IB Physics HLTopic 4 — Force FieldsPaper 1 & 2Etotal = −GMm/2r~17 min read
Orbits: Speed & Energy
A satellite is not “beyond gravity”. It is falling — falling hard, all the time — and missing the Earth because it is moving sideways fast enough. That single idea gives you its speed in one line. Push a little further and you get its kinetic energy, its potential energy, and a total energy that is always negative. Negative energy sounds alarming. It is simply the mathematics of being trapped.
📘 What you need to know
The gravitational force provides the centripetal force keeping a satellite in orbit
vorbital = √(GM/r) — independent of the satellite’s mass
Kinetic energy: Ek = +GMm/2r (positive)
Potential energy: Ep = −GMm/r (negative)
Total energy: Etotal = −GMm/2r, and it is constant for a given orbit
Three relations worth memorising: Etotal = −Ek = ½Ep
As rincreases: v decreases, Ek decreases, Ep increases, Etotal increases (all become less negative), T increases
A negative total energy means the satellite is bound to the planet
vesc = √2 × vorbital at the same radius
Orbital speed
Nothing holds a satellite up. The only force on it is gravity, pointing straight at the planet’s centre — and that is precisely the direction a centripetal force must point. So gravity is the centripetal force. Set them equal.
Gravity supplies the centripetal forceGMm / r2 = mv2 / r
Orbital speedv2 = GM/r → vorbital = √( GM / r )M = mass of the central body (kg) • r = orbital radius from its centre (m)
The velocity is tangential; the force is radial. They are perpendicular, which is why gravity does no work on a circular orbit and the speed stays constant.
The satellite’s mass has cancelled again. Every object at the same orbital radius travels at the same speed — the ISS, a bolt that fell off it, and an astronaut on a spacewalk all sail along together at 7.67 km s−1. That is why they float alongside each other. They are not weightless; they are all falling at exactly the same rate.
The energy of an orbiting satellite
Now take that result, v² = GM/r, and feed it into the kinetic energy.
Kinetic energy in orbitEk = ½mv2 = ½m · GM/r = + GMm / 2r
The potential energy you already know — it is the energy of a mass m sitting at distance r in the field of M:
Potential energy in orbitEp = − GMm / r
Add them. The kinetic energy is exactly half the size of the potential energy, and opposite in sign, so half of the potential energy survives the addition:
Total energy in orbitEtotal = GMm/2r − GMm/r = − GMm / 2rconstant for a given orbit • always negative • and equal to −Ek
Three relations worth memorisingEtotal = −Ek = ½Epknow any one of the three energies and you instantly know the other two
Look at the green and teal curves: at every radius they are the same distance from the zero line, on opposite sides. That is Etotal = −Ek, drawn.
Higher orbit, or lower?
Every quantity on this page depends on r, and it pays to know which way each one moves.
The trap: a satellite in a higher orbit moves slower. Being further out costs energy overall, but that energy goes into potential, not into speed.
Orbit X (smaller r)
Orbit Y (larger r)
Gravitational force on it
Larger
Smaller
Orbital speed v
Faster
Slower
Kinetic energy Ek
Greater
Lower
Potential energy Ep
Lower (more negative)
Greater (less negative)
Total energy Etotal
Lower (more negative)
Greater (less negative)
Orbital period T
Shorter
Longer
Here is the sentence that catches everyone: a higher orbit has more total energy but less speed. Both are true, and they are not in conflict. Climbing to orbit Y costs you energy — you must fire an engine to get there — but all of that energy, and more, is banked as potential energy. The kinetic energy actually goes down. Ask a student “which satellite is going faster?” and half will say the one with more energy. Don’t be that half.
The link to escape speed
We promised this last page. Compare the two speeds at the same radius:
Orbit versus escapevorbital = √(GM/r) vesc = √(2GM/r)vesc = √2 × vorbitalgo about 41% faster than orbital speed and you are gone for good
Gravity = centripetal force
gives
v = √(GM/r)
substitute into ½mv²
Ek = GMm/2r
add Ep
Etotal = −GMm/2r
🛰️ Working an orbit question
Find r from the centre. Altitude plus planet radius. Every single time.
Speed?v = √(GM/r). Period?T = 2πr/v, or Kepler’s third law.
Energies? Get one of them, then use Etotal = −Ek = ½Ep for the rest.
Comparing two orbits? Use proportionality: v ∝ r−½, Ek ∝ r−1, T ∝ r3/2.
Sign check.Ek positive, Ep and Etotal negative. Always.
WE 1
The International Space Station orbits at an altitude of 400 km above the Earth’s surface. Calculate (a) its orbital speed and (b) its orbital period in minutes. (ME = 5.97 × 10²⁴ kg, RE = 6.37 × 10⁶ m)
Step 1 — find r from the centrer = 6.37 × 10⁶ + 4.00 × 10⁵ = 6.77 × 10⁶ m(a) Step 2 — orbital speedv = √(GM/r) = √(3.98 × 10¹⁴ / 6.77 × 10⁶)v = √(5.88 × 10⁷)v = 7.67 × 10³ m s⁻¹ = 7.67 km s⁻¹(b) Step 3 — period from speedT = 2πr / v = 2π × (6.77 × 10⁶) / 7669T = 5546 sT = 92 minutesThe astronauts see a sunrise every 92 minutes — sixteen a day. And the ISS is not beyond gravity: at that altitude g is still about 8.7 N kg⁻¹, nearly 90% of its surface value.
WE 2
A satellite of mass 1500 kg orbits the Earth at a radius of 1.2 × 10⁷ m. Determine its kinetic energy, its gravitational potential energy and its total energy. (GME = 3.98 × 10¹⁴ N m² kg⁻¹)
Step 1 — kinetic energyEk = GMm / 2r = (3.98 × 10¹⁴) × 1500 / (2 × 1.2 × 10⁷)Ek = +2.5 × 10¹⁰ JStep 2 — potential energy (twice as big, negative)Ep = −GMm / r = −2 × EkEp = −5.0 × 10¹⁰ JStep 3 — total energyEtotal = Ek + Ep = 2.5 × 10¹⁰ − 5.0 × 10¹⁰Etotal = −2.5 × 10¹⁰ JOnly Step 1 needed a calculator. Once you have Ek, the relations Ep = −2Ek and Etotal = −Ek hand you the rest for free. The negative total confirms the satellite is bound.
WE 3
Two identical satellites X and Y orbit the same planet, at radii R and 3R respectively. Determine the ratios vY/vX, EkY/EkX and TY/TX. State which satellite has the greater total energy.
Step 1 — speed goes as r to the power −½v ∝ 1/√r → vY/vX = 1/√3vY/vX = 0.58Step 2 — kinetic energy goes as 1/rEk = GMm/2r → ratio = 1/3EkY/EkX = 0.33Step 3 — period from Kepler’s third lawT² ∝ r³ → (TY/TX)² = 3³ = 27TY/TX = √27TY/TX = 5.2Step 4 — which has the greater total energy?Etotal = −GMm/2r, and Y has the larger r, so its Etotal is less negative.
Y — the outer, slower satelliteSlower, yet greater total energy. If that still feels wrong, look again at the graph: the green curve fell, but the red one rose faster.
💡 Top tips
Start every derivation from “gravitational force provides the centripetal force“. It’s a marked statement.
Memorise Etotal = −Ek = ½Ep. It converts a three-part question into a one-part calculation.
Say “increases” and “less negative” together, so the examiner knows you mean it.
A satellite in a higher orbit is slower but has more total energy. Never mix these up.
Gravity does no work on a circular orbit — force and velocity are perpendicular. Hence constant speed.
vesc = √2 vorbital. A quick way to check either one.
The satellite’s mass cancels from v and T, but stays in all three energies.
⚠ Common mistakes
Thinking a satellite in a higher orbit moves faster. It moves slower
Writing Ek = −GMm/2r. Kinetic energy is never negative
Forgetting the 2 in Etotal = −GMm/2r — that’s Ep, not the total
Saying astronauts float because “there is no gravity”. g is nearly 90% of its surface value up there; they are in free fall
Using v = √(2GM/r) for an orbit. That’s the escape speed
Cancelling the satellite’s mass out of the energies. It only cancels from v and T
Using an altitude as r, or a period in minutes in a formula
Quick recap: Gravity supplies the centripetal force, so vorbital = √(GM/r), independent of the satellite’s mass. Feeding that into ½mv² gives Ek = +GMm/2r, and with Ep = −GMm/r the total comes to Etotal = −GMm/2r — constant, negative, and equal to −Ek. Move to a higher orbit and the satellite goes slower, loses kinetic energy, but gains total energy. And vesc = √2 vorbital.
Everything above assumed a perfect orbit: no air, no friction, total energy fixed forever. But the atmosphere does not stop dead at some tidy boundary — it thins away, and a satellite below about 600 km is ploughing through the last wisps of it. It loses energy. Its orbit shrinks. And then something genuinely counter-intuitive happens: it speeds up. Losing energy makes it go faster. The final page of this section explains why: Effects of Drag on Orbital Motion.
Orbital energy signs tying you in knots?
Book a free meeting and we’ll drill the derivations, the ratios and the “slower but more energy” paradox until it clicks.
