IB Physics HLTopic 4 — Force FieldsPaper 1 & 2v = √(2GM/r)~16 min read
Escape Speed
Throw a ball upwards and it comes back. Throw it harder and it goes higher, but it still comes back. Is there a speed — one single number — at which it simply never returns? There is. Fire a stone from Earth’s surface at 11.2 kilometres every second, cut the engine, and gravity will slow it forever without ever quite stopping it. That number is the escape speed, and it falls out of a single line of energy bookkeeping.
📘 What you need to know
Escape speed: the minimum speed that will allow an object to escape a gravitational field with no further energy input
It is the speed at which all the object’s kinetic energy is exactly used up paying off its gravitational potential energy
Found by equating: ½mv² = GMm/r, giving vesc = √(2GM/r)
The escaping mass cancels — escape speed is the same for a feather, a tennis ball and a rocket
It depends only on the mass and radius of the body being escaped from
Escape speed is not the speed to escape the planet, but to escape its gravitational field altogether
Rockets do not need to reach escape speed to orbit, because they keep burning fuel — the “no further energy input” condition is broken
It takes less energy to reach orbit than to escape
Earth: 11.2 km s−1. Moon: 2.37 km s−1. Jupiter: 60 km s−1
What “escape” actually means
Escape speed — definitionthe minimum speed that will allow an object to escape a gravitational field with no further energy input
Three phrases, and every one of them is load-bearing.
Minimum. Go faster and you still escape — you just arrive with speed to spare.
Gravitational field, not “the planet”. Gravity has infinite range, so escaping means climbing all the way out to infinity. Never write “escape the planet’s gravity” or “beat the gravitational pull” — the examiner is waiting for that.
No further energy input. One shove, then nothing. Everything after launch is coasting.
The energy argument
Give the object kinetic energy at the surface. As it climbs, gravity converts that kinetic energy into gravitational potential energy. The object just escapes when its kinetic energy runs out at exactly the same moment as it reaches infinity — arriving there with a speed of zero.
Step 1 — all the kinetic energy is spent climbing out½mv2esc = GMm / rthe KE at the surface equals the depth of the potential well
Step 2 — the escaping mass cancels½v2esc = GM / r
Step 3 — multiply by two, take the square rootvesc = √( 2GM / r )M = mass of the body being escaped (kg) • r = distance from its centre (m)
The red line hits the curve, and that crossing point is the highest the object gets. The green line never does — it runs alongside the curve forever, closing the gap but never touching.
The bookkeeping, drawn out
At launch, the object has positive kinetic energy and negative potential energy. Escape speed is the speed that makes those two exactly cancel. Total energy zero. And since energy is conserved, it stays zero all the way to infinity, where both terms have separately dwindled to nothing.
The green bar and the red bar are drawn the same height on purpose. Setting them equal is the derivation: ½mv² = GMm/R.
Watch the little m disappear between step 1 and step 2. It is the same vanishing act you saw with g = GM/r² and with Kepler’s third law, and it means something remarkable: a feather, a tennis ball and a fully fuelled rocket all need exactly 11.2 km s−1 to leave Earth. Heavier things need more energy, certainly — but not more speed. Both sides of the equation grow with mass, so the mass has no say.
How escape speed varies between worlds
Since vesc = √(2GM/r), a body that is more massive or more compact is harder to leave. Compare a few:
Escape speed governs whether a world can keep an atmosphere. Gas molecules move faster when it is hot — if they routinely exceed vesc, the air simply leaks away into space.
Escape speed from density
Sometimes you are given a body’s density rather than its mass. Substitute M = ρ(4/3)πr³ into the escape speed formula and things simplify beautifully:
Escape speed from densityvesc = √( 2G · ρ(4/3)πr3 / r ) = r √( 8πGρ / 3 )so for a fixed density, vesc ∝ r — a bigger ball of the same rock is harder to leave
So why don’t rockets reach 11.2 km s⁻¹?
A satellite launch never comes close to that speed on the pad, and never needs to. Two reasons, and both are marks.
A rocket keeps burning fuel. Escape speed assumes one shove and then coasting — no further energy input. A rocket breaks that assumption on purpose, feeding in energy the whole way up.
Reaching orbit is not the same as escaping. An orbiting satellite is still firmly bound to the Earth — it is falling around it forever. It takes considerably less energy to get into orbit than to leave the field altogether.
Question
Answer
Does escape speed depend on the escaping mass?
No. The m cancels
Does the escaping energy depend on that mass?
Yes.E = GMm/r keeps its m
Does it depend on the direction of launch?
No. It is a speed, not a velocity
Does it depend on where you launch from?
Yes — on r. Higher up, it is smaller
Must a rocket reach it to orbit?
No. It has engines, and orbit is cheaper
Escape the planet, or escape the field?
The field. That is the phrase to write
All the KE ½mv²
is spent paying off
The energy debt GMm/r
the m cancels
vesc = √(2GM/r)
🚀 Calculating an escape speed
Identify the body being escaped.M is its mass, never the rocket’s.
Find r from the centre. Usually the surface radius. Halve any diameter you are given.
Given a density instead? Use M = ρ(4/3)πr³, or go straight to v = r√(8πGρ/3).
Substitute into v = √(2GM/r). Do the division before the square root.
Sanity check. A few km s−1 for a moon or planet. If you get metres per second, you dropped a power of ten.
WE 1
Calculate the escape speed at the surface of the Earth. Take ME = 5.97 × 10²⁴ kg and RE = 6.37 × 10⁶ m.
Step 1 — write the equationvesc = √(2GM / r)Step 2 — work out the inside of the root2GM = 2 × (6.67 × 10⁻¹¹) × (5.97 × 10²⁴) = 7.96 × 10¹⁴2GM / r = 7.96 × 10¹⁴ / 6.37 × 10⁶ = 1.25 × 10⁸Step 3 — square rootv = √(1.25 × 10⁸)vesc = 1.12 × 10⁴ m s⁻¹ = 11.2 km s⁻¹Forty thousand kilometres per hour. And it is the same whether you are launching a satellite or flicking a paperclip — the mass never entered the calculation.
WE 2
A spherical asteroid has a radius of 480 km and a mean density of 2600 kg m⁻³. Determine the escape speed at its surface.
Step 1 — we have density, not massM = ρV = ρ (4/3) π r³Step 2 — substitute into the escape speed equation and simplifyv = √(2G × ρ(4/3)πr³ / r) = r √(8πGρ / 3)Step 3 — evaluate the root8πGρ/3 = 8π × (6.67 × 10⁻¹¹) × 2600 / 3 = 1.45 × 10⁻⁶√(1.45 × 10⁻⁶) = 1.21 × 10⁻³Step 4 — multiply by the radius, in metresv = (4.8 × 10⁵) × (1.21 × 10⁻³)vesc = 5.8 × 10² m s⁻¹ ≈ 0.58 km s⁻¹Slower than a rifle bullet. On an asteroid this size you could, in principle, jump hard enough to never come down. Note the neat result: at fixed density, vesc ∝ r.
WE 3
(a) Explain why the escape speed from a planet does not depend on the mass of the escaping object. (b) A rocket is launched from the Earth’s surface into a low orbit, never exceeding 8 km s⁻¹. Explain why it does not need to reach the escape speed of 11.2 km s⁻¹.
(a) why the mass drops out
Escape happens when ½mv² = GMm/r.
The object’s mass m appears on both sides and cancels.
v depends only on M and r of the planet
A heavier object needs more energy, but not more speed.
(b) why the rocket gets away with less
Escape speed assumes no further energy input after launch.
A rocket burns fuel continuously, so it is being given energy all the way up.
Also, reaching orbit is not escaping — the rocket stays bound to the Earth.
less energy is needed to orbit than to escapePart (b) is worth two separate marks, and students usually give only one of them. Say both: continuous thrust, and orbit ≠ escape.
💡 Top tips
In the definition, write “escape a gravitational field” — never “escape gravity” or “the gravitational pull”.
Include “minimum speed” and “no further energy input“. Both are marked.
Be ready to derive it from ½mv² = GMm/r. The equation is in the data booklet; the derivation is not.
It is a speed, not a velocity — direction is irrelevant, so long as you don’t hit the ground.
Escape speed does not depend on the escaping mass. Escape energy does.
Given a density? vesc = r√(8πGρ/3) saves you a step.
Learn 11.2 km s−1 for Earth. It’s a useful check on any answer.
⚠ Common mistakes
Saying escape speed lets you “escape the planet”. It lets you escape the planet’s field, which extends to infinity
Thinking a heavier rocket needs a higher escape speed. It needs more energy, at the same speed
Forgetting the 2 in √(2GM/r), and quoting the orbital speed instead
Using the mass of the rocket as M
Using an altitude, or a diameter, in place of r
Claiming rockets must reach 11.2 km s−1 to get into orbit. They have engines, and orbit is not escape
Forgetting to take the square root, or rooting before dividing
Quick recap:Escape speed is the minimum speed allowing an object to escape a gravitational field with no further energy input. Set the kinetic energy equal to the depth of the potential well — ½mv² = GMm/r — and the escaping mass cancels, leaving vesc = √(2GM/r). It is the speed at which the total energy is exactly zero. It is the same for a feather and a rocket, it depends only on M and r of the body you are leaving, and rockets never need it, because they keep their engines running and orbit is cheaper than escape.
Notice we have just said something strange: it is cheaper to orbit than to escape. How much cheaper, exactly? A satellite in orbit has kinetic energy and potential energy, and there turns out to be a fixed, elegant relationship between the two — and between orbital speed and the escape speed you have just found. (A hint: the ratio is √2.) Next page: Orbital Motion, Speed & Energy.
Escape speed derivation not sticking?
Book a free meeting and we’ll work through the energy argument, the density shortcut and the classic “why not 11.2?” question.
