Year 12 Physics Module 5 · IQ3: Gravitational Fields 45 min Practice bank · 5 MC Lesson 12 of 18

Energy in Orbits

For a circular orbit, moving higher gives a satellite more total energy but less kinetic energy. A signed energy ledger explains this result and distinguishes the instantaneous effect of a burn from the properties of the final circular orbit.

Today's hook: A prograde engine burn immediately makes a spacecraft faster, yet after it reaches and circularises in a higher orbit it moves more slowly than before. Where did the added energy go?
0/6TASKS
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Connect and orient

Warm up first

Three quick questions from earlier lessons. Pulling old material back to mind before you learn something new makes the new material stick better, so this is not busywork.

Worksheets

Practise this lesson

One printable worksheet for foundations, signs and circular-orbit energy relationships.

Build Foundations & guided practice
Think First: Bound or Free?
warm-up

A satellite in circular orbit has kinetic energy from its speed and gravitational potential energy from its position. Predict the signs of KE, U and total energy using the standard $U=0$ reference at infinity.

Learning Intentions
goals

Know

  • $KE = \frac{1}{2}\frac{GMm}{r}$ for any circular orbit.
  • $U = -\frac{GMm}{r}$, always negative, zero at infinity.
  • $E_\text{total} = -\frac{1}{2}\frac{GMm}{r}$ for a circular orbit.

Understand

  • Why negative total energy represents a gravitationally bound system.
  • Why the reference point for GPE is set at infinity, not Earth's surface.
  • Why a higher final circular orbit has less KE but greater total energy.

Can Do

  • Calculate KE, GPE, and total energy for a satellite at a given orbital radius.
  • Read signed KE, U and total-energy graphs against orbital radius.
  • Calculate the energy change between two circular orbits.
  • Compare total energies of satellites at different orbital radii.
Scan these before reading
vocab
Gravitational potential energy$U = -GMm/r$, always negative for finite separation; zero is defined at infinite distance. Work must be done against gravity to reach infinity.
Kinetic energy (orbital)$KE = \frac{1}{2}GMm/r$, derived from the orbital velocity formula. Always positive; decreases as orbital radius increases.
Total mechanical energy$E_\text{total} = KE + U = -\frac{1}{2}GMm/r$ for a circular orbit. A negative value means the system is gravitationally bound relative to $E=0$ at infinity.
Orbital energyEquivalent to total mechanical energy. More negative = lower orbit = more tightly bound satellite.
Circular energy relationFor a circular orbit, $E_\text{total}=-KE=U/2$. Do not apply these instantaneous-$r$ expressions unchanged to a non-circular transfer trajectory.
Lesson boundary: Lesson 11 supplied circular speed and period. Lesson 12 owns circular-orbit KE, gravitational PE, total energy and energy changes between circular orbits. Lesson 16 owns the full escape-speed derivation; Lesson 17 owns non-circular Kepler orbits and periapsis/apoapsis analysis.

Core Content

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Build the circular-orbit kinetic-energy model

Key Point
Module 5's energy topic is about the interplay between kinetic energy (always positive), gravitational potential energy (always negative in orbit), and total energy (always negative for bound systems). The signs are not arbitrary, they tell you whether a satellite is trapped or free.
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Kinetic Energy in Orbit
+5 XP

Positive energy from orbital motion

Fire a thruster on the Hubble Space Telescope (535 km orbit, launched 1990) to add energy. The orbit expands, Hubble moves to a higher altitude. At that higher altitude, the orbital speed is lower ($v = \sqrt{GM/r}$ decreases as $r$ increases). You added energy, yet the satellite slowed down. This counter-intuitive result follows directly from calculating the kinetic energy of an orbiting satellite.

From the circular orbit condition $\frac{GMm}{r^2} = \frac{mv^2}{r}$, we get $v^2 = \frac{GM}{r}$. Substituting into $KE = \frac{1}{2}mv^2$:

$$KE = \frac{1}{2}m \cdot \frac{GM}{r} = \frac{GMm}{2r}$$
Key insight: $KE \propto 1/r$, kinetic energy decreases as orbital radius increases. Satellites in higher orbits move slower.
Energy in Orbits diagram showing KE, GPE, and total energy as functions of orbital radius

Energy quantities for a circular orbit as a function of radius $r$. Note that $|U| = 2KE$ and $E_\text{total} = -KE$ for all circular orbits.

Worked Example, Kinetic Energy of a Satellite

A 500 kg satellite orbits Earth at $r = 7.0 \times 10^6$ m from Earth's centre. Calculate its kinetic energy. ($G = 6.67 \times 10^{-11}$ N m²/kg², $M_E = 5.97 \times 10^{24}$ kg)

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Formula: $KE = \frac{GMm}{2r}$
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Substitute: $$KE = \frac{6.67 \times 10^{-11} \times 5.97 \times 10^{24} \times 500}{2 \times 7.0 \times 10^6}$$
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Calculate: $KE = 1.42 \times 10^{10}$ J
Answer: $KE = 1.42 \times 10^{10}$ J (positive, as expected)

Circular-orbit KE: $KE = GMm/(2r)$ (positive), derived by substituting $v^2=GM/r$ into $KE=\tfrac12mv^2$. It applies to the stated circular model. Higher circular orbit → lower speed → less KE.

Pause, copy the highlighted orbital KE formula and the $1/r$ trend into your book before moving on.

A satellite moves to a higher orbit. Its kinetic energy will:

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Define gravitational potential energy and its reference

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Gravitational Potential Energy
+5 XP

Always negative, zero at infinity

We just saw that orbital KE is always positive. That raises a question: what about gravitational potential energy, why is the standard formula $U = -GMm/r$ negative? This card answers it → the reference is set at infinity ($U = 0$); all finite orbits are below that, so $U < 0$.

For radial gravitational fields, the standard reference is $U=0$ at infinite separation. With that convention, every finite separation has negative gravitational potential energy.

$$U = -\frac{GMm}{r}$$
Always negative for finite $r$. Work must be done against gravity to separate the two masses to infinity (where $U = 0$).

The negative sign reflects gravity being an attractive force. To bring a satellite from distance $r$ to infinity, you must do positive work on it, raising $U$ from a negative value up to zero.

Reference point
Near Earth's surface, a convenient local zero may be chosen when only $\Delta U$ matters. The radial expression $U=-GMm/r$ specifically uses $U=0$ at infinity. State the reference before interpreting a sign.

More negative = deeper in the gravitational well = closer to the central mass. A satellite at $r = 7.0 \times 10^6$ m has more negative $U$ than one at $r = 10.0 \times 10^6$ m. It is harder to move to infinity from a lower orbit.

Worked Example, Gravitational Potential Energy

Calculate the gravitational potential energy of the same 500 kg satellite at $r = 7.0 \times 10^6$ m.

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Formula: $U = -\frac{GMm}{r}$
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Substitute: $$U = -\frac{6.67 \times 10^{-11} \times 5.97 \times 10^{24} \times 500}{7.0 \times 10^6}$$
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Calculate: $U = -2.84 \times 10^{10}$ J
Answer: $U = -2.84 \times 10^{10}$ J (negative, as expected for bound orbit)

Gravitational PE: $U = -GMm/r$, always negative at finite $r$; zero at infinity. Lower orbit → more negative $U$ (deeper in gravitational well). Cannot use $U = mgh$ for orbital problems, different reference.

Add the highlighted GPE formula and the reference-point rule to your notes before the check below.

Gravitational potential energy is negative for a satellite at any finite distance from a planet.

In orbital mechanics, gravitational potential energy is defined as zero at Earth's surface.

A satellite at a smaller orbital radius has a more negative gravitational potential energy than one at a larger radius.

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Combine KE and U for a circular orbit

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Total Energy in Bound Orbits
+5 XP

The sum that determines whether an object remains bound

We just saw that KE is positive and $U$ is negative. That raises a question: when you add them together, what does the sign of the total energy tell you? This card answers it → $E_\text{total} = -GMm/(2r)$ (always negative = bound); binding energy $= |E_\text{total}|$.

Adding KE and U for a circular orbit gives a result that is both simple and physically profound. The total energy is always negative, which tells us everything about whether the satellite is trapped or free.

$$E_\text{total} = KE + U = \frac{{G}{M}{m}}{2r} + \left(-\frac{{G}{M}{m}}{r}\right) = -\frac{{G}{M}{m}}{2r}$$
Critical result: $E_\text{total} = -KE = U/2$, the total energy is the negative of the kinetic energy, and half the potential energy. This is the Virial theorem for circular orbits.

The total energy of this circular orbit is negative relative to $E=0$ at infinity, so the system is gravitationally bound. For a non-circular bound orbit, total energy remains constant but the general expression uses semi-major axis rather than the instantaneous radius; that belongs in Lesson 17.

Energy boundary preview
The magnitude $|E_\text{total}|$ is the energy gap from the circular orbit to the $E=0$ boundary. Lesson 16 uses this boundary to derive escape speed; this lesson does not derive or assess that speed.
Worked Example, Total Energy and Binding Energy

Calculate the total energy and binding energy of the 500 kg satellite at $r = 7.0 \times 10^6$ m. (Use previous results: $KE = 1.42 \times 10^{10}$ J, $U = -2.84 \times 10^{10}$ J)

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Total energy: $E_\text{total} = KE + U = 1.42 \times 10^{10} + (-2.84 \times 10^{10}) = -1.42 \times 10^{10}$ J
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Virial check: $E_\text{total} = -KE = -1.42 \times 10^{10}$ J ✓ and $E_\text{total} = U/2 = -2.84 \times 10^{10}/2$ ✓
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Binding energy: $|E_\text{total}| = 1.42 \times 10^{10}$ J
Answer: $E_\text{total} = -1.42 \times 10^{10}$ J; binding energy $= 1.42 \times 10^{10}$ J

Circular-orbit total energy: $E_\text{total}=-GMm/(2r)<0$. Circular relation: $E_\text{total}=-KE=U/2$. More negative $E$ means a smaller circular radius and a more tightly bound system relative to the $E=0$ reference.

Pause, copy the highlighted total energy formula and the Virial theorem into your book before moving on.

A satellite has total mechanical energy $E = -5.0 \times 10^{10}$ J. Compared to a satellite with $E = -2.0 \times 10^{10}$ J (orbiting the same planet), the first satellite is:

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Interpret signed energy–radius graphs

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Energy Against Circular-Orbit Radius
+5 XP

Read value, sign and gradient trend without confusing energy with force

  • $KE=+GMm/(2r)$ lies above zero and approaches zero as $r$ increases.
  • $U=-GMm/r$ lies below zero and approaches zero from below.
  • $E=-GMm/(2r)$ also lies below zero, halfway between $U$ and zero at every circular-orbit radius.
Graph discipline
The vertical coordinate is energy in joules, not force. All three curves flatten as $r$ increases. At a fixed $r$, compare signed values: $U=2E=-2KE$.

Open energy demonstrator fullscreen ↗

As circular-orbit radius increases: KE decreases; U increases toward zero; total energy increases toward zero. The final orbit is less tightly bound even though its circular speed is smaller.

On an $E$-against-$r$ graph for circular orbits, which quantity approaches zero from below and always has half the magnitude of $U$?

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Analyse energy changes between circular orbits

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Burns, Transfer Motion and Final Circular Orbits

A prograde burn does positive work and immediately increases the spacecraft's speed and total energy. At that instant the craft is not yet in the higher circular orbit; it is on a transfer trajectory. A second appropriately directed burn at the new radius circularises the orbit.

Only after circularisation may the final circular-orbit formulas be compared directly:

$$\Delta E=E_f-E_i=-\frac{GMm}{2r_f}+\frac{GMm}{2r_i}$$
For $r_f>r_i$, $\Delta E>0$: total energy increases (becomes less negative), while the final circular-orbit KE is smaller.
✗ “The first burn instantly places the craft in the higher circular orbit.”
✓ The burn changes velocity at one position and begins a transfer trajectory. Circularisation requires a later burn at the destination radius.
✗ “Adding energy must make the final circular speed larger.”
✓ The added energy mainly raises gravitational potential energy. At the larger final radius, $v=\sqrt{GM/r}$ and KE are smaller.

Activities

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Respond using a signed energy ledger

Activity 1, Orbital Energy Calculations
ApplyBand 4

A 600 kg satellite orbits Earth at $r = 7.5 \times 10^6$ m from Earth's centre. ($G = 6.67 \times 10^{-11}$ N m²/kg², $M_E = 5.97 \times 10^{24}$ kg)

  1. Calculate the satellite's kinetic energy using $KE = GMm/(2r)$.
  2. Calculate its gravitational potential energy using $U = -GMm/r$.
  3. Calculate the total mechanical energy $E_\text{total}$.
  4. Verify that $|E_\text{total}| = KE$ and $U = 2E_\text{total}$ (Virial theorem). Show all three values explicitly.
  5. The satellite is boosted to $r = 1.5 \times 10^7$ m. Calculate the change in total energy and explain whether energy was added to or removed from the satellite's orbit.
Activity 2, Understanding Negative Total Energy
EvaluateBand 5

A satellite has total mechanical energy $E = -2.0 \times 10^{10}$ J. A second satellite (same mass, same planet) has $E = -5.0 \times 10^{10}$ J.

  1. Explain what the negative value of total energy means physically. Why does negative total energy indicate the satellite is bound?
  2. How much energy must be supplied to move the first satellite from its orbit to infinity?
  3. Which satellite is more tightly bound? Explain your reasoning in terms of orbital radius and binding energy.
Copy Into Books, Key Formulas

Orbital KE

  • $KE = \dfrac{GMm}{2r}$, positive; decreases as $r$ increases.

Gravitational PE

  • $U = -\dfrac{GMm}{r}$, always negative; zero at infinity.

Total orbital energy

  • $E_\text{total} = -\dfrac{GMm}{2r}$, always negative for bound orbits.
  • Virial: $E_\text{total} = -KE = U/2$

Orbit change

  • $\Delta E=E_f-E_i$
  • Higher final circle: $\Delta E>0$, smaller KE, less-negative $U$ and $E$.
Revisit Your Thinking

Now that you know $E_\text{total}=-GMm/(2r)$ for a circular orbit, revisit your initial response. A higher final circular orbit has greater total energy (less negative), less KE and less-negative $U$. Explain why this does not contradict the immediate speed increase caused by the first prograde burn.

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Apply circular-orbit energy independently

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Multiple Choice
+5 XP

A fresh set drawn from this lesson's question bank, feedback shown immediately. +5 XP per correct · +25 XP all correct

Pick your answer, then rate your confidence, that tells the system what to drill next.

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Multiple Choice, In-Lesson
+5 XP

Five targeted questions on Energy in Orbits

1. The total energy of a bound orbit is:

APositive, the satellite has kinetic energy
BZero, the satellite is in equilibrium
CNegative, $E_\text{total} = -\frac{GMm}{2r}$
DIt depends on the reference point chosen

2. A prograde burn begins a transfer from a lower circular orbit to a higher one. Immediately after the first burn, the craft is:

AAlready in the higher circular orbit
BOn a transfer trajectory with increased total energy
CIn a lower circular orbit with unchanged energy
DOutside the gravitational field

3. Gravitational potential energy in orbital mechanics is defined as zero at:

AEarth's surface
BThe centre of the planet
CInfinite separation
DLow Earth orbit altitude

4. For a circular orbit, the Virial theorem states:

A$E_\text{total} = KE + U$ only when the orbit is geostationary
B$E_\text{total} = -KE = U/2$
C$E_\text{total} = U - KE$
D$E_\text{total} = 0$ always

5. Doubling the orbital radius of a satellite makes its total energy:

ATwice as negative (more bound)
BHalf as negative (less bound, closer to escape)
CFour times as negative
DUnchanged, total energy is conserved
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Short Answer, 10 marks
+5 XP

ApplyBand 4(3 marks) 1. An 800 kg satellite orbits Earth at an altitude of 500 km. Calculate its kinetic energy, gravitational potential energy, and total energy. ($G = 6.67 \times 10^{-11}$ N m²/kg², $M_E = 5.97 \times 10^{24}$ kg, $R_E = 6.37 \times 10^6$ m)

ApplyBand 5(3 marks) 2. A 500 kg satellite moves between circular Earth orbits of radii $7.0\times10^6$ m and $1.4\times10^7$ m. Calculate $\Delta E$ and explain why the final circular speed is lower even though energy was added.

EvaluateBand 5–6(4 marks) 3. Evaluate the statement: "A satellite with total energy $E = -3 \times 10^{10}$ J is more tightly bound than one with $E = -1 \times 10^{10}$ J." Explain what "more tightly bound" means physically and calculate the energy required to move each satellite to infinity.

Show all answers

Multiple choice (in-lesson)

MC1: C, Negative ($E_\text{total} = -GMm/(2r)$)

MC2: B, on a transfer trajectory with increased total energy

MC3: C, Infinite separation

MC4: B, $E_\text{total} = -KE = U/2$

MC5: B, Half as negative (less bound)

Activity 1, Orbital Energy Calculations

At $r = 7.5 \times 10^6$ m: $KE = GMm/(2r) = (6.67 \times 10^{-11} \times 5.97 \times 10^{24} \times 600)/(2 \times 7.5 \times 10^6) = 1.59 \times 10^{10}$ J

$U = -GMm/r = -3.18 \times 10^{10}$ J

$E_\text{total} = KE + U = 1.59 \times 10^{10} - 3.18 \times 10^{10} = -1.59 \times 10^{10}$ J

Virial check: $|E_\text{total}| = KE = 1.59 \times 10^{10}$ J ✓; $U/2 = -3.18 \times 10^{10}/2 = -1.59 \times 10^{10}$ J = $E_\text{total}$ ✓

At $r = 1.5 \times 10^7$ m: $E_\text{new} = -0.795 \times 10^{10}$ J. $\Delta E = +7.95 \times 10^9$ J (energy was added; orbit is higher and less bound).

Activity 2, Negative Total Energy

Negative total energy means the satellite does not have enough energy to reach infinity (where $E = 0$). It is gravitationally bound, gravity holds it in orbit.

Energy to move satellite 1 ($E = -2.0 \times 10^{10}$ J) to infinity: $|E_\text{total}| = 2.0 \times 10^{10}$ J must be supplied.

Satellite 2 ($E = -5.0 \times 10^{10}$ J) is more tightly bound because its energy is more negative, it is in a lower orbit (smaller $r$) and $5.0 \times 10^{10}$ J must be supplied to free it. It requires 2.5 times more energy to escape to infinity.

Short Answer Model Responses

Q1 (3 marks): $r = 6.37 \times 10^6 + 5.00 \times 10^5 = 6.87 \times 10^6$ m [1]

$KE = GMm/(2r) = (6.67 \times 10^{-11} \times 5.97 \times 10^{24} \times 800)/(2 \times 6.87 \times 10^6) = 2.32 \times 10^{10}$ J [1]

$U = -GMm/r = -4.64 \times 10^{10}$ J; $E_\text{total} = 2.32 \times 10^{10} - 4.64 \times 10^{10} = -2.32 \times 10^{10}$ J [1]

Q2 (3 marks): $E_i=-GMm/(2r_i)=-1.42\times10^{10}$ J [1]. $E_f=-GMm/(2r_f)=-7.11\times10^9$ J, so $\Delta E=E_f-E_i=+7.11\times10^9$ J [1]. Energy was added because total energy became less negative. The final circular speed is lower because $v=\sqrt{GM/r}$ decreases as the final radius increases; the gain in $U$ is larger than the decrease in KE [1].

Q3 (4 marks): The statement is TRUE [1]. "More tightly bound" means the satellite is in a lower orbit (more negative $E_\text{total} = -GMm/(2r)$ implies smaller $r$) and more energy is required to move it to infinity [1]. Satellite 1 ($E = -3 \times 10^{10}$ J): binding energy $= 3 \times 10^{10}$ J [1]. Satellite 2 ($E = -1 \times 10^{10}$ J): binding energy $= 1 \times 10^{10}$ J. Satellite 1 requires 3× more energy to escape, it is deeper in the gravitational well [1].

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Retrieve and reflect

Check what actually stuck
RAPID REVIEW
The big ideas in four tiles

KE in orbit

$KE = GMm/(2r)$, positive; decreases as $r$ increases. Higher orbit = slower satellite.

GPE in orbit

$U = -GMm/r$, always negative; zero at infinity. Lower orbit = more negative $U$.

Total energy

$E_\text{total} = -GMm/(2r)$, always negative for bound orbits. Virial: $E_\text{total} = -KE = U/2$.

Orbit change

Higher final circular orbit: $\Delta E>0$; total energy and U increase, while KE and circular speed decrease.

Asteroid Blaster, Energy in Orbits
boss

Rapid-fire questions on kinetic energy, gravitational potential energy, total orbital energy and energy changes between circular orbits.

How did your thinking change?

Return to your Think First response and explain why a circular satellite has positive KE, negative U and negative total energy relative to $U=0$ at infinity.

Then distinguish two moments in an orbit raise: immediately after the prograde burn, speed and total energy have increased and the craft is on a transfer trajectory; after circularisation at the larger radius, total energy remains greater but circular speed and KE are smaller.

← Previous: Gravitational Orbits Next: Circular Motion and Orbits Consolidation →

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