Is escape velocity on the HSC data sheet?

No. The 2019 data sheet gives F = GMm/r^2, v = 2πr/T, r^3/T^2 = GM/4π^2 and U = −GMm/r, but not vesc = √(2GM/r), v = √(GM/r), g = GM/r^2 or E = −GMm/2r. Learn to derive each one in a few lines.

Why is gravitational potential energy negative?

Because zero is defined at an infinite distance, where the attraction is zero. Any closer, you would need to add energy to pull the object back out to infinity, so its potential energy is below zero.

What is the difference between a geostationary and a geosynchronous orbit?

A geosynchronous orbit has a period of one day. A geostationary orbit is the special geosynchronous orbit that is circular, above the equator and in the direction Earth spins, so the satellite stays over one point. A tilted geosynchronous orbit drifts north and south each day.

Does a satellite's mass affect its orbit?

Not its speed or period: the satellite's mass cancels in GMm/r^2 = mv^2/r. It does affect the gravitational force on it, its kinetic and potential energy, and the energy needed to launch it or change its orbit.

What is the difference between g and G?

G = 6.67 × 10^−11 N m^2 kg^−2 is the universal gravitational constant, the same everywhere. g is the gravitational field strength at a place, g = GM/r^2, which is 9.8 m s^−2 at Earth's surface and smaller further out.

At the surface of Earth g = 9.8 m s^−2. What is g at a height of one Earth radius above the surface?

zero, because it is above the atmosphere

Which statement about a geostationary satellite is correct?

It orbits above the equator with a period of one day, moving in the direction Earth spins.

Any satellite with a period of one day is geostationary.

A heavier geostationary satellite must orbit at a smaller radius.

It orbits lower than the International Space Station so it can stay over one place.

Why do astronauts on the International Space Station feel weightless?

They and the station are in free fall together, with gravity providing the centripetal force.

There is no gravity at that altitude.

The Moon's gravity cancels Earth's gravity there.

The station's speed creates an outward force that balances gravity.

HSC Physics · Module 5 · Advanced Mechanics · 13 min read

Gravitational fields and orbits

Satellites, moons and planets stay in orbit because gravity pulls them towards the body they circle, and that pull is exactly the centripetal force they need. In HSC Physics almost every orbit question starts from one line: F_g = F_c, or GMm/r² = mv²/r. From it come orbital speed, Kepler's third law and the orbital energy equations.

This guide covers gravitational field strength, orbital velocity, Kepler's laws, geostationary and low Earth orbits, gravitational potential energy, orbital energy and escape velocity. It finishes with a quick quiz and a full worked example. This topic has carried more HSC marks than any other in recent papers.

What you need to know

Use Newton's law of universal gravitation, F = GMm/r², to calculate the force two objects exert on each other, and use g = GM/r² to find what sets gravitational field strength and predict it anywhere in a field, including at a planet's surface.
Apply the links between the force of gravity, centripetal force and acceleration, the masses, the orbital radius, the orbital velocity and the period to explain how planets and artificial satellites orbit.
Predict the orbital properties of satellites and planets in many situations, including geostationary and near-Earth orbits, and link each orbit to what it is used for.
Using v = 2πr/T and r³/T² = GM/4π², relate the forces on a planet and its total energy, in circular or non-circular orbits, to Kepler's laws of planetary motion.
For radial gravitational fields, derive and apply gravitational force and potential energy, including escape velocity v_esc = √(2GM/r), potential energy U = −GMm/r, total orbital energy U + K = −GMm/2r, the energy change when a satellite changes orbit, and Kepler's three laws.

The equations (from the 2019 HSC data sheet unless marked):

F = GMm/r²Gravitational force between masses M and m whose centres are r apart. G = 6.67 × 10⁻¹¹ N m² kg⁻².

v = 2πr/TOrbital speed from the radius and the period.

r³/T² = GM/4π²Kepler's third law. M is the mass being orbited (the planet or star), not the satellite.

U = −GMm/rGravitational potential energy, zero at infinity and negative everywhere else.

g = GM/r² not on the data sheetField strength at distance r from the centre: F = GMm/r² divided by m.

v = √(GM/r) not on the data sheetOrbital speed, from GMm/r² = mv²/r.

E = K + U = −GMm/2r not on the data sheetTotal energy of a circular orbit (K = GMm/2r).

v_esc = √(2GM/r) not on the data sheetEscape velocity, from K + U = 0.

Dot points: PH-M5-10, PH-M5-11, PH-M5-12, PH-M5-13, PH-M5-14 (2017 Physics syllabus, HSC 2026 and 2027), in our own words.

Mini lesson

Gravitational force and field strength

Every mass attracts every other mass with F = GMm/r², where r is the distance between their centres. The force on each body is the same size, in opposite directions (Newton's third law).

Gravitational field strength g is the force per kilogram: g = F/m = GM/r², in N kg⁻¹ (the same as m s⁻²). It depends only on the mass of the planet and the distance from its centre, not on the object in the field. Because of the r², doubling the distance from the centre cuts g to a quarter. At the height of the International Space Station (about 400 km up) g is still around 8.7 m s⁻², so astronauts are not 'out of gravity'. They feel weightless because they and the station are both in free fall.

g = GM/r²: double the distance from the centre and g falls to a quarter; triple it and g falls to a ninth. r is measured from the centre, not the surface.

Gravity is 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. Set them equal and the satellite's mass m cancels:

GMm/r² = mv²/r, so v = √(GM/r). A lower orbit is a faster orbit.
Put v = 2πr/T into that: GM/r = 4π²r²/T², which rearranges to r³/T² = GM/4π² (Kepler's third law). Practise this four-line derivation; it is often asked.
The orbital radius r is measured from the centre of the planet: r = r_E + h, where h is the height above the surface.
Because m cancels, a 5 kg satellite and a 5000 kg satellite at the same radius have the same speed and period. Mass does change the force and the energy.

Every orbit equation uses r from the centre of the planet. Add Earth's radius (6.371 × 10⁶ m on the data sheet) to any height you are given.

Kepler's three laws

First law: planets move in ellipses with the Sun at one focus. A circle is the special case HSC calculations usually use.
Second law: the line from the Sun to a planet sweeps out equal areas in equal times. So a planet moves fastest when it is closest to the Sun. In energy terms: closer means lower (more negative) U, so higher K, with the total staying constant.
Third law: r³/T² is the same for every body orbiting the same central mass, and equals GM/4π². Four times the radius means 4^3/2 = 8 times the period. Plot r³ against T² and the gradient is GM/4π², which lets you 'weigh' the central body.

Geostationary and low Earth orbits

Kepler's third law fixes the radius for any given period, so each job needs its own orbit.

Geostationary: period one day (strictly 23 h 56 min, one turn of Earth relative to the stars), directly above the equator, moving in the same direction Earth spins. That fixes r ≈ 4.2 × 10⁷ m (about 36 000 km up) and v ≈ 3.1 km s⁻¹. The satellite stays above one spot, so a dish can point at it without moving: used for TV broadcasting, communications and weather images of a whole hemisphere.
Low Earth orbit: roughly 200 km to 2000 km up, with periods of about 90 to 130 minutes and speeds of about 6.9 to 7.8 km s⁻¹. Close to the ground, so good for detailed imaging, Earth observation and crewed stations like the ISS. Each one passes over a given place only briefly, so a full service needs many satellites.
Trade-off: higher orbit means a longer period and a slower speed, but a wider view of Earth and longer signal delays.

Potential energy and orbital energy

Gravitational potential energy is U = −GMm/r. It is zero at infinite distance and negative everywhere else, because you would have to do work to pull the object out to infinity. Moving away from a planet makes U less negative, which is an increase.

For a circular orbit, K = ½mv² = GMm/2r (using v² = GM/r). Adding them gives the total energy E = K + U = −GMm/2r. Raising a satellite to a higher orbit makes K smaller (it moves slower) but U and E larger (less negative). The energy the rockets must supply is the change in total energy, ΔE = E_final − E_initial.

Green K = GMm/2r, blue E = −GMm/2r, orange U = −GMm/r. In a higher orbit K is smaller, but U and E are larger (closer to zero). E is always half of U.

Escape velocity

To escape completely, an object needs enough kinetic energy to reach infinity (where U = 0) with nothing left over. So at launch K + U = 0: ½mv² − GMm/r = 0, giving v_esc = √(2GM/r). With the data sheet values, Earth's surface escape velocity is about 11.2 km s⁻¹.

The mass of the escaping object cancels, and energy is a scalar, so the launch direction doesn't matter (as long as the path doesn't hit the planet). At any radius, v_esc = √2 × the circular orbital speed there.

Common mistakes

Using the height above the surface as r in F = GMm/r², v = √(GM/r) or Kepler's third law.r is from the centre of the planet: r = r_E + h. Convert km to m first.
Putting the satellite's mass into r³/T² = GM/4π².M is the mass of the body being orbited. The satellite's mass cancels out.
Saying potential energy decreases as a satellite moves further from Earth (because the number gets smaller).U = −GMm/r becomes less negative as r grows, which is an increase. Zero (at infinity) is the highest it gets.
Thinking a satellite in a higher orbit moves faster or has more kinetic energy.Higher orbits are slower: K = GMm/2r falls. But the total energy −GMm/2r rises, so moving up still costs energy.
Saying astronauts float because there is no gravity in orbit.g at the ISS is about 8.7 m s⁻². They feel weightless because they and the station fall together, with gravity providing the centripetal force.
Leaving the period in minutes, hours or days in Kepler's third law.Convert to seconds first (days to s, then square). NESA markers flagged this unit slip in 2021 and again in 2024.

Worked example

A 720 kg Earth-observation satellite is placed in a circular orbit around Earth with a period of 105 minutes.

Calculate the satellite's height above Earth's surface, and its total mechanical energy in this orbit. Use the data sheet values for G, M_E and r_E.

(5 marks)

I'll think out loud, the way you should in the exam, using KNOW-UD: Knowns, Not known, Organise, Work it out, Units, Direction.

Prediction first. The ISS, about 400 km up, takes roughly 90 minutes per orbit. 105 minutes is longer, so this orbit is higher: I'd guess about 1000 km. For the energy, r ≈ 7.4 × 10⁶ m, so E = −GMm/2r ≈ −(4 × 10¹⁴ × 720) ÷ (1.5 × 10⁷) ≈ −2 × 10¹⁰ J.

K · Knowns

I sketch Earth, the orbit and the satellite, and put the values on it. The period is in minutes, so I convert it to seconds now, before it can go into any equation.

T = 105 min = 105 × 60 = 6300 s
m = 720 kg
M_E = 6.0 × 10²⁴ kg, r_E = 6.371 × 10⁶ m, G = 6.67 × 10⁻¹¹ N m² kg⁻²

Knowns on the sketch: T = 105 min = 6300 s, m = 720 kg, M_E and r_E from the data sheet. Unknowns: the height h and the total energy E.

N · Not known

Two answers: a height in metres and an energy in joules. The height comes from the orbital radius, so r is the first thing to find.

h = ? m (on the way: r = ? m), E = ? J

O · Organise

I know the period, so Kepler's third law gives the radius. r is from Earth's centre, so I subtract Earth's radius to get the height. For the energy, −GMm/2r isn't on the data sheet, so I build it: gravity is the centripetal force, so mv² = GMm/r, which makes K = GMm/2r. Then I add U.

r³/T² = GM/4π² → r = ∛(GMT²/4π²)
h = r − r_E
K = ½mv² = GMm/2r, U = −GMm/r
E = K + U = −GMm/2r

W · Work it out

I work out GM once and keep it in full. Then r³, then the cube root. I keep r unrounded for the energy line.

GM = 6.67 × 10⁻¹¹ × 6.0 × 10²⁴ = 4.002 × 10¹⁴ N m² kg⁻¹
r³ = 4.002 × 10¹⁴ × 6300² ÷ 4π² = 1.5884 × 10²² ÷ 39.478 = 4.0234 × 10²⁰ m³
r = ∛(4.0234 × 10²⁰) = 7.3824 × 10⁶ m
h = 7.3824 × 10⁶ − 6.371 × 10⁶ = 1.0114 × 10⁶ m
E = −(4.002 × 10¹⁴ × 720) ÷ (2 × 7.3824 × 10⁶) = −1.9516 × 10¹⁰ J

U · Units

Metres for the height, joules for the energy. The data are to two or three significant figures, so I give two.

h ≈ 1.0 × 10⁶ m (about 1000 km)
E ≈ −2.0 × 10¹⁰ J

D · Direction

Energy is a scalar, so there's no direction, but the sign matters: a negative total energy means the satellite is bound to Earth, which it must be if it is in orbit. As a sense check on the radius, the speed is v = 2πr/T = 2π × 7.3824 × 10⁶ ÷ 6300 = 7.36 × 10³ m s⁻¹, inside the usual 6.9 to 7.8 km s⁻¹ for low Earth orbit.

E is negative: the satellite is bound. v = 7.4 km s⁻¹, a typical low Earth orbit speed ✓

The satellite orbits about 1.0 × 10⁶ m (about 1000 km) above Earth's surface, and its total mechanical energy is about −2.0 × 10¹⁰ J.

Check against the prediction. I predicted about 1000 km and about −2 × 10¹⁰ J, from comparing with the ISS. The answers, 1.0 × 10⁶ m and −2.0 × 10¹⁰ J, match.

HSC advice

What markers keep saying

Be ready to derive Kepler's third law: set the gravitational force equal to the centripetal force, substitute v = 2πr/T and rearrange, showing every line. When the law comes from a graph, use a large gradient triangle on the line of best fit, convert km³ to m³ and days² to s², and rearrange the gradient for the central mass.NESA marking feedback, 2024, summarised
Remember that U = −GMm/r increases (becomes less negative) as r increases. In energy calculations, label each quantity (U_P, K_Q and so on), write each equation, then substitute, then give the unrounded result.NESA marking feedback, 2023, summarised
A satellite launched from rest on the surface into orbit gains both kinetic and gravitational potential energy. Convert hours and minutes to seconds before using the period.NESA marking feedback, 2021, summarised
For an elliptical orbit, total mechanical energy is conserved: nearer the star U is lower and K is higher, so the planet is faster. Explain the whole orbit, not just one point. For two bodies orbiting each other, reason with forces, not energy.NESA marking feedback, 2020, summarised
Derive escape velocity from K + U = 0 rather than quoting it, and use conservation laws (energy, and momentum where something splits apart) to describe what happens next.NESA marking feedback, 2025, summarised
Measure every orbital radius from the centre of the planet. If a question gives a height, add the planet's radius before substituting.CortexHSC teaching notes

Summaries in our own words. Read NESA's own feedback with the exam packs: 2020 · 2021 · 2023 · 2024 · 2025.

How often it is examined

Gravitational fields and orbits questions were worth 70 marks across the seven HSC Physics exams from 2019 to 2025 (about 10.0 marks a paper), and appeared in 7 of 7 papers.

Marks per paper (Module 5, Motion in Gravitational Fields). Source: CortexHSC's own tagging of the NESA HSC Physics exams 2019–2025, each question counted under its main topic, so treat the numbers as close, not exact.

FAQ

Is escape velocity on the HSC data sheet?: No. The 2019 data sheet gives F = GMm/r², v = 2πr/T, r³/T² = GM/4π² and U = −GMm/r, but not v_esc = √(2GM/r), v = √(GM/r), g = GM/r² or E = −GMm/2r. Learn to derive each one in a few lines.
Why is gravitational potential energy negative?: Because zero is defined at an infinite distance, where the attraction is zero. Any closer, you would need to add energy to pull the object back out to infinity, so its potential energy is below zero.
What is the difference between a geostationary and a geosynchronous orbit?: A geosynchronous orbit has a period of one day. A geostationary orbit is the special geosynchronous orbit that is circular, above the equator and in the direction Earth spins, so the satellite stays over one point. A tilted geosynchronous orbit drifts north and south each day.
Does a satellite's mass affect its orbit?: Not its speed or period: the satellite's mass cancels in GMm/r² = mv²/r. It does affect the gravitational force on it, its kinetic and potential energy, and the energy needed to launch it or change its orbit.
What is the difference between g and G?: G = 6.67 × 10⁻¹¹ N m² kg⁻² is the universal gravitational constant, the same everywhere. g is the gravitational field strength at a place, g = GM/r², which is 9.8 m s⁻² at Earth's surface and smaller further out.

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