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Year 12 Physics Module 5 ⏱ ~50 min 5 MC · 3 Short Answer Lesson 18 of 18 IQ3: Gravitational Fields Consolidation

Phase 3 Consolidation

On 13 April 1970, an oxygen tank explosion on Apollo 13 occurred at 321,860 km from Earth. NASA engineers calculated a free-return trajectory using gravitational slingshot around the Moon — simultaneously applying escape velocity, orbital energy, and gravitational potential concepts under time pressure. Splashdown occurred on 17 April 1970. Today's checkpoint consolidates exactly those tools.

Today's challenge: On 13 April 1970, the Apollo 13 oxygen tank exploded at 321,860 km from Earth. NASA had less than 4 days to calculate a free-return trajectory — using the Moon's gravity to slingshot the crew back to Earth without engine power. Before starting today's consolidation, write down: which gravitational formula is hardest to remember, and what is Apollo 13's scenario testing about gravitational potential and orbital energy?
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Worksheets

Practise this lesson

Four printable worksheets that build from the foundations up to exam-style questions — start at whatever level suits you.

Build Foundations & guided practice Apply Application & data response Master Mastery tasks Exam Exam-style questions
Before you begin — reflect

Which gravitational formula is hardest to remember? Write it down three times along with when to use it.

Warm-up — what is the correct expression for gravitational potential energy at distance $r$ from a planet of mass $M$?

Learning Intentions
goals

Know — Recall all Phase 3 Formulae

  • State and explain all five Phase 3 gravitational field formulae
  • Identify the trap in each formula and when it applies

Understand — Identify and Correct Common Errors

  • Diagnose each of the six common exam errors
  • Distinguish gravitational potential $V$ from potential energy $U$

Can Do — Solve Mixed Problems Under Exam Conditions

  • Complete ten mixed practice questions (Band 3–6)
  • Answer three timed extended-response questions (4 marks each)
Key Terms — Gravitational Fields
vocab
Universal gravitation$F = GMm/r^2$; all masses attract all other masses.
Field strength$g = GM/r^2$; force per unit mass at a point in the field.
Potential energy$U = -GMm/r$; always negative, defined as zero at infinity.
Gravitational potential$V = -GM/r$; potential energy per unit mass (J/kg), not to be confused with $U$.
Escape velocity$v_e = \sqrt{2GM/r}$; minimum speed to escape a gravitational field permanently.
Kepler's Third Law$T^2 \propto r^3$; exact for any closed orbit using the semi-major axis.
Cross-lesson links: L14–L17 built the gravitational field toolkit. L18 consolidates the entire gravitation strand — the Apollo 13 scenario tests escape velocity (L16), orbital energy (L12, L14), and Kepler's Third Law (L17) simultaneously, exactly as HSC extended response questions do.
Misconceptions to fix before you review
Wrong: Using $g = 9.8 \text{ m/s}^2$ in orbital or altitude problems.
Right: $g = 9.8 \text{ m/s}^2$ is only valid at Earth's surface. At altitude $h$, use $g = GM/(R+h)^2$ or the ratio $g' = g(R/(R+h))^2$.
Wrong: Confusing gravitational potential $V$ (J/kg) with potential energy $U$ (J).
Right: $V = -GM/r$ is per unit mass; $U = -GMm/r$ is total energy. They are related by $U = mV$ and $\Delta U = m\Delta V$.

The formula $g = 9.8 \text{ m/s}^2$ can be used to calculate gravitational field strength at any altitude above Earth's surface.

Gravitational potential $V$ and gravitational potential energy $U$ have the same SI units.

IQ3: Gravitational Fields Formula Sprint

1
Phase 3 Formula Sheet & Sprint Cards
+5 XP

Five formulae — when to use each, key variables, and common traps. Cover the right side, recall, then click to check.

Full Module Summary — all Phase 3 gravitational field formulae

Full Module Summary — all Phase 3 gravitational field formulae at a glance.

Newton's law of gravitation

$F = \dfrac{GMm}{r^2}$

Use: Attractive force between two masses. $G = 6.67 \times 10^{-11} \text{ N m}^2/\text{kg}^2$. Always: $r$ = centre-to-centre distance.

Gravitational field strength

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

Use: Force per unit mass at distance $r$ from a mass $M$. At altitude $h$: $r = R + h$. Ratio form: $g' = g\!\left(\dfrac{R}{R+h}\right)^2$.

Gravitational potential energy

$U = -\dfrac{GMm}{r}$   ·   $W = \Delta U = GMm\!\left(\dfrac{1}{r_1} - \dfrac{1}{r_2}\right)$

Use: Total potential energy of a two-body system. Negative sign is mandatory — $U = 0$ at $r = \infty$. Never use $mgh$ for orbital problems.

Gravitational potential

$V = -\dfrac{GM}{r}$   ·   $g = -\dfrac{dV}{dr}$

Use: Potential energy per unit mass (J/kg). Related to $U$ by $U = mV$. Equipotential surfaces are spherical.

Escape velocity

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

Use: Minimum speed to escape a gravitational field permanently. Factor of $\sqrt{2}$ is mandatory — from energy conservation. $v_e = \sqrt{2}\,v_{\text{orbital}}$.

Sprint Cards — click to reveal traps

$F = \dfrac{GMm}{r^2}$
Click to reveal when to use, trap, and connection
Use whenTwo masses, distance between centres known. Finding force of gravitational attraction.
Trap$r$ is centre-to-centre, not surface-to-surface. Always write $r = R + h$ explicitly.
Connects to$g = F/m = GM/r^2$; for circular orbits, equate with $F = mv^2/r$.
$U = -\dfrac{GMm}{r}$
Click to reveal when to use, trap, and connection
Use whenEnergy calculations, work done moving between radii, total orbital energy.
TrapThe negative sign is mandatory. $U = 0$ at $r = \infty$. Never use $U = mgh$ for orbital problems.
Connects to$W = \Delta U = GMm(1/r_1 - 1/r_2)$; total energy $E = -GMm/2r$ for circular orbits.
$V = -\dfrac{GM}{r}$
Click to reveal when to use, trap, and connection
Use whenPotential per unit mass (J/kg). Equipotential surfaces. Relating $g$ and $V$.
Trap$V$ is potential (J/kg), NOT potential energy $U$ (J). Do not confuse units. To get energy: $U = mV$ and $\Delta U = m\Delta V$.
Connects to$g = -dV/dr$; $\Delta U = m\Delta V$.
$v_e = \sqrt{\dfrac{2GM}{r}}$
Click to reveal when to use, trap, and connection
Use whenMinimum speed to escape a gravitational field permanently (total energy = 0).
TrapFactor of 2 is mandatory (from energy conservation). $v_e = \sqrt{2} \times v_{\text{orbital}}$. Omitting gives ~41% error.
Connects toTotal orbital energy $E = -GMm/2r$; binding energy $|E|$.
$T^2 = \dfrac{4\pi^2}{GM}\,r^3$
Click to reveal when to use, trap, and connection
Use whenRelating orbital period to orbital radius for any orbiting body.
TrapValid for elliptical orbits if you use the semi-major axis $a$ in place of $r$. Derived from equating $GMm/r^2 = mv^2/r$ with $v = 2\pi r/T$.
Connects to$E_{\text{total}} = -GMm/2r$ for circular orbits; Kepler's Third Law is exact for any conic orbit.

Full module formula set: $F = GMm/r^2$; $g = GM/r^2$ (use $r = R+h$); $U = -GMm/r$ (negative, mandatory); $W = GMm(1/r_1 - 1/r_2)$; $V = -GM/r$ (J/kg); $g = -dV/dr$; $v_e = \sqrt{2GM/r}$ ($\sqrt{2}$ mandatory); $T^2 = (4\pi^2/GM)r^3$; $E_\text{total} = -GMm/2r$.

Pause — copy the highlighted formula set into your book before moving on.

A satellite orbits at height $h = 2R_E$ above Earth's surface. The gravitational field strength there compared to the surface is:

Error Clinic

2
Six Common Gravitational Errors
+5 XP

Each card describes a common error that costs marks in exams. Find the fix, then reveal the explanation.

We just saw the complete gravitational fields formula set. That raises a question: which of these formulae do students most often apply incorrectly? This card answers it → six common errors: wrong $g$, missing negative, confusing $V$ and $U$, wrong $r$, missing $\sqrt{2}$, and misapplying Kepler's Third Law.

E1
Using $g = 9.8$ for orbital problems
▾
Student working: A satellite orbits 400 km above Earth. Student uses $F = mg = m \times 9.8$ to find gravitational force. Incorrect.
Fix: $g = 9.8 \text{ m/s}^2$ is only valid at Earth's surface. At altitude $h$, always use $g = GM/(R+h)^2$ or the ratio $g' = g(R/(R+h))^2$. Treating $g$ as constant in orbital calculations is the most common error in HSC exams.
E2
Forgetting the negative sign on $U = -GMm/r$
▾
Student working: Work to move satellite from $r_1$ to $r_2$: $W = GMm(1/r_1 - 1/r_2)$, but student uses $U = +GMm/r$, giving the wrong sign for work. Incorrect.
Fix: The negative sign is physically meaningful — gravity is attractive and $U = 0$ is defined at $r = \infty$. All bound systems have negative total energy. Omitting it gives incorrect energy rankings and wrong signs for work calculations.
E3
Confusing $V$ (J/kg) with $U$ (J)
▾
Student working: Question asks for gravitational potential energy of a 2000 kg satellite. Student calculates $V = -GM/r$ and writes that as the answer in joules. Incorrect.
Fix: $V = -GM/r$ has units of J/kg (potential per unit mass). $U = -GMm/r$ has units of J (total potential energy). To get energy from potential: $U = mV$ and $\Delta U = m\Delta V$. Always check units in your final answer.
E4
Using $r = R$ instead of $r = R + h$
▾
Student working: A satellite orbits at 500 km altitude. Student substitutes $r = R_E = 6.37 \times 10^6$ m into $F = GMm/r^2$. Incorrect.
Fix: $r$ is always centre-to-centre distance. For a satellite at height $h$ above the surface, $r = R_{\text{planet}} + h$. Write this explicitly in every problem. Never substitute $R$ when the question gives you an altitude.
E5
Forgetting the $\sqrt{2}$ in escape velocity
▾
Student working: Escape velocity from Earth's surface: student writes $v_e = \sqrt{GM/R}$. Incorrect.
Fix: $v_e = \sqrt{2GM/r}$, not $\sqrt{GM/r}$. The factor of 2 comes from energy conservation: $\tfrac{1}{2}mv_e^2 = |U| = GMm/r$. $\sqrt{2} \approx 1.414$, so the error is about 41% if you forget it.
E6
Thinking Kepler's Third Law requires circular orbits
▾
Student working: Student refuses to apply $T^2 \propto r^3$ to a comet with an elliptical orbit. Incorrect reasoning.
Fix: Kepler's Third Law $T^2 \propto a^3$ is exact for any closed orbit if you use the semi-major axis $a$. For circles, $a = r$. For ellipses, $a$ is half the longest diameter. Newton derived this from his law of gravitation for all conic sections.

Six error rules: E1 — use $g = GM/r^2$ at altitude (not 9.8); E2 — $U = -GMm/r$, negative sign mandatory; E3 — $V$ (J/kg) $\neq$ $U$ (J); convert: $U = mV$; E4 — always $r = R + h$; E5 — $v_e = \sqrt{2GM/r}$, not $\sqrt{GM/r}$; E6 — Kepler's 3rd Law applies to ellipses using semi-major axis $a$.

Add the highlighted six error checklist to your notes before the check below.

Three of these statements about gravitational fields are correct. Pick the odd one out (the incorrect statement).

Mixed Practice

3
Mixed Practice Questions
+5 XP

Ten questions from Band 3 to Band 6. Worked solutions in the Answers section of the Practice phase.

We just saw the six common errors in gravitational fields problems. That raises a question: can you now apply the full formula set correctly across all question types? This card answers it → ten mixed practice questions from Band 3 (recall) to Band 6 (derivation and analysis).

Apply Band 3 2 marks

Q1. Calculate the gravitational force between two 50 kg masses separated by 0.50 m.

Apply Band 3 2 marks

Q2. Calculate the gravitational field strength at a point 2000 km above Earth's surface. $(M_E = 5.97 \times 10^{24} \text{ kg},\ R_E = 6.37 \times 10^6 \text{ m})$

Understand Band 3 2 marks

Q3. State the zero-reference point for gravitational potential energy and explain why this point was chosen.

Apply Band 4/5 3 marks

Q4. A 2000 kg satellite orbits Earth at $r = 8.5 \times 10^6$ m from Earth's centre. Calculate its kinetic energy, gravitational potential energy, and total mechanical energy. $(M_E = 5.97 \times 10^{24} \text{ kg})$

Apply Band 4/5 3 marks

Q5. Calculate the escape velocity from the Moon's surface and compare it with Earth's escape velocity. $(M_M = 7.35 \times 10^{22} \text{ kg},\ R_M = 1.74 \times 10^6 \text{ m},\ M_E = 5.97 \times 10^{24} \text{ kg},\ R_E = 6.37 \times 10^6 \text{ m})$

Apply Band 4/5 3 marks

Q6. Calculate the gravitational potential $V$ and the gravitational field strength $g$ at $r = 2.0 \times 10^7$ m from Earth's centre. $(M_E = 5.97 \times 10^{24} \text{ kg})$

Apply Band 4/5 4 marks

Q7. Calculate the work required to move a 1000 kg satellite from Earth's surface to an orbital radius of $r = 1.5 \times 10^7$ m. $(M_E = 5.97 \times 10^{24} \text{ kg},\ R_E = 6.37 \times 10^6 \text{ m})$

Analyse Band 6 4 marks

Q8. Derive the relationship $g = -\dfrac{dV}{dr}$ starting from $V = -\dfrac{GM}{r}$.

Analyse Band 6 4 marks

Q9. A planet has 4 times Earth's mass and 2 times Earth's radius. Find the surface gravitational field strength and escape velocity as multiples of Earth's values ($g_E$ and $v_{eE}$).

Analyse Band 6 5 marks

Q10. Jupiter's moon Io orbits with a period of 1.77 days. Using Kepler's Third Law, find its orbital radius and orbital speed. $(M_J = 1.90 \times 10^{27} \text{ kg})$

Gravitational problem method: (1) identify $r = R + h$ (centre-to-centre); (2) select formula; (3) preserve negative sign on $U$. Total orbital energy: $E_\text{total} = -GMm/2r$ (negative; equals $-KE$). Work done: $W = GMm(1/r_1 - 1/r_2)$ — positive moving outward.

Add the highlighted problem-solving method to your notes before the check below.

For Q4 above, the total mechanical energy of a satellite in a circular orbit is:

Timed Exam Block

4
Extended Response Practice
+5 XP

Three exam-style questions. Each: 4 marks, 8 minutes recommended. Solutions in the Answers section of the Practice phase.

We just saw ten mixed practice questions across all gravitational field formula types. That raises a question: can you write structured, exam-quality extended responses under timed conditions? This card answers it → three 4-mark questions covering orbital energy, equipotentials, and Kepler's Third Law analysis.

Exam strategy

For 4-mark questions: spend ~1 min planning, ~5 min writing, ~2 min checking units and significant figures. Each question should have at least 4 distinct steps or points to earn full marks.

08:00
Question 11 (4 marks, ~8 min)

A 500 kg probe is in circular orbit around Earth at $r = 9.0 \times 10^6$ m from Earth's centre. $(M_E = 5.97 \times 10^{24} \text{ kg})$

  1. Calculate the total mechanical energy of the probe. (2 marks)
  2. Determine the minimum additional energy required for the probe to escape Earth's gravitational field. (2 marks)
08:00
Question 12 (4 marks, ~8 min)

At $r = 1.2 \times 10^7$ m from Earth's centre. $(M_E = 5.97 \times 10^{24} \text{ kg})$

  1. Calculate the gravitational potential $V$ and the gravitational field strength $g$. (2 marks)
  2. Explain why equipotential surfaces around a spherical mass are spherical and describe the relationship between equipotentials and field lines. (2 marks)
08:00
Question 13 (4 marks, ~8 min)

An exoplanet orbits its star with a period of 45 days. The star has mass $2.0 \times 10^{30}$ kg.

  1. Using Kepler's Third Law, calculate the orbital radius of the exoplanet. (2 marks)
  2. A student claims: "If the star's mass doubled, the planet's orbital speed would stay the same because the planet is still the same distance from the star." Assess this claim. (2 marks)

Extended response strategy: state formula, define all variables, write $r = R + h$ explicitly, preserve negative sign on $U$, explain sign of $\Delta E$ in context. For Kepler problems: convert $T$ to seconds, cube-root the result for $r$, then use $v = \sqrt{GM/r}$. Show each step clearly — examiners award method marks.

Pause — write the highlighted exam strategy into your book before moving on.

Activity 1 — Gravitational Fields Consolidation Drills
ApplyBand 3

Practise the key Phase 3 concepts from this lesson.

  1. Without looking at your notes, write down all five Phase 3 formulae: Newton's law of gravitation, gravitational field strength, gravitational potential energy, gravitational potential, and escape velocity. State the SI unit for each quantity.
  2. A 1200 kg satellite orbits Earth at $r = 9.0 \times 10^6$ m. Calculate: (a) gravitational potential energy $U$, (b) kinetic energy $K = GMm/2r$, (c) total mechanical energy $E$. Verify that $E = U/2$. ($G = 6.67 \times 10^{-11}$ N m²/kg², $M_E = 5.97 \times 10^{24}$ kg)
  3. Explain why the negative sign on $U = -GMm/r$ is mandatory. What physical meaning does the negative sign carry, and what happens to energy calculations if you omit it?
Activity 2 — Concept Check
UnderstandBand 5

Explain the reasoning behind a key gravitational fields principle.

A student claims that "a satellite in a lower orbit has more energy because it is moving faster." Evaluate this claim carefully using the formulae $KE = GMm/2r$, $U = -GMm/r$, and $E_{\text{total}} = -GMm/2r$. Is the student correct, partially correct, or incorrect? Provide a real-world example (e.g., comparing LEO and GEO satellites) to support your answer.

Fill the gap. For a satellite in a circular orbit, the total energy is $E = -GMm/2r$. If the orbital radius doubles, the total energy becomes _____ times its original value. (Give your answer as a fraction, e.g. 1/2)

Wrap-up — Misconceptions & Summary

Misconceptions — final check

Wrong: "A lower satellite orbit has more total energy because the satellite moves faster."
Right: A lower orbit has more negative total energy ($E = -GMm/2r$). Although kinetic energy is greater, the potential energy is even more negative — the satellite is more tightly bound. Higher orbits have less negative (greater) total energy.
Wrong: "Gravitational potential is zero at Earth's surface."
Right: Gravitational potential is zero at $r = \infty$ by definition. At Earth's surface, $V = -GM/R \approx -6.25 \times 10^7$ J/kg — a large negative value. The convention of $V = 0$ at infinity is the only physically natural reference point.

Copy into your books

The Five Formulae

  • $F = GMm/r^2$ · $g = GM/r^2$
  • $U = -GMm/r$ · $V = -GM/r$
  • $v_e = \sqrt{2GM/r}$

Derived Results

  • $W = GMm(1/r_1 - 1/r_2)$
  • $E_{\text{total}} = -GMm/2r$
  • $g = -dV/dr$ · $U = mV$

Six Common Errors

  • Using $g = 9.8$ at altitude (E1)
  • Forgetting negative on $U$ (E2)
  • Confusing $V$ and $U$ units (E3)

Constants

  • $G = 6.67 \times 10^{-11}$ N m²/kg²
  • $M_E = 5.97 \times 10^{24}$ kg
  • $R_E = 6.37 \times 10^6$ m

A satellite in a lower orbit has a higher total mechanical energy than one in a higher orbit.

The escape velocity from any radius is exactly $\sqrt{2}$ times the circular orbital speed at that radius.

Gravitational potential $V$ is zero at Earth's surface.

Quick recall — Phase 3 Consolidation
+5 XP

A fresh five-question set drawn from this lesson's 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.

Short Answer — 10 marks
+5 XP

ApplyBand 4(3 marks) 1. A 1500 kg satellite orbits Earth at $r = 1.0 \times 10^7$ m from Earth's centre. Calculate (a) its orbital speed, (b) its kinetic energy, and (c) its total mechanical energy. $(M_E = 5.97 \times 10^{24} \text{ kg})$

1 mark: correct orbital speed using $v = \sqrt{GM/r}$ · 1 mark: kinetic energy · 1 mark: total energy using $E = -GMm/2r$ or $E = -KE$

AnalyseBand 5(3 marks) 2. Explain why the gravitational potential $V$ is always negative and why it approaches zero as $r \rightarrow \infty$. Include the definition of $V$ in your explanation.

1 mark: defines $V = -GM/r$ · 1 mark: explains negative sign (attractive field, work done by field is negative) · 1 mark: explains $V \to 0$ as $r \to \infty$ (zero reference point convention)

EvaluateBand 6(4 marks) 3. A student argues: "Since $g = GM/r^2$ and $V = -GM/r$, a point where $V$ is more negative must also have a larger $g$." Evaluate this statement, using a specific numerical example at two different radii to support your reasoning. $(M_E = 5.97 \times 10^{24} \text{ kg})$

1 mark: identifies the trend is correct but the relationship is not proportional · 1 mark: $g \propto 1/r^2$ while $V \propto 1/r$ — different power laws · 1 mark: numerical example at $R_E$ and $2R_E$ for both $g$ and $V$ · 1 mark: correct conclusion ($g = |dV/dr|$, not $g \propto V$)

Show all answers

Mixed Practice Answers (Q1–Q10)

Q1 (2 marks): $F = GMm/r^2 = (6.67 \times 10^{-11})(50)(50)/(0.50)^2 = 1.668 \times 10^{-7}/0.25 = \mathbf{6.7 \times 10^{-7} \text{ N}}$.

Q2 (2 marks): $r = R_E + h = 6.37 \times 10^6 + 2.0 \times 10^6 = 8.37 \times 10^6$ m. $g = GM_E/r^2 = (6.67 \times 10^{-11})(5.97 \times 10^{24})/(8.37 \times 10^6)^2 = 3.98 \times 10^{14}/7.01 \times 10^{13} = \mathbf{5.68 \text{ m/s}^2}$.

Q3 (2 marks): The zero-reference point is $r = \infty$ (infinite separation). Chosen because: (1) it is the only natural, unambiguous reference where gravitational force between masses becomes exactly zero; (2) it ensures all bound systems have negative total energy, making it easy to identify bound vs. unbound orbits.

Q4 (3 marks): $K = GMm/2r = (6.67 \times 10^{-11})(5.97 \times 10^{24})(2000)/(2 \times 8.5 \times 10^6) = \mathbf{4.68 \times 10^{10} \text{ J}}$. $U = -2K = \mathbf{-9.36 \times 10^{10} \text{ J}}$. $E_{\text{total}} = -K = \mathbf{-4.68 \times 10^{10} \text{ J}}$.

Q5 (3 marks): Moon: $v_e = \sqrt{2GM_M/R_M} = \sqrt{2(6.67 \times 10^{-11})(7.35 \times 10^{22})/(1.74 \times 10^6)} = \mathbf{2.37 \text{ km/s}}$. Earth: $v_e = \sqrt{2(6.67 \times 10^{-11})(5.97 \times 10^{24})/(6.37 \times 10^6)} = \mathbf{11.2 \text{ km/s}}$. Moon's $v_e$ is about one-fifth of Earth's.

Q6 (3 marks): $V = -GM_E/r = -(6.67 \times 10^{-11})(5.97 \times 10^{24})/(2.0 \times 10^7) = \mathbf{-1.99 \times 10^7 \text{ J/kg}}$. $g = GM_E/r^2 = \mathbf{0.995 \approx 1.0 \text{ m/s}^2}$.

Q7 (4 marks): $W = GM_Em(1/R_E - 1/r_2) = (6.67 \times 10^{-11})(5.97 \times 10^{24})(1000)(1/6.37 \times 10^6 - 1/1.5 \times 10^7) = 3.98 \times 10^{17} \times 9.03 \times 10^{-8} = \mathbf{3.59 \times 10^{10} \text{ J}}$.

Q8 (4 marks): Start: $V = -GM/r = -GM \cdot r^{-1}$. Differentiate: $dV/dr = -GM(-1)r^{-2} = GM/r^2$. Therefore $g = -dV/dr = -GM/r^2$. The negative sign indicates the field points in the direction of decreasing $r$ (inward), which is correct since $g$ is directed toward the mass. The magnitude is $|g| = GM/r^2$.

Q9 (4 marks): $M_P = 4M_E$, $R_P = 2R_E$. $g_P = G(4M_E)/(2R_E)^2 = 4GM_E/4R_E^2 = GM_E/R_E^2 = \mathbf{g_E}$ (same). $v_{eP} = \sqrt{2G(4M_E)/2R_E} = \sqrt{2} \cdot \sqrt{2GM_E/R_E} = \sqrt{2} \cdot v_{eE}$. So $v_{eP} = \mathbf{\sqrt{2}\,v_{eE} \approx 1.41\,v_{eE}}$.

Q10 (5 marks): $T = 1.77 \times 24 \times 3600 = 1.529 \times 10^5$ s. $r^3 = GM_JT^2/4\pi^2 = (6.67 \times 10^{-11})(1.90 \times 10^{27})(1.529 \times 10^5)^2/39.478 = 7.50 \times 10^{25}$. $r = \mathbf{4.22 \times 10^8 \text{ m}}$. $v = 2\pi r/T = 2\pi(4.22 \times 10^8)/(1.529 \times 10^5) = \mathbf{1.73 \times 10^4 \text{ m/s}}$.

Timed Exam Answers (Q11–Q13)

Q11 (4 marks): (a) $E_{\text{total}} = -GM_Em/2r = -(6.67 \times 10^{-11})(5.97 \times 10^{24})(500)/(2 \times 9.0 \times 10^6) = \mathbf{-1.11 \times 10^{10} \text{ J}}$ (2 marks: formula + answer). (b) To escape: total energy must $\geq 0$. Minimum additional energy $= |E_{\text{total}}| = \mathbf{1.11 \times 10^{10} \text{ J}}$ (2 marks: reasoning + answer).

Q12 (4 marks): (a) $V = -GM_E/r = -(6.67 \times 10^{-11})(5.97 \times 10^{24})/(1.2 \times 10^7) = \mathbf{-3.32 \times 10^7 \text{ J/kg}}$. $g = GM_E/r^2 = \mathbf{2.77 \text{ m/s}^2}$ (2 marks). (b) $V = -GM/r$ depends only on $r$ and not on direction — every point at the same $r$ from the centre has the same $V$. The locus of constant $V$ is a sphere. Field lines are radial (perpendicular to equipotentials), pointing inward (2 marks).

Q13 (4 marks): (a) $T = 45 \times 86400 = 3.888 \times 10^6$ s. $r^3 = GM_\star T^2/4\pi^2 = (6.67 \times 10^{-11})(2.0 \times 10^{30})(3.888 \times 10^6)^2/39.478 = 5.11 \times 10^{31}$. $r = \mathbf{3.71 \times 10^{10} \text{ m}}$ (2 marks). (b) The claim is incorrect. $v = \sqrt{GM/r}$; if $M$ doubles while $r$ stays the same, $v$ would increase by $\sqrt{2}$. In reality orbital parameters adjust — doubling $M$ with the same $r$ would require a larger orbital speed and a shorter period. The student confuses fixed $r$ with fixed orbital parameters (2 marks).

Multiple Choice — Key

MC answers and full explanations are shown inline as you complete each question. Use the retry button to attempt a fresh set drawn from the lesson bank.

Short Answer — Model Answers

SA1 (3 marks): (a) $v = \sqrt{GM_E/r} = \sqrt{(6.67 \times 10^{-11})(5.97 \times 10^{24})/(1.0 \times 10^7)} = \mathbf{6.31 \times 10^3 \text{ m/s}}$ (1 mark). (b) $KE = \tfrac{1}{2}mv^2 = 0.5 \times 1500 \times (6310)^2 = \mathbf{2.99 \times 10^{10} \text{ J}}$ (1 mark). (c) $E_{\text{total}} = -KE = \mathbf{-2.99 \times 10^{10} \text{ J}}$ (1 mark; or equivalently $-GMm/2r$).

SA2 (3 marks): $V = -GM/r$ (1 mark). Gravitational potential is defined as the work done per unit mass to bring a test mass from infinity to distance $r$ — since gravity is attractive, the field does the work, so $V$ is negative (1 mark). At $r = \infty$, no work is done by or against the field, so $V = 0$. The zero is at infinity because it is the only natural reference where the gravitational influence of $M$ becomes zero (1 mark).

SA3 (4 marks): The statement is partially correct but misleading (1 mark). Both $g \propto 1/r^2$ and $V \propto 1/r$ do increase in magnitude as $r$ decreases, so a more negative $V$ does correspond to a larger $g$ in direction — but not in proportion (1 mark). Numerical example: at $r = R_E = 6.37 \times 10^6$ m: $V = -6.25 \times 10^7$ J/kg, $g = 9.81$ m/s². At $r = 2R_E = 1.274 \times 10^7$ m: $V = -3.13 \times 10^7$ J/kg (halved), $g = 2.45$ m/s² (quartered) (1 mark). When $r$ doubles, $V$ halves but $g$ quarters, because $g \propto 1/r^2$ while $V \propto 1/r$. The correct relationship is $g = |dV/dr|$, not $g \propto V$ (1 mark).

Boss Battle — Full Module Quiz
boss

Five timed questions covering all three phases of Module 5: Advanced Mechanics. Beat the final boss to bank a tier — gold (perfect + fast), silver (80%+), or bronze (cleared).

⚔ Enter the arena
Arcade Practice · Asteroid Blaster

Gravitational orbits — apply your knowledge in the final module challenge. Lighter than the boss — pure practice that hammers home the gravitational field relationships.

How did your thinking change?

At the start you were asked about the Apollo 13 emergency on 13 April 1970 — an oxygen tank explosion at 321,860 km from Earth, requiring NASA engineers to calculate a free-return trajectory around the Moon under time pressure, with splashdown achieved on 17 April 1970.

The free-return trajectory calculation required: (a) gravitational potential ($U = -GMm/r$) at 321,860 km to determine available energy; (b) escape velocity check — was Apollo 13 still gravitationally bound to Earth? (At that distance, $v_e = \sqrt{2GM/r} \approx 1.4 \text{ km/s}$, and the spacecraft speed was ~1.0 km/s, so yes, bound); (c) orbital energy to determine whether the Moon's gravity could redirect the trajectory back to Earth. The free-return path was essentially a figure-8 — an elegant application of the same tools you have consolidated today. After working through the practice questions and error clinic, reflect again:

  • Which of the six errors (E1–E6) do you need to watch for most in the exam?
  • Can you now distinguish $V$ (J/kg) from $U$ (J) without looking at your notes?
  • What is your plan to avoid your most common error?
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Module overview · Physics · Checkpoint 3 · Module quiz

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