas in equal times is conservation of angular momentum, and it describes the same speed change as the energy argument.
6The other focus: empty
An ellipse has two foci. Nothing sits at the second one. It matters only for drawing the shape: the sum of the distances from any point on the ellipse to both foci is constant.

Both shaded regions are swept out in the same time. The one near the Sun is short and wide; the far one is long and thin.

The third law is the one you calculate with. Because $r^{3} / T^{2}$ depends only on the central mass, measuring one orbit around a body measures that body's mass. It is how every planet with a moon has been weighed.

r³/T² constant

Six planets, a range of eighty in period, one line. A moved point shows the period any orbit would have.

The numbers behind this graph

Planet	r (AU)	T (years)	r³/T² (AU³ per year²)
Mercury	0.387	0.241	0.998
Venus	0.723	0.615	0.999
Earth	1	1	1.000
Mars	1.524	1.881	1.000
Jupiter	5.203	11.86	1.001
Saturn	9.537	29.46	0.999

Orbital radius marked2.80 AU

Period at the marked radius: 4.69 years
r³/T² in SI units: 3.36 × 10¹⁸ m³ s⁻²
Sun mass from the gradient: 1.99 × 10³⁰ kg

Every slider is a normal range input, so the arrow keys move it one step and Home and End jump to the extremes.

Worked example3 marks

Weighing the Earth with the Moon

The Moon orbits the Earth at a radius of $3.84 \times 1 0^{8}$ m with a period of $27.3$ days. Find the mass of the Earth.

Convert the period

G is in SI units, so time must be in seconds.

$T = 27.3 \times 86400 = 2.36 \times 1 0^{6} s$

Rearrange Kepler's third law for the central mass

The Moon's mass cancelled when the law was derived, so it is neither given nor needed.

$M = \frac{4 π ^{2} r ^{3}}{G T ^{2}} = \frac{4 π ^{2} \times ( 3.84 \times 1 0 ^{8} ) ^{3}}{6.67 \times 1 0 ^{- 11} \times ( 2.36 \times 1 0 ^{6} ) ^{2}} = 6.0 \times 1 0^{24} kg$

Answer

The Earth's mass is $6.0 \times 1 0^{24}$ kg.

Is that answer sensible?

This matches the data sheet value, and it had to: the data sheet value was found in essentially this way. A result many orders of magnitude out almost always means the period was left in days.

CheckpointAnswer before reading on.

Io, one of Jupiter's moons, has a nearly circular orbit of radius $4.22 \times 1 0^{8}$ m and a period of $1.77$ days.

Calculate the mass of Jupiter.

Your answerkg

Give it to 2 significant figures.

Hint 1Kepler's third law contains the central mass.

Hint 2Convert the period to seconds: $1.77 \times 24 \times 3600$ .

Hint 3 $M = \frac{4 π ^{2} r ^{3}}{G T ^{2}}$ .

Exam question

Harder · about 4 min

3 marks

A comet follows an elliptical orbit around the Sun. Using energy, explain why it moves fastest when it is closest to the Sun, and relate this to Kepler's second law.

Write your response, then mark it against the criteria. Writing it out first is the part that works.

Hint 1With gravity the only force, the comet and Sun form a closed system. What is conserved?

Hint 2How does the potential energy $U = - GM m / r$ change as $r$ decreases?

Hint 3Kepler's second law says equal areas in equal times. What must happen to the speed when the line to the Sun is short?

Where students lose marks on this one

Saying the comet speeds up because gravity is stronger when it is closer.
Why it happens: A stronger force is associated with more speed.
A stronger force means a larger acceleration, not a larger speed. The speed is set by energy: how much potential energy has been converted to kinetic energy.

Written for this site.

Energy in a radial field

Near the surface, $m g h$ works because the field hardly changes over the heights involved. Over orbital distances the field changes a lot, so a different expression is needed:

$U = - \frac{GM m}{r}$

Its zero is at infinite separation, where the two bodies no longer interact. Bringing them together releases energy, so every finite separation has less than zero. That is why the sign is negative. As $r$ grows, $U$ rises towards zero, which is an increase, even though the number shrinks in size.

Gravitational potential energy in a radial fieldon the NESA formulae sheet

U = - \frac{GM m}{r}

Symbols: $U$ gravitational potential energy of the two body system, zero when they are infinitely far apartJ
$G$ universal gravitational constantN m² kg⁻²
$M$ mass of the central bodykg
$m$ mass of the smaller bodykg
$r$ distance between their centresm
Valid when: The zero of potential energy is chosen at infinite separation. Every finite separation then has a negative potential energy, rising towards zero as the bodies move apart.
Not valid when: It is mixed with mgh, whose zero is at a chosen surface. The two are consistent only for changes in energy, and mgh itself only holds for heights small compared with the planet's radius.
Rearranged: for r: $r = - \frac{GM m}{U}$
Where it turns up: Finding the energy needed to move a satellite between orbits
Deriving the escape speed by setting the kinetic energy equal to the size of U
Explaining why a planet in an ellipse speeds up as it approaches the Sun
Where marks go missing: Dropping the minus sign, which reverses every conclusion about energy changes
Squaring r, confusing potential energy with force
Thinking a larger negative number means more energy

Derivation

Total energy of a circular orbit

Starts from

the kinetic energy of the orbit plus its potential energy

Ends at

E = - \frac{GM m}{2 r}

Holds only if

The orbit is circular
Gravity is the only force

Total energy of a circular orbit

3 steps

1
Kinetic energy from the orbit condition
Gravity as the centripetal force gives mv² directly, without needing v itself.
$\frac{GM m}{r ^{2}} = \frac{m v ^{2}}{r} \Rightarrow K = \frac{1}{2} m v^{2} = \frac{GM m}{2 r}$
2
Add the potential energy
The total energy of the satellite and Earth is the sum of the two.
$E = K + U = \frac{GM m}{2 r} - \frac{GM m}{r} = - \frac{GM m}{2 r}$
3
Read off the pattern
These three relations make every energy change question quick.
$K = - E U = 2 E$
The kinetic energy is half the size of the potential energy. The total is negative, so the satellite is bound: it cannot reach infinity without more energy.

Total energy of a circular orbitnot on the formulae sheet, learn it

U + K = - \frac{GM m}{2 r}

Symbols: $U$ gravitational potential energyJ
$K$ kinetic energy of the orbiting bodyJ
$G$ universal gravitational constantN m² kg⁻²
$M$ mass of the central bodykg
$m$ mass of the orbiting bodykg
$r$ orbital radiusm
Valid when: The orbit is circular, so K = GMm/2r from setting gravity equal to the centripetal force. The total is negative, which is the condition for a bound orbit.
Not valid when: The orbit is elliptical, where K and U trade off around the path and r must be replaced by the semi major axis, or the body is not in orbit at all.
Rearranged: for r: $r = - \frac{GM m}{2 E}$
Where it turns up: Finding the energy needed to lift a satellite from a low orbit to a higher one
Showing that a satellite losing energy to drag spirals inwards and speeds up
Checking that K = −U/2 and E = −K for any circular orbit
Where marks go missing: Losing the minus sign and concluding that a higher orbit has less energy
Adding the full orbital speed kinetic energy to mgh
Assuming a satellite in a higher orbit is faster because it has more total energy

Where this comes from

Kinetic energy K
Potential energy U
Total energy K + U

Moving to a higher orbit raises the total energy but lowers the kinetic energy. The satellite ends up slower, even though work had to be done on it.

Satellite mass1000 kg

Lower orbit altitude400 km

Higher orbit altitude35786 km

Total energy, lower orbit: -2.955 × 10¹⁰ J
Total energy, higher orbit: -4.747 × 10⁹ J
Energy that must be supplied: 2.481 × 10¹⁰ J
Change in kinetic energy: -2.481 × 10¹⁰ J
Change in potential energy: 4.961 × 10¹⁰ J

Every slider is a normal range input, so the arrow keys move it one step and Home and End jump to the extremes.

Higher orbits: more energy, less speed

To move to a higher circular orbit a satellite's engine must do work, so the total energy rises, from a larger negative number to a smaller one. Yet the new orbit is slower, so the kinetic energy falls. The potential energy rises by twice the work done, taking the difference from the kinetic energy. Stating all three changes, with their directions, is what separates a full mark answer on orbit transfers.

Worked example4 marks

A satellite losing height to drag

A $1200$ kg satellite in a circular orbit at $400$ km altitude loses energy to atmospheric drag and settles into a circular orbit at $350$ km. Find the changes in its total, kinetic and potential energy, and describe what happens to its speed.

The two radii

All three energies depend on distance from the centre.

$r_{1} = 6.771 \times 1 0^{6} m r_{2} = 6.721 \times 1 0^{6} m$

Change in total energy

The total energy is the one quantity drag changes directly, so it comes first.

$Δ E = - \frac{GM m}{2} (\frac{1}{r _{2}} - \frac{1}{r _{1}}) = - \frac{4.00 \times 1 0 ^{14} \times 1200}{2} (\frac{1}{6.721 \times 1 0 ^{6}} - \frac{1}{6.771 \times 1 0 ^{6}}) = - 2.6 \times 1 0^{8} J$

Use the pattern for the other two

In any circular orbit K = −E and U = 2E, so their changes follow without further substitution.

$Δ K = - Δ E = + 2.6 \times 1 0^{8} J Δ U = 2Δ E = - 5.3 \times 1 0^{8} J$

Answer

The total energy falls by $2.6 \times 1 0^{8}$ J, the energy lost to drag. The potential energy falls by twice that, $5.3 \times 1 0^{8}$ J, and the kinetic energy rises by $2.6 \times 1 0^{8}$ J. The satellite speeds up, from $7.69$ to $7.72$ km s⁻¹, even though drag opposes its motion.

Is that answer sensible?

A force opposing the motion that makes the object faster sounds wrong, and it is the famous satellite drag paradox. The resolution is that the satellite drops to a lower orbit, and gravity does positive work on it as it falls, more than drag removes. The numbers show exactly that: potential energy lost, $5.3 \times 1 0^{8}$ J, is split equally between kinetic energy and heat.

CheckpointAnswer before reading on.

A satellite is moved from a low circular orbit to a higher circular orbit. Its engines do positive work on it during the transfer.

Which statement about the satellite in its new orbit is correct?

AIts total energy is greater and its speed is lower.
BIts total energy is greater, so its speed is greater.
CIts total energy is lower, because it is further from the Earth.
DIts speed is unchanged, because orbital speed depends only on the Earth's mass.

Hint 1Compare the circular orbit speeds at the two radii using $v = GM / r$ .

Hint 2Kinetic energy follows speed. Total energy follows $- GM m /2 r$ .

Hint 3Where did the work, and the lost kinetic energy, go?

CheckpointAnswer before reading on.

A $1000$ kg satellite is moved from a circular orbit of radius $6.771 \times 1 0^{6}$ m to a geostationary orbit of radius $4.22 \times 1 0^{7}$ m. The Earth's mass is $6.0 \times 1 0^{24}$ kg.

Calculate the increase in the satellite's total orbital energy.

Your answerJ

Give it to 3 significant figures.

Hint 1The total energy of a circular orbit is $E = - \frac{GM m}{2 r}$ .

Hint 2Find $E$ for each orbit, keeping the signs.

Hint 3The increase is $E_{final} - E_{initial}$ .

Exam question

Harder · about 4 min

3 marks

Show that the total energy of a satellite of mass $m$ in a circular orbit of radius $r$ around a planet of mass $M$ is $E = - \frac{GM m}{2 r}$ .

Write your response, then mark it against the criteria. Writing it out first is the part that works.

Hint 1Total energy is $K + U$ , with $U = - GM m / r$ .

Hint 2Find $v^{2}$ from gravity providing the centripetal force.

Hint 3Then $K = \frac{1}{2} m v^{2}$ .

Where students lose marks on this one

Using $U=mgh$ for the potential energy.
Why it happens: mgh is the potential energy formula met first.
mgh assumes a constant field and a zero at a chosen surface. Over orbital distances the field changes, so the radial form with its zero at infinity is needed.

Written for this site.

Escape velocity

An object launched upwards slows as it rises. If it is fast enough, it slows forever but never stops, and never comes back. The least launch speed that achieves this is the escape velocity.

Derivation

Escape velocity

Starts from

energy conservation between launch and infinite separation

Ends at

v_{esc} = \frac{2 GM}{r}

Holds only if

No air resistance
No thrust after launch
The planet is not rotating, or its rotation is ignored

Escape velocity

3 steps

1
State the condition
Just escaping means arriving at infinity with zero speed, where U is also zero. So the total energy at infinity is zero.
$E_{\infty} = 0$
2
Conserve energy from launch
Gravity is the only force, so the total energy at launch equals the total energy at infinity.
$\frac{1}{2} m v^{2} - \frac{GM m}{r} = 0$
3
Solve for v
The object's mass cancels, so escape velocity is a property of the planet and the launch radius only.
$v_{esc} = \frac{2 GM}{r}$

Escape velocityon the NESA formulae sheet

v_{esc} = \frac{2 GM}{r}

Symbols: $v_{esc}$ least launch speed that lets an unpowered object reach infinite distancem s⁻¹
$G$ universal gravitational constantN m² kg⁻²
$M$ mass of the body being escapedkg
$r$ distance from its centre at launch, the radius when launched from the surfacem
Valid when: The object is given its speed at the start and then moves under gravity alone. It follows from setting the kinetic energy equal to the size of the potential energy, so that the total energy is zero.
Not valid when: Air resistance matters, or the craft keeps firing its engine, in which case it can leave at any speed. The direction of launch does not matter, provided the path does not hit the planet.
Rearranged: for M: $M = \frac{v _{esc}^{2} r}{2 G}$
Where it turns up: Finding the escape speed from the surface of a planet or moon
Comparing it with the circular orbit speed at the same radius, which is smaller by a factor of the square root of 2
Explaining why the Moon has no atmosphere
Where marks go missing: Including the object's mass, which cancels
Omitting the factor of 2 and finding the circular orbit speed instead
Thinking escape velocity must be directed straight up

Where this comes from

Comparing with the orbital speed, $v = GM / r$ , the escape velocity at any radius is $2$ times the circular orbit speed there. From the Earth's surface that is $11.2$ km s⁻¹, against $7.9$ km s⁻¹ for an orbit skimming the surface. Launch direction does not matter, because energy is a scalar, as long as the path does not run into the planet.

Speed

Launch altitude2000 km

Speed as a multiple of circular1.15 ×

Satellite mass1000 kg

Circular orbit speed

The same situation drawn as a graph, shown because this page has no JavaScript to animate it.

The numbers behind this graph

Altitude (km)	Speed (km s⁻¹)	Period (h)
400	7.69	1.54
2000	6.91	2.11
10000	4.94	5.78
20200	3.88	11.95
35786	3.08	23.88

Watch the spokes on an ellipse. They bunch together far from the Earth and spread out close to it, because the satellite moves fastest nearest the Earth. That is Kepler's second law.

Try the speed multiple at $1.00$ for a circle, then $1.15$ and $0.90$ for ellipses. Watch the spokes, drawn at equal time intervals, crowd together at the far end of each ellipse. Push the multiple to $1.42$ , just past $2$ , and the total energy turns positive: the orbit opens up and the satellite does not return.

CheckpointAnswer before reading on.

The Moon has a mass of $7.35 \times 1 0^{22}$ kg and a radius of $1.74 \times 1 0^{6}$ m.

Calculate the escape velocity from the Moon's surface.

Your answerm s⁻¹

Give it to 3 significant figures.

Hint 1The escape velocity formula is on the data sheet.

Hint 2At the surface, $r$ is the Moon's radius.

Hint 3 $v = 2 \times 6.67 \times 1 0^{- 11} \times 7.35 \times 1 0^{22} /1.74 \times 1 0^{6}$

Common mistake

Using the altitude of a satellite as r in any of these formulae.

Why it happens: Questions give heights above the surface, because that is how orbits are described in practice.

Every formula in this topic comes from the law of gravitation, where $r$ is the distance between the centres of the two bodies. Add the planet's radius to the altitude before substituting, every time. The error is not small: for a low orbit it changes the answer by a factor of hundreds.

Derivetypically 3 to 5 marks

Demands: Reach the stated result from stated starting points, by steps that each follow.
Shape: Name the principle you start from, then algebra, with the reason for each non-obvious step.
Loses marks: Working backwards from the answer, which shows only that the answer is consistent.

Through a marker’s eyes

3 marks

Derive the expression for the escape velocity from the surface of a planet of mass M and radius R. (3 marks)

The attempt

1 out of 3

Escape velocity is when kinetic energy equals potential energy, so $\frac{1}{2} m v^{2} = GM m / R$ , which gives $v = 2 GM / R$ .

What the marker sees

The final result is right and earns a mark, but a derivation is marked on its reasoning. The response never says why the energies should be equal. Potential energy is negative, so as written the statement is not even consistent with $U = - GM m / R$ . There is no mention of energy conservation or of the condition at infinity, which is the physics the question is testing.

The same answer, fixed

3 out of 3

With gravity the only force, the total mechanical energy is conserved. At launch, $E = \frac{1}{2} m v^{2} - \frac{GM m}{R}$ .
To just escape, the object reaches infinite separation with zero speed. There $K = 0$ and $U = 0$ , so the total energy is zero.
Conservation gives $\frac{1}{2} m v^{2} - \frac{GM m}{R} = 0$ , so $v_{esc} = \frac{2 GM}{R}$ , independent of the object's mass.

Where the negative potential energy comes from

The potential energy expression is the work done by gravity as a mass comes in from infinity. Since the force changes with distance, the work is an integral:

$W = \int_{\infty}^{r} \frac{GM m}{x ^{2}} d x = \frac{GM m}{r}$

Gravity does positive work, $GM m / r$ , on the way in, so the potential energy falls by that much from its zero at infinity, to $- GM m / r$ . Near the surface, the difference between two radii $R$ and $R + h$ with $h ≪ R$ is

$Δ U = GM m (\frac{1}{R} - \frac{1}{R + h}) \approx \frac{GM m}{R ^{2}} h = m g h$

so the familiar formula is the small height limit of the radial one.

Elliptical orbits and energy

For an ellipse the total energy is still $- GM m /2 a$ , where $a$ is the semi major axis, half the longest diameter. So the energy of an orbit depends only on its size, not on how stretched it is. A satellite launched sideways at the circular speed has $a = r$ . Faster, and it has more energy, so a larger $a$ , and the launch point becomes the closest point of an ellipse. That is the basis of a transfer orbit: a short engine burn in a low orbit puts the satellite on an ellipse that reaches out to a higher orbit, and a second burn at the far point makes the new orbit circular.

Why launch sites are near the equator

The Earth's surface at the equator already moves east at about $0.46$ km s⁻¹. A rocket launched east from there starts with that speed, which reduces the energy the rocket must supply. It is one reason launch sites are placed as close to the equator as a country's territory allows, and why geostationary satellites, which must end up above the equator anyway, benefit most.

Where this reappears

Module 6 uses the same structure with electric fields: a field around a charge, an inverse square force, and an energy picture of moving charges. The orbital condition $F = m v^{2} / r$ returns for charged particles in magnetic fields. Module 8 uses Newton's gravitational law implicitly in describing the evolution of stars.