Physics • Year 12 • Module 5 • Lesson 16
Escape Velocity
Apply ve = √(2GM/r) to calculate escape velocities from different bodies, interpret data, and explain orbital-versus-escape-velocity relationships.
1. Interpret a data table — escape velocities across the Solar System
The table below gives data for five Solar System bodies. Some escape velocity values are missing. Use G = 6.67 × 10−11 N m2 kg−2. 8 marks
| Body | Mass (kg) | Radius (m) | Escape velocity (km/s) |
|---|---|---|---|
| Earth | 5.97 × 1024 | 6.37 × 106 | 11.2 |
| Moon | 7.34 × 1022 | 1.74 × 106 | |
| Mars | 6.42 × 1023 | 3.40 × 106 | |
| Jupiter | 1.90 × 1027 | 7.15 × 107 | 59.5 |
| Sun | 1.99 × 1030 | 6.96 × 108 |
1.1 Calculate the escape velocity from the surface of the Moon. Show full working with units. 2 marks
1.2 Calculate the escape velocity from the surface of Mars. 2 marks
1.3 Calculate the escape velocity from the surface of the Sun. 2 marks
1.4 Using the completed table, identify the trend: does a greater mass always mean a greater escape velocity? Explain with reference to the formula. 2 marks
2. Interpret a graph — escape velocity versus launch radius
The graph below shows how escape velocity varies with the distance from Earth’s centre (r), plotted from r = RE to 5RE. Take RE = 6.37 × 106 m and ve,surface = 11.2 km/s. 6 marks
Figure 2. Escape velocity from Earth versus launch radius. Illustrative data.
2.1 Using the graph, estimate the escape velocity at r = 3RE and compare it to the calculated value ve(3RE) = 11.2 / √3. 2 marks
2.2 Describe the shape of the curve and explain why escape velocity decreases as r increases, with reference to the formula. 2 marks
2.3 A student says: “At a large enough distance from Earth, escape velocity reaches zero, so you never truly need to provide energy to escape.” Critique this statement. 2 marks
3. Predict and justify — escape velocity and orbital velocity
6 marks
Context. The International Space Station (ISS) orbits Earth in a circular orbit at an altitude of 400 km above the surface (r ≈ 6.77 × 106 m from Earth’s centre). A news article states: “If the ISS were to fire its thrusters and speed up to escape velocity at its current altitude, it would need to double its speed.”
3.1 Calculate the orbital speed of the ISS at its current altitude. Use ME = 5.97 × 1024 kg, G = 6.67 × 10−11 N m2 kg−2. 2 marks
3.2 Calculate the escape velocity at the same altitude. 2 marks
3.3 Evaluate the claim in the news article. Is the factor “double” correct? Use your calculated values and the general relationship ve = √2 × vorbital. 2 marks
Q1.1 — Escape velocity from the Moon
ve = √(2 × 6.67 × 10−11 × 7.34 × 1022 / 1.74 × 106) = √((9.79 × 1012) / (1.74 × 106)) = √(5.626 × 106) = 2.37 × 103 m/s = 2.37 km/s.
Marking notes: 1 mark for correct substitution with units; 1 mark for correct final answer in km/s.
Q1.2 — Escape velocity from Mars
ve = √(2 × 6.67 × 10−11 × 6.42 × 1023 / 3.40 × 106) = √((8.57 × 1013) / (3.40 × 106)) = √(2.52 × 107) = 5.02 × 103 m/s = 5.02 km/s.
Marking notes: 1 mark for correct substitution; 1 mark for answer ~5.0 km/s.
Q1.3 — Escape velocity from the Sun
ve = √(2 × 6.67 × 10−11 × 1.99 × 1030 / 6.96 × 108) = √((2.655 × 1020) / (6.96 × 108)) = √(3.81 × 1011) = 6.18 × 105 m/s = 618 km/s.
Marking notes: 1 mark for correct substitution; 1 mark for answer ~618 km/s.
Q1.4 — Trend: does more mass always mean higher ve?
Not necessarily [1]. ve = √(2GM/r) depends on both mass and radius. Jupiter has greater mass and larger radius than Earth; the net result is still a much higher escape velocity (59.5 vs 11.2 km/s) because mass increases faster than radius. However, the Moon has far less mass and a smaller radius; both reduce ve [1]. You must consider the ratio M/r, not mass alone.
Q2.1 — Graph reading at 3RE
From graph, ve(3RE) ≈ 6.5 km/s [1]. Calculated: 11.2 / √3 = 11.2 / 1.732 = 6.47 km/s. The values are consistent [1].
Q2.2 — Shape and explanation
The curve is an inverse-square-root shape (ve ∝ r−1/2): it decreases steeply at first then levels off [1]. As r increases, the gravitational potential energy at that point becomes less negative (closer to zero), so less kinetic energy is needed to reach zero-energy at infinity, hence ve is smaller [1].
Q2.3 — Critique of the student’s claim
The student is partially right but misleading [1]. While ve → 0 as r → ∞, you still need to get to large r in the first place, which requires energy. Launching from Earth’s surface at a lower speed than escape velocity means you will eventually run out of kinetic energy and fall back — you can never “coast” to infinity without providing the full escape energy at launch [1].
Q3.1 — ISS orbital speed
vorbital = √(GM/r) = √((6.67 × 10−11 × 5.97 × 1024) / (6.77 × 106)) = √((3.982 × 1014) / (6.77 × 106)) = √(5.882 × 107) = 7.67 × 103 m/s = 7.67 km/s.
Marking notes: 1 mark for correct formula and substitution; 1 mark for correct answer.
Q3.2 — Escape velocity at ISS altitude
ve = √(2GM/r) = √2 × vorbital = √2 × 7.67 = 1.085 × 104 m/s = 10.85 km/s. (Or compute directly: √(2 × 5.882 × 107) = √(1.176 × 108) = 1.085 × 104 m/s.)
Marking notes: 1 mark for correct method; 1 mark for correct answer ~10.8 km/s.
Q3.3 — Evaluate the news claim
The claim is incorrect. The ISS would need to increase its speed by a factor of √2 (≈ 1.41), not 2 [1]. The relationship ve = √2 × vorbital always holds at the same radius; the ISS would need to increase from 7.67 km/s to 10.85 km/s, a factor of 1.41, not a doubling to 15.3 km/s [1].