Physics • Year 12 • Module 5 • Lesson 16

Escape Velocity

Lock in the derivation of escape velocity from energy conservation, the key formula, and its core properties before tackling harder questions.

Build · Vocab & Recall

1. Term–definition match

Match each term on the left with its correct definition on the right. Write the letter of the definition in the blank. 6 marks (1 each)

 Term Definition
1.Escape velocity
A. The boundary at radius Rs beyond which nothing, not even light, can escape.
B. The minimum speed needed to escape a gravitational field permanently with no further propulsion.
C. The path taken by an object launched at exactly escape velocity; it reaches infinity with zero residual speed.
D. The path taken by an object with more than escape energy; it escapes with surplus speed at infinity.
E. The radius at which escape velocity equals the speed of light; the size of a black hole’s event horizon.
F. The phenomenon where firing a rocket engine at closest approach to a body produces the greatest energy gain.
2.Parabolic trajectory
3.Hyperbolic trajectory
4.Event horizon
5.Schwarzschild radius
6.Oberth effect
Stuck? Revisit the Key Terms panel in the lesson and Card 3 (Black Holes & Schwarzschild Radius).

2. True or false — with correction

Circle T or F. If the statement is false, write the correct version on the line below it. 10 marks (1 T/F + 1 correction each)

2.1 Escape velocity depends on the mass of the object being launched.    T  /  F

2.2 The formula for escape velocity is ve = √(2GM / r).    T  /  F

2.3 At launch, the total mechanical energy of a projectile moving at exactly escape velocity equals zero.    T  /  F

2.4 Escape velocity from the surface of Jupiter is smaller than from Earth because Jupiter is further from the Sun.    T  /  F

2.5 Escape velocity at a distance 2R from a planet’s centre is ve / √2, where ve is the surface escape velocity.    T  /  F

Stuck? Revisit Cards 1 and 2 of the lesson and the Key Formula panel.

3. Derivation scaffold — step by step

Complete the derivation of escape velocity from conservation of energy. Fill in each blank. 6 marks (1 per step)

Step 1. Write the total mechanical energy at launch (surface, radius r, speed ve):

Elaunch = ½mve2 + _______________

Step 2. Write the total mechanical energy “at infinity” (object just barely escapes):

E∞ = _______________ + _______________

Step 3. Apply conservation of energy (Elaunch = E∞):

½mve2 − GMm / r = _______________

Step 4. Rearrange to isolate ve2:

ve2 = _______________

Step 5. Take the positive square root:

ve = _______________

Step 6. State one reason why the factor of 2 appears in the formula (not 1):

Stuck? Revisit Card 1 “Deriving Escape Velocity” and the worked example.

4. Fill-in-the-blank paragraph

Use the word bank to complete the passage. Each word is used once. 8 marks (1 per blank)

Word bank:

cancels  ·  conservation  ·  event horizon  ·  gravitational  ·  hyperbolic  ·  infinity  ·  parabolic  ·  Schwarzschild

Escape velocity is derived from ___________ of energy. A projectile launched at exactly escape velocity follows a ___________ trajectory and arrives at ___________ with zero kinetic energy. If the launch speed exceeds escape velocity, the trajectory becomes ___________ and the projectile retains surplus speed at infinity. Escape velocity is independent of the projectile’s mass because that mass ___________ from both sides of the energy equation. If a body is compressed until its escape velocity equals the speed of light, its radius equals the ___________ radius. At this boundary, called the ___________, no signal can escape the body’s ___________ field.

Stuck? Revisit Cards 1, 2 and 3 of the lesson and the Key Terms panel.
Answers — Do not peek before attempting

Q1 — Term–definition match

1 → B • 2 → C • 3 → D • 4 → A • 5 → E • 6 → F

Q2 — True / false with correction

2.1 False. Escape velocity is independent of the projectile’s mass; the mass m cancels in the energy derivation. A feather and a rocket need the same launch speed to escape.

2.2 True.

2.3 True. At escape velocity, KE = +½mve2 and GPE = −GMm/r; these sum to zero, so total mechanical energy = 0.

2.4 False. Jupiter’s escape velocity (~59.5 km/s) is much larger than Earth’s (~11.2 km/s) because Jupiter has far greater mass (M) and a much larger radius (r). Distance from the Sun is irrelevant to a planet’s surface escape velocity.

2.5 True. ve(2R) = √(2GM / 2R) = √(1/2) × √(2GM/R) = ve,surface / √2.

Q3 — Derivation scaffold

Step 1: Elaunch = ½mve2 + (−GMm/r)  [gravitational potential energy is negative]

Step 2: E∞ = 0 + 0  [both KE and GPE are zero at infinity at escape threshold]

Step 3: ½mve2 − GMm/r = 0

Step 4: ve2 = 2GM/r

Step 5: ve = √(2GM/r)

Step 6: The factor of 2 comes from the kinetic energy formula ½mv2; it means the projectile must provide enough KE to fully overcome the magnitude of the (negative) gravitational potential energy, requiring ve2 = 2GM/r rather than GM/r.

Q4 — Cloze paragraph

In order: conservation · parabolic · infinity · hyperbolic · cancels · Schwarzschild · event horizon · gravitational

HSCScience • Physics • Year 12 • Module 5 • Lesson 16 • Worksheet 1 (Build)