1
Learn
2
Questions
3
Game
01

Calculate KE, GPE, and total energy in bound orbits

02

Derive and use escape velocity

03

Explain the significance of negative total energy

Think First

CONCEPTUAL — A satellite in orbit has kinetic energy. Is its total energy positive, negative, or zero? What does this tell you about whether it can escape?

Autosaved
Year 12 Physics Module 5: Advanced Mechanics 45 min Lesson 12 of 18 IQ2: Circular Motion

⚡ Energy in Orbits

Calculate kinetic energy, gravitational potential energy, and total energy in bound orbits. Derive escape velocity using energy conservation.

1

Kinetic Energy in Orbit

Positive energy from orbital motion

From our derivation of orbital velocity $v = \sqrt{GM/r}$, we can find the kinetic energy of a satellite in circular orbit. The satellite mass $m$ cancels in the orbital velocity formula, but it remains in the energy expression.

Energy In Orbits

Energy In Orbits

Σ
Deriving orbital kinetic energy
$$KE = \frac{1}{2}mv^2$$
$$KE = \frac{1}{2}m \left(\sqrt{\frac{GM}{r}}\right)^2$$
$$KE = \frac{1}{2}\frac{GMm}{r}$$
Key insight: KE is positive and decreases as $r$ increases — satellites in larger orbits move slower.

Since $KE \propto 1/r$, a satellite at a higher altitude (larger $r$) has less kinetic energy than one in a lower orbit. This may seem counterintuitive at first, but remember: orbital speed decreases with altitude, and this decrease in $v^2$ outweighs any constant mass term.

Worked Example — Kinetic Energy of a Satellite
A 500 kg satellite orbits Earth at a centre-to-centre distance of $r = 7.0 \times 10^6 \text{ m}$. Calculate its kinetic energy. ($G = 6.67 \times 10^{-11} \text{ N m}^2\text{/kg}^2$, $M_E = 5.97 \times 10^{24} \text{ kg}$)
1
GIVEN
$m = 500 \text{ kg}$, $r = 7.0 \times 10^6 \text{ m}$, $G = 6.67 \times 10^{-11} \text{ N m}^2\text{/kg}^2$, $M_E = 5.97 \times 10^{24} \text{ kg}$
2
FIND
Kinetic energy $KE$
3
METHOD
Use $KE = \frac{1}{2}GMm/r$
$$KE = \frac{1}{2} \times \frac{6.67 \times 10^{-11} \text{ N m}^2\text{/kg}^2 \times 5.97 \times 10^{24} \text{ kg} \times 500 \text{ kg}}{7.0 \times 10^6 \text{ m}}$$
$$KE = \frac{1}{2} \times \frac{1.99 \times 10^{17} \text{ J m}}{7.0 \times 10^6 \text{ m}}$$
$$KE = \frac{1}{2} \times 2.84 \times 10^{10} \text{ J} = 1.42 \times 10^{10} \text{ J}$$
ANSWER: $KE = 1.42 \times 10^{10} \text{ J}$ (or $1.42 \times 10^{10} \text{ J} = 14.2 \text{ GJ}$)
2

Gravitational Potential Energy

Always negative in bound systems — zero at infinity

Gravitational potential energy in orbital mechanics is defined with a specific convention: zero at infinite separation. This means all finite orbits have negative potential energy.

Σ
Gravitational potential energy
$$U = -\frac{GMm}{r}$$
U = gravitational potential energy (J) — always negative for finite $r$
Negative sign: Work must be done against gravity to move a satellite to infinity (where U = 0)

The negative sign is essential. It reflects the fact that gravity is an attractive force. To move a satellite from distance $r$ to infinity, you must do positive work on it — increasing its potential energy from a negative value up to zero.

Reference point matters: Unlike near-Eurface problems where we often set U = 0 at ground level, in orbital mechanics we define U = 0 at infinity. This is the only convention that makes the virial theorem and energy conservation work cleanly for orbits.

More negative = lower energy = closer to the central mass. A satellite at $r = 7.0 \times 10^6 \text{ m}$ has a more negative U than one at $r = 10.0 \times 10^6 \text{ m}$. The satellite closer to Earth is deeper in the gravitational well.

Worked Example — Gravitational Potential Energy of a Satellite
Calculate the gravitational potential energy of the same 500 kg satellite at $r = 7.0 \times 10^6 \text{ m}$ from Earth's centre.
1
GIVEN
$m = 500 \text{ kg}$, $r = 7.0 \times 10^6 \text{ m}$, $G = 6.67 \times 10^{-11} \text{ N m}^2\text{/kg}^2$, $M_E = 5.97 \times 10^{24} \text{ kg}$
2
FIND
Gravitational potential energy $U$
3
METHOD
Use $U = -GMm/r$
$$U = -\frac{6.67 \times 10^{-11} \text{ N m}^2\text{/kg}^2 \times 5.97 \times 10^{24} \text{ kg} \times 500 \text{ kg}}{7.0 \times 10^6 \text{ m}}$$
$$U = -\frac{1.99 \times 10^{17} \text{ J m}}{7.0 \times 10^6 \text{ m}}$$
$$U = -2.84 \times 10^{10} \text{ J}$$
ANSWER: $U = -2.84 \times 10^{10} \text{ J}$ (or $-28.4 \text{ GJ}$)
3

Total Energy in Bound Orbits

The sum that determines whether an object remains bound

The total mechanical energy of an orbiting satellite is the sum of its kinetic and gravitational potential energies. For a circular orbit, this sum simplifies to an elegant result.

Σ
Total energy derivation
$$E_{\text{total}} = KE + U$$
$$E_{\text{total}} = \frac{1}{2}\frac{GMm}{r} + \left(-\frac{GMm}{r}\right)$$
$$E_{\text{total}} = -\frac{1}{2}\frac{GMm}{r}$$
Critical result: $E_{\text{total}} = -\frac{1}{2}\frac{GMm}{r} = -KE = \frac{U}{2}$

The total energy is always negative for any bound orbit. This is the defining characteristic of a gravitationally bound system: the satellite does not have enough energy to escape to infinity.

Binding energy: The energy needed to move a satellite from its orbit to infinity (where $E = 0$) is $|E_{\text{total}}| = \frac{1}{2}GMm/r$. This is called the binding energy — you must supply at least this much energy to free the satellite from Earth's gravity.
Worked Example — Total Energy and Binding Energy
Calculate the total energy and binding energy of the 500 kg satellite at $r = 7.0 \times 10^6 \text{ m}$.
1
GIVEN
$KE = 1.42 \times 10^{10} \text{ J}$, $U = -2.84 \times 10^{10} \text{ J}$ (from previous examples)
2
FIND
Total energy $E_{\text{total}}$ and binding energy
3
METHOD
Use $E_{\text{total}} = KE + U = -\frac{1}{2}GMm/r$
$$E_{\text{total}} = 1.42 \times 10^{10} \text{ J} + (-2.84 \times 10^{10} \text{ J})$$
$$E_{\text{total}} = -1.42 \times 10^{10} \text{ J}$$
$$\text{Binding energy} = |E_{\text{total}}| = 1.42 \times 10^{10} \text{ J}$$
ANSWER: $E_{\text{total}} = -1.42 \times 10^{10} \text{ J}$; binding energy = $1.42 \times 10^{10} \text{ J}$
4

Escape Velocity Derivation

The minimum speed to break free from gravity

Escape velocity is the minimum initial speed needed for an object to travel from a given point to infinity, with just enough kinetic energy to overcome gravitational binding. At infinity, both the velocity and potential energy approach zero.

Σ
Escape velocity derivation
$$\text{Condition: } KE + U \geq 0 \text{ at launch point}$$
$$\frac{1}{2}mv_e^2 - \frac{GMm}{r} = 0$$
$$\frac{1}{2}mv_e^2 = \frac{GMm}{r}$$
$$v_e = \sqrt{\frac{2GM}{r}}$$
Key relation: $v_e = \sqrt{2} \times v_{\text{orbital}}$ at the same radius — the factor of 2 under the square root is mandatory

Notice that the mass of the escaping object $m$ cancels out — escape velocity depends only on the central body's mass and the starting distance from its centre. A spacecraft and a single proton launched from the same point need the same initial speed to escape (ignoring atmospheric drag).

HSC Tip: The factor of 2 inside the square root is essential. A common error is writing $v_e = \sqrt{GM/r}$, which is actually the orbital velocity. Escape velocity is $\sqrt{2} \approx 1.41$ times greater than orbital velocity at the same radius.

For Earth, at the surface ($r = R_E = 6.37 \times 10^6 \text{ m}$):

$$v_e = \sqrt{\frac{2 \times 6.67 \times 10^{-11} \text{ N m}^2\text{/kg}^2 \times 5.97 \times 10^{24} \text{ kg}}{6.37 \times 10^6 \text{ m}}}$$
$$v_e = \sqrt{1.25 \times 10^8 \text{ m}^2\text{/s}^2} = 1.12 \times 10^4 \text{ m/s} = 11.2 \text{ km/s}$$
Worked Example — Escape Velocity from Mars
Calculate the escape velocity from the surface of Mars. ($M_M = 6.42 \times 10^{23} \text{ kg}$, $R_M = 3.40 \times 10^6 \text{ m}$)
1
GIVEN
$M_M = 6.42 \times 10^{23} \text{ kg}$, $R_M = 3.40 \times 10^6 \text{ m}$, $G = 6.67 \times 10^{-11} \text{ N m}^2\text{/kg}^2$
2
FIND
Escape velocity $v_e$ from Mars's surface
3
METHOD
Use $v_e = \sqrt{2GM_M/R_M}$
$$v_e = \sqrt{\frac{2 \times 6.67 \times 10^{-11} \text{ N m}^2\text{/kg}^2 \times 6.42 \times 10^{23} \text{ kg}}{3.40 \times 10^6 \text{ m}}}$$
$$v_e = \sqrt{\frac{8.56 \times 10^{13} \text{ m}^3\text{/s}^2}{3.40 \times 10^6 \text{ m}}}$$
$$v_e = \sqrt{2.52 \times 10^7 \text{ m}^2\text{/s}^2} = 5.02 \times 10^3 \text{ m/s}$$
ANSWER: $v_e = 5.02 \times 10^3 \text{ m/s} = 5.02 \text{ km/s}$ (approximately half Earth's escape velocity)
Σ
Key Formulas — Energy in Orbits
$$KE = \frac{1}{2}\frac{GMm}{r}$$
Kinetic energy of an orbiting body (J). Positive.
$$U = -\frac{GMm}{r}$$
Gravitational potential energy (J). Negative; zero at infinity.
$$E_{\text{total}} = -\frac{1}{2}\frac{GMm}{r}$$
Total orbital energy (J). Negative for all bound orbits.
$$v_e = \sqrt{\frac{2GM}{r}}$$
Escape velocity (m/s). Factor of 2 is mandatory.
$$v_e = \sqrt{2} \times v_{\text{orbital}} \text{ at same } r$$
Escape velocity is $\sqrt{2} \approx 1.41$ times orbital speed at the same radius.
Common Misconceptions
×
"A satellite has zero potential energy in orbit"

$U = -GMm/r$ is negative everywhere at finite distance from a mass. Zero potential energy is defined at infinity only. A satellite in orbit is deep in a gravitational well — its potential energy is significantly negative.

×
"Escape velocity depends on the mass of the escaping object"

$v_e = \sqrt{2GM/r}$ — the projectile mass $m$ cancels in the derivation. A feather and a rocket require the same initial speed to escape from the same point (neglecting atmospheric drag).

×
"You need to maintain thrust at escape velocity"

Once escape velocity is reached, no further energy input is needed. The object will coast to infinity, slowing down asymptotically toward zero speed (in the absence of other masses and atmosphere). Escape velocity is the initial speed required at the starting point — not a speed that must be maintained.

Real World: Spacecraft Propulsion

Voyager 1 and the Oberth Effect

Voyager 1 escaped the solar system using gravity assists (slingshot manoeuvres) from Jupiter and Saturn. These increased its speed without burning fuel by transferring orbital energy from the planets to the spacecraft. The planets lost an infinitesimal amount of orbital energy; the spacecraft gained a transformative boost.

The Oberth effect explains why burning engines at periapsis (closest approach) is most fuel-efficient: because orbital speed is highest at the closest point, the same $\Delta v$ produces the greatest change in kinetic energy. The relationship $\Delta E = \frac{1}{2}m(v + \Delta v)^2 - \frac{1}{2}mv^2$ shows that for a given $\Delta v$, the energy gain is maximised when $v$ is already large. This is why spacecraft fire their engines at closest approach during flybys.

A

Activities

Q1. Find the KE, U, and $E_{\text{total}}$ for a 1000 kg satellite orbiting at $r = 10{,}000 \text{ km}$ from Earth's centre. ($G = 6.67 \times 10^{-11} \text{ N m}^2\text{/kg}^2$, $M_E = 5.97 \times 10^{24} \text{ kg}$)

[3 marks]

Q2. Calculate the escape velocity from the Moon's surface. ($M_M = 7.35 \times 10^{22} \text{ kg}$, $R_M = 1.74 \times 10^6 \text{ m}$)

[2 marks]

Q3. A satellite in low Earth orbit has $E_{\text{total}} = -4.5 \times 10^{10} \text{ J}$. How much energy must be supplied to move it to geostationary orbit, where its total energy is $-1.2 \times 10^{10} \text{ J}$?

[2 marks]

Copy Into Books

Show notes to copy
Kinetic energy in orbit:

$$KE = \frac{1}{2}\frac{GMm}{r}$$

Derived from $v = \sqrt{GM/r}$. Always positive. Decreases as $r$ increases.

Gravitational potential energy:

$$U = -\frac{GMm}{r}$$

Always negative for finite $r$. Zero at infinity.

Total energy (bound orbits):

$$E_{\text{total}} = -\frac{1}{2}\frac{GMm}{r}$$

Always negative. Binding energy = $|E_{\text{total}}|$.

Escape velocity:

$$v_e = \sqrt{\frac{2GM}{r}}$$

Factor of 2 is mandatory. $v_e = \sqrt{2} \times v_{\text{orbital}}$ at same $r$. Independent of projectile mass.

Revisit Your Thinking

Now that you have learned the energy equations, revisit your initial answer. For a bound satellite: $E_{\text{total}} = -\frac{1}{2}GMm/r$, which is always negative. This negative total energy is what keeps the satellite bound — it does not have enough energy to reach infinity (where $E = 0$). The more negative the total energy, the more tightly bound the satellite is. Only if $E_{\text{total}} \geq 0$ could the satellite escape.

Autosaved
Interactive: Energy Orbits Interactive
Key Terms
Kinetic energy (orbit)
$KE = \frac{1}{2}GMm/r$ — positive energy from orbital motion
Gravitational potential energy
$U = -GMm/r$ — negative; zero defined at infinity
Total orbital energy
$E_{\text{total}} = -\frac{1}{2}GMm/r$ — negative for all bound orbits
Binding energy
$|E_{\text{total}}| = \frac{1}{2}GMm/r$ — energy needed to escape to infinity
Escape velocity
$v_e = \sqrt{2GM/r}$ — minimum speed to escape from distance $r$
Centre-to-centre distance
$r = R + h$ — always measure from the centre of the central body
Multiple Choice

1. The total energy of a bound orbit is:

A Positive
B Zero
C Negative
D Depends on the reference point

2. Escape velocity at distance $r$ from a mass $M$ is:

A $\sqrt{GM/r}$
B $\sqrt{2GM/r}$
C $2\sqrt{GM/r}$
D $GM/r$

3. Compared to orbital velocity at the same radius, escape velocity is:

A The same
B $\sqrt{2}$ times greater
C Twice as great
D Half as much

4. Gravitational potential energy is defined as zero at:

A Earth's surface
B The centre of Earth
C Infinity
D Low Earth orbit

5. Doubling the orbital radius of a satellite:

A Doubles its total energy
B Halves its total energy (makes it less negative)
C Quarters its total energy
D Does not change total energy
Short Answer

Q1. An 800 kg satellite orbits Earth at an altitude of 500 km. Calculate its kinetic energy, gravitational potential energy, and total energy.

($G = 6.67 \times 10^{-11} \text{ N m}^2\text{/kg}^2$, $M_E = 5.97 \times 10^{24} \text{ kg}$, $R_E = 6.37 \times 10^6 \text{ m}$)

[3 marks] Apply Band 4-5

Q2. Calculate the escape velocity from the surface of Earth. Show algebraically why this result is independent of the mass of the projectile.

[3 marks] Apply Band 5-6

Q3. Evaluate the statement: "A satellite with total energy $E = -3 \times 10^{10} \text{ J}$ is more tightly bound than one with $E = -1 \times 10^{10} \text{ J}$." Explain what "more tightly bound" means physically and calculate the energy required to move each satellite to infinity.

[4 marks] Evaluate Band 6
Model Answers

Q1. Satellite at 500 km altitude — 3 marks

Step 1 — Find orbital radius (centre-to-centre):

$$r = R_E + h = 6.37 \times 10^6 \text{ m} + 5.00 \times 10^5 \text{ m} = 6.87 \times 10^6 \text{ m}$$

Step 2 — Calculate kinetic energy:

$$KE = \frac{1}{2}\frac{GMm}{r} = \frac{1}{2} \times \frac{6.67 \times 10^{-11} \text{ N m}^2\text{/kg}^2 \times 5.97 \times 10^{24} \text{ kg} \times 800 \text{ kg}}{6.87 \times 10^6 \text{ m}}$$

$$KE = 2.32 \times 10^{10} \text{ J}$$

Step 3 — Calculate gravitational potential energy:

$$U = -\frac{GMm}{r} = -\frac{6.67 \times 10^{-11} \text{ N m}^2\text{/kg}^2 \times 5.97 \times 10^{24} \text{ kg} \times 800 \text{ kg}}{6.87 \times 10^6 \text{ m}}$$

$$U = -4.64 \times 10^{10} \text{ J}$$

Step 4 — Calculate total energy:

$$E_{\text{total}} = KE + U = 2.32 \times 10^{10} \text{ J} + (-4.64 \times 10^{10} \text{ J}) = -2.32 \times 10^{10} \text{ J}$$

Award 1 mark for correct radius with centre-to-centre, 1 mark for correct KE and U, 1 mark for correct total energy.

Q2. Escape velocity from Earth — 3 marks

Step 1 — Set total energy to zero at escape threshold:

$$\frac{1}{2}mv_e^2 - \frac{GM_E m}{R_E} = 0$$

Step 2 — Solve for $v_e$:

$$\frac{1}{2}mv_e^2 = \frac{GM_E m}{R_E}$$

$$v_e^2 = \frac{2GM_E}{R_E}$$

Step 3 — The projectile mass $m$ cancels:

$$v_e = \sqrt{\frac{2GM_E}{R_E}} = \sqrt{\frac{2 \times 6.67 \times 10^{-11} \text{ N m}^2\text{/kg}^2 \times 5.97 \times 10^{24} \text{ kg}}{6.37 \times 10^6 \text{ m}}}$$

$$v_e = \sqrt{1.25 \times 10^8 \text{ m}^2\text{/s}^2} = 1.12 \times 10^4 \text{ m/s} = 11.2 \text{ km/s}$$

Independence from projectile mass: The mass $m$ appears on both sides of the energy equation and cancels out completely. Therefore, $v_e$ depends only on the central body's mass $M_E$ and the starting radius $R_E$, not on what is being launched. A spacecraft and a marble require the same initial speed to escape from Earth's surface (neglecting atmospheric effects).

Award 1 mark for correct derivation with factor of 2, 1 mark for correct numerical answer, 1 mark for showing mass cancellation.

Q3. Evaluate "more tightly bound" — 4 marks

The statement is TRUE.

From $E_{\text{total}} = -\frac{1}{2}GMm/r$, a more negative total energy means a smaller orbital radius $r$ — the satellite is closer to the central body.

"More tightly bound" means:

  • The satellite is in a lower orbit (smaller $r$)
  • More energy is required to move it to infinity (break the gravitational binding)
  • It is deeper in the gravitational well

Energy to move to infinity (binding energy):

For $E_1 = -3 \times 10^{10} \text{ J}$: escape energy = $|E_1| = 3 \times 10^{10} \text{ J}$

For $E_2 = -1 \times 10^{10} \text{ J}$: escape energy = $|E_2| = 1 \times 10^{10} \text{ J}$

The first satellite requires three times as much energy to free it from Earth's gravity. This confirms it is more tightly bound — it is deeper in the gravitational well and harder to dislodge.

Award 1 mark for correct evaluation of the statement, 1 mark for explaining "more tightly bound," 1 mark for correct escape energies, 1 mark for comparative reasoning.

Boss Battle

🎯 Energy in Orbits Boss Battle

Test your orbital energy and escape velocity calculation skills under pressure!

Game iframe will be embedded here.

Boss Battle: Energy in Orbits