Orbital Motion and Energy Revision Notes | A Level AQA Physics
Orbital speed is v = √(GM/r), found by equating gravitational and centripetal force; the satellite's own mass cancels. Free AQA notes and 12 questions answered.
Updated 12 Sept 2026 · Revise Right Team
Orbital Motion and Energy is part of Gravitational fields (A-level only) in A Level AQA Physics.
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What this lesson covers
Orbital speed is v = √(GM/r), found by equating gravitational and centripetal force; the satellite's own mass cancels. Free AQA notes and 12 questions answered.
Derive the relationship $T^2 \propto r^3$ for a satellite in circular orbit using gravitational and centripetal force equations
Calculate orbital period and orbital speed for a satellite at a given orbital radius
Describe the gravitational potential energy and kinetic energy of an orbiting satellite and how they vary with orbital radius
Explain the total mechanical energy of an orbiting satellite and why it is negative
Key ideas to keep in view
Use this lesson to stay inside Gravitational fields (A-level only) while connecting the detail back to the wider Fields and their consequences (A-level only) topic.
Orbital period: The time taken for a satellite to complete one full orbit around a central body. Measured in seconds (s).
Gravitational potential energy (in orbit): The energy a satellite possesses due to its position in a gravitational field, given by $E_p = -\frac{GMm}{r}$. Always negative, becoming less negative as orbital radius increases.
Total orbital energy: The sum of kinetic energy and gravitational potential energy for an orbiting satellite, equal to $-\frac{GMm}{2r}$. Always negative for a bound orbit.
Revision notes
Read the core explanations from Orbital Motion and Energy before testing yourself from memory.
1. Why Satellites Stay Up — and What Keeps Them Moving
Here is the puzzle: a satellite is falling towards Earth at every moment, yet it never arrives. The reason is that it is also moving sideways fast enough that the curved surface of the Earth falls away beneath it at the same rate. Gravity is not fighting the satellite's motion — it is the satellite's motion. That single insight unlocks everything in this topic.
For a circular orbit, the gravitational force provides exactly the centripetal force needed to keep the satellite on its circular path. Setting these two forces equal is the starting point for deriving how orbital period and speed depend on orbital radius.
2. Deriving $T^2 \propto r^3$ — Kepler's Third Law from First Principles
The gravitational force on a satellite of mass $m$ orbiting a central body of mass $M$ at radius $r$ is $F = \frac{GMm}{r^2}$. For circular motion, the centripetal force required is $F = \frac{mv^2}{r}$. Because gravity supplies the centripetal force, these two expressions must be equal.
Cancelling $m$ from both sides and rearranging gives $v^2 = \frac{GM}{r}$. This is the key result for orbital speed — notice that the satellite's own mass cancels out entirely, so orbital speed depends only on the central body's mass and the orbital radius.
Now substitute the expression for orbital speed in terms of period. The satellite travels a circumference $2\pi r$ in one period $T$, so $v = \frac{2\pi r}{T}$. Substituting into $v^2 = \frac{GM}{r}$ gives the derivation of Kepler's Third Law.
Worked example: Deriving $T^2 \propto r^3$
Show that the orbital period $T$ and orbital radius $r$ satisfy $T^2 \propto r^3$.
$$\frac{GMm}{r^2} = \frac{mv^2}{r}$$ — Gravity provides the centripetal force for circular orbit. Mass $m$ of the satellite appears on both sides.
$$v^2 = \frac{GM}{r}$$ — The satellite's own mass cancels — orbital speed is independent of satellite mass. [M1]
$$v = \frac{2\pi r}{T} \implies v^2 = \frac{4\pi^2 r^2}{T^2}$$ — The satellite travels one circumference $2\pi r$ in time $T$.
$$\frac{4\pi^2 r^2}{T^2} = \frac{GM}{r}$$ — Replace $v^2$ on the left with the period expression. [M1]
$$T^2 = \frac{4\pi^2}{GM} r^3$$ — Multiply both sides by $T^2$ and by $r$, then divide by $GM$. [A1]
$$T^2 \propto r^3 \quad \text{(since } \frac{4\pi^2}{GM} \text{ is constant for a given central body)}$$ — This is Kepler's Third Law. The constant $\frac{4\pi^2}{GM}$ depends only on the central body — so the same relationship holds for every satellite orbiting the same body.
The result $T^2 = \frac{4\pi^2}{GM} r^3$ is one of the most powerful equations in gravitational physics. It lets you find the mass of any body that has a satellite orbiting it — from Earth to the Sun to a distant exoplanet — by measuring $T$ and $r$ alone.
3. Orbital Speed — How Fast Must a Satellite Travel?
From Step 2 of the derivation, orbital speed is $v = \sqrt{\frac{GM}{r}}$. There is a counterintuitive result buried here: satellites in lower orbits move faster, not slower. Halving the orbital radius increases the speed by a factor of $\sqrt{2}$.
You can also find orbital speed from $v = \frac{2\pi r}{T}$ if you know the period instead. Both routes give the same answer — choose whichever quantities the question provides.
Worked example: Calculating Orbital Period and Speed
The International Space Station orbits Earth at an altitude of $4.0 \times 10^5$ m above the surface. The radius of Earth is $6.4 \times 10^6$ m and its mass is $6.0 \times 10^{24}$ kg. Calculate (a) the orbital speed and (b) the orbital period of the ISS.
$$r = R_{\text{Earth}} + h = 6.4 \times 10^6 + 4.0 \times 10^5$$
$$r = 6.4 \times 10^6 + 0.40 \times 10^6 = 6.80 \times 10^6 \text{ m}$$ — Orbital radius is measured from the centre of Earth, not from the surface. Always add the altitude to Earth's radius. [M1]
Answer: (a) $v \approx 7700$ m s⁻¹; (b) $T \approx 5570$ s (approximately 93 minutes)
4. Energy of an Orbiting Satellite — Kinetic and Potential
Now that the motion is clear, look at the energy. An orbiting satellite has two contributions to its total mechanical energy: kinetic energy from its motion, and gravitational potential energy from its position in the field. The interplay between these two is where the real physics lives.
The negative sign is not a quirk — it carries physical meaning. A satellite bound in orbit has less energy than one at infinity (where $E_p = 0$). The more negative $E_p$ is, the more tightly bound the satellite is to the central body.
Kinetic energy follows directly from the orbital speed result. Since $v^2 = \frac{GM}{r}$, substituting into $E_k = \frac{1}{2}mv^2$ gives a clean expression that links kinetic energy directly to orbital radius.
5. Total Energy — Why It Is Always Negative
The total mechanical energy of the satellite is the sum of its kinetic and potential energies. Combining the two expressions derived above gives a remarkably tidy result.
Worked example: Deriving Total Orbital Energy
Show that the total mechanical energy of a satellite in circular orbit is $E_{\text{total}} = -\frac{GMm}{2r}$.
$$E_k = +\frac{GMm}{2r}, \quad E_p = -\frac{GMm}{r}$$ — $E_k$ is always positive. $E_p$ is always negative for a bound orbit. Note that $|E_p| = 2E_k$ — the potential energy magnitude is always exactly twice the kinetic energy.
$$E_{\text{total}} = E_k + E_p = \frac{GMm}{2r} + \left(-\frac{GMm}{r}\right)$$ — Combine the two terms over a common denominator of $2r$.
$$E_{\text{total}} = \frac{GMm}{2r} - \frac{2GMm}{2r} = -\frac{GMm}{2r}$$ — The total energy is negative. This is the hallmark of a bound system — the satellite cannot escape to infinity without additional energy being supplied. [A1]
Answer: $$E_{\text{total}} = -\frac{GMm}{2r}$$
The negative total energy is not a mathematical accident. It tells you that the satellite is gravitationally bound — it would need a positive amount of energy equal to $\frac{GMm}{2r}$ to escape to infinity. This is the binding energy of the orbit.
There is also a striking relationship between the three energy quantities: $E_p = 2E_{\text{total}}$ and $E_k = -E_{\text{total}}$. These follow directly from the algebra above and are worth remembering for multi-step exam questions.
6. Energy Changes When a Satellite Changes Orbit
Connecting back to the derivation in Section 2: raising a satellite to a higher orbit increases $r$, which makes $E_{\text{total}} = -\frac{GMm}{2r}$ less negative — so total energy increases. Yet kinetic energy decreases. Both statements are true simultaneously, and understanding why is a classic exam challenge.
When a rocket boosts a satellite into a higher orbit, it does positive work on the satellite. This energy goes into the gravitational potential energy, which increases by twice as much as the kinetic energy decreases. The net result is that total energy increases (becomes less negative), even though the satellite slows down.
Reading the chart: as orbital radius increases, $E_k$ decreases (satellite slows), $E_p$ increases (becomes less negative), and $E_{\text{total}}$ increases (becomes less negative). The magnitude of $E_p$ is always exactly twice $E_k$, and $E_{\text{total}}$ is always exactly $-E_k$.
Worked example: Calculating Energy Changes Between Orbits
A satellite of mass $500$ kg orbits Earth ($M = 6.0 \times 10^{24}$ kg) at an orbital radius of $7.0 \times 10^6$ m. It is boosted to a new orbit of radius $1.4 \times 10^7$ m. Calculate the change in total energy of the satellite.
$$E_{\text{total}} = -\frac{GMm}{2r}$$ — This applies to any circular orbit. We need to evaluate it at both radii.
$$E_2 = -\frac{2.001 \times 10^{17}}{2 \times 1.4 \times 10^7} = -\frac{2.001 \times 10^{17}}{2.8 \times 10^7} = -7.146 \times 10^9 \text{ J}$$ — The numerator $GMm$ is the same as before. Only the denominator changes: $2 \times 1.4 \times 10^7 = 2.8 \times 10^7$. [M1]
$$\Delta E = E_2 - E_1 = -7.146 \times 10^9 - (-1.429 \times 10^{10})$$
$$= -7.146 \times 10^9 + 1.429 \times 10^{10}$$
$$= -7.146 \times 10^9 + 14.29 \times 10^9 = +7.14 \times 10^9 \text{ J}$$ — The change is positive — the satellite gains energy when moved to a higher orbit. This energy must be supplied by the rocket engine. [A1]
$$r_2 = 2r_1 \implies E_2 = \frac{E_1}{2} = \frac{-1.429 \times 10^{10}}{2} = -7.15 \times 10^9 \text{ J} \checkmark$$ — Since $E_{\text{total}} \propto \frac{1}{r}$ and $r_2 = 2r_1$, the total energy at $r_2$ should be half as negative as at $r_1$. This confirms the calculation.
Answer: $\Delta E \approx +7.1 \times 10^9$ J (the satellite gains approximately $7.1$ GJ of total mechanical energy)
7. Exam Technique — Getting Full Marks on Orbital Energy Questions
Orbital energy questions reward students who are precise about signs and who can explain the direction of energy changes without just quoting a formula. The following guidance targets the most common mark losses in AQA exam scripts.
Every spec requirement in this topic connects back to the same chain of reasoning: gravity provides centripetal force, which fixes orbital speed and period as functions of radius, which in turn determines all three energy quantities. Master that chain and the individual results follow naturally rather than needing to be memorised in isolation.
Sources and content standards
Published by Revise Right. Use the official specification reference to check the requirements for your course. Content updated: 2026-09-12.
Orbital speed is v = √(GM/r), found by equating gravitational and centripetal force; the satellite's own mass cancels. Lower orbits are faster: halving the radius increases the speed by a factor of √2. The ISS, at r = 6.80 × 10⁶ m around Earth (M = 6.0 × 10²⁴ kg), moves at about 7700 m s⁻¹ with a period of about 93 minutes.
How do you derive T² ∝ r³ for a satellite?
Set gravitational force equal to centripetal force, GMm/r² = mv²/r, so v² = GM/r. Substituting v = 2πr/T gives 4π²r²/T² = GM/r, which rearranges to T² = (4π²/GM)r³. Since 4π²/GM is constant for a given central body, T² ∝ r³, which is Kepler's third law.
What is the total energy of a satellite in orbit?
The total energy of a satellite in circular orbit is −GMm/2r, the sum of its kinetic energy, GMm/2r, and its potential energy, −GMm/r. It is negative because the satellite is gravitationally bound: it would need an extra GMm/2r of energy to escape to infinity, and its potential energy is always twice its total energy.
What happens to a satellite's energy when it moves to a higher orbit?
Moving to a higher orbit increases the total energy, which becomes less negative, but the kinetic energy decreases because the satellite slows down. The gravitational potential energy increases by twice as much as the kinetic energy decreases, so a rocket must do positive work. Boosting 500 kg from 7.0 × 10⁶ m to 1.4 × 10⁷ m needs about 7.1 × 10⁹ J.
How can you find the mass of a planet from a satellite's orbit?
Rearranging T² = (4π²/GM)r³ gives M = 4π²r³ ÷ (G × T²), so measuring the orbital period T and orbital radius r of any satellite gives the mass of the central body it orbits. This works for Earth, the Sun or a distant exoplanet, because the constant 4π²/GM depends only on the central body.
Does a satellite's mass affect its orbital speed?
A satellite's own mass does not affect its orbital speed, because it cancels when gravitational force is set equal to centripetal force: GMm/r² = mv²/r gives v² = GM/r. Orbital speed depends only on the central body's mass and the orbital radius.
Why do lower satellites move faster?
Satellites in lower orbits move faster because orbital speed is v = √(GM/r), so a lower orbit, with less radius, gives a higher speed. Halving the orbital radius increases the speed by a factor of √2.
What is the orbital period of the ISS?
The International Space Station orbits at an altitude of 4.0 × 10⁵ m, a radius of 6.80 × 10⁶ m from the centre. With Earth's mass of 6.0 × 10²⁴ kg, its orbital speed is about 7700 m s⁻¹ and its period about 5570 s, approximately 93 minutes.
What is the kinetic energy of a satellite in orbit?
The kinetic energy of a satellite in circular orbit is GMm/2r, found by substituting v² = GM/r into ½mv². It is always positive, and its magnitude is exactly half that of the potential energy, −GMm/r.
Is orbital radius measured from the centre or the surface?
Orbital radius is measured from the centre of the planet, not from its surface, so the altitude is always added to the planet's radius. For the ISS, r = 6.4 × 10⁶ + 4.0 × 10⁵ = 6.80 × 10⁶ m.
Why don't satellites fall to Earth?
A satellite is falling towards Earth at every moment, but it also moves sideways fast enough that the curved surface falls away beneath it at the same rate, so it never arrives. Gravity provides exactly the centripetal force needed to keep it on its circular path.
What is the binding energy of an orbit?
The binding energy of an orbit is GMm/2r, the positive energy a satellite would need to escape to infinity from a circular orbit of radius r. It equals the magnitude of the satellite's negative total energy, −GMm/2r.