ROCKETS · chemical energy → controlled motion
START HERE | Energy is stored in the propellant’s chemical bonds.
Stored chemical energy begins the chain
burning makes hot gas
nozzle speeds gas backward
🔥 Propellant burns
💨 Nozzle accelerates exhaust
🚀 Exhaust momentum pushes rocket
Chemical energy → hot gas
Fast exhaust flows backward
Exhaust carries momentum away
Takeaway: burning supplies energy, expelled mass supplies reaction force, and steering shapes the path—altitude alone is not orbit.
Illustrative assumptions: 100 t at ignition; 80 t propellant; 20 t final vehicle; exhaust speed 3.0 km/s.
Ideal rocket equation: Δv = vₑ ln(mᵢ/m_f) = 3.0 ln(100/20) ≈ 4.83 km/s
Each small propellant packet leaves backward at about 3.0 km/s relative to the vehicle; as mass falls, later propellant accelerates less mass. Combining those increments gives the logarithm.
100 t → 20 t remaining
80 t expelled backward
Ideal Δv only—not a launch-performance prediction: ignores gravity, drag, steering losses, engine limits and actual mass distribution.
Clean colored-marker technical illustration, transparent background. Side view of a small launch rocket above a curved Earth horizon, entire Earth limb visible as a broad arc across the lower portion. Rocket is in a gravity-turn trajectory: its nose tilts increasingly toward the right, no longer pointing straight up; draw a smooth curved flight path rising from near the horizon, curving over to become nearly horizontal at high altitude. Show the rocket's exhaust plume pointing backward and down-left, opposite the rocket's forward direction. Include a thin, clearly curved orbital path around Earth at a modest distance above the surface and a tiny satellite moving tangentially along it, making the needed sideways motion visually distinct from vertical altitude. Keep all components complete and clearly separated, no text, labels, symbols, measurements, or scenery beyond Earth, rocket, exhaust, path and satellite. Bold clean outlines with restrained blue, teal and gold accents; modest safety margins; the illustration must not imply the rocket remains pointed upward to orbit.
combustion chamber
throat
exhaust momentum pushes forward
shed spent mass; steer sideways
🧩 Staging sheds empty mass
🛰️ Steering builds sideways speed
Less empty mass to accelerate
Orbit requires sideways speed
Overview step: 🔥 burns (orange) → 💨 nozzle speeds exhaust (blue)
Fuel + carried oxidizer
Combustion
reacts; bonds rearrange
chemical energy → heat; products expand
A rocket brings both fuel and oxidizer; burning heats the gas and raises its pressure inside the chamber.
Converging: subsonic gas speeds up
Divergent bell: supersonic gas expands and accelerates
Throat: narrowest area; may choke at Mach 1
The widening section accelerates supersonic flow by expansion; performance depends on pressure ratio and nozzle shape. A nozzle redirects reaction energy into a fast jet—it does not push on surrounding air.
From the nozzle: exhaust leaves backward relative to the rocket
thrust on rocket ←
rocket + propellant
Equal-and-opposite momentum changes: exhaust gains rightward momentum; the rocket gains leftward momentum. No contact with surrounding air is required.
expanding nozzle
From section 2: the nozzle sends fast exhaust backward.
rocket momentum + thrust →
exhaust momentum ←
Rocket + expelled gas: total momentum is conserved (ideal external-force-free view). Opposite momentum changes give the rocket forward motion—no air is needed to push against.
Δp⃗rocket = −Δp⃗exhaust | thrust is the exhaust’s momentum carried away per second
T = 750 kN; m = 100,000 kg T/m = 7.5 m/s² thrust-only
thrust T ↑
Hot, high-pressure gas
Vertical snapshot, drag ignored: a_net = T/m − g = 7.5 − 9.8 = −2.3 m/s² (still downward)
vehicle
weight mg ↓
Takeaway: force sets acceleration; it does not directly specify speed or direction traveled.
Tiny burn step: rocket sheds a small mass packet; exhaust leaves backward at speed vₑ relative to rocket.
rocket gains +δv →
rocket m
ejected δm
Momentum balance (no external force): m δv = vₑ δmₑ, where δmₑ > 0 is expelled mass and rocket mass changes by dm = −δmₑ. So m dv = −vₑ dm. Integrate from mass m₀ to m_f:
exhaust velocity is backward relative to rocket: −vₑ
Δv = vₑ ln(m₀ / m_f)
Ideal only: constant exhaust speed; all expelled mass counted; external forces neglected. This gives the speed change from mass ratio—not a real-flight prediction.
Real vertical flight instead: a = (T − mg − D) / m. T is thrust, mg weight, D drag; m and T vary, g changes slightly, and the path may tilt. As propellant burns, less mass means the same thrust can produce more acceleration.
Example caveat: 7.5 − 9.8 = −2.3 m/s² is the unconstrained free-flight snapshot. On the launch pad, the support force keeps a sub-weight-thrust rocket from accelerating downward; liftoff starts only when upward thrust exceeds weight (with drag accounted for once airborne).
exhaust momentum carried away →
m
rocket
On the pad, N + T balances mg while T < mg: no downward acceleration. At liftoff N falls to zero; upward acceleration begins when T > mg.
pad force N ↑
Force balance changes at liftoff; thrust alone is not the net force beforehand.
At fixed thrust, less mass → more thrust-only acceleration
15
T/m (m/s²)
7.5
Illustration holds T = 750 kN fixed: at 100 t, T/m = 7.5 m/s²; at 50 t, 15 m/s². Real thrust, gravity and drag also change, so net acceleration need not follow this curve exactly.
vehicle mass m (kg)
m dv = −vₑ dm → dv = −vₑ (dm / m)
∫ᵥ₀ᵛᶠ dv = −vₑ ∫ₘ₀ᵐᶠ (dm / m)
v_f − v₀ = −vₑ(ln m_f − ln m₀) = vₑ ln(m₀ / m_f)
Since m_f < m₀, ln(m₀/m_f) > 0: the rocket’s ideal speed gain is forward. External forces are excluded in this derivation.
Illustrative free-flight snapshot · drag set to zero
T = 750 kN ↑
Net force ↓ = 230 kN So a = 2.3 m/s² downward
mg = 980 kN ↓
Because T < mg here, the unconstrained snapshot accelerates downward; on the pad, support force prevents that fall.
Teal upper stage + payload continue upward
Spent purple first stage separates and falls away
From 🚀 exhaust momentum: continued thrust speeds the vehicle; staging stops hauling an already-spent stage.
Liftoff needs upward net force: T > mg (and, in air, enough to overcome drag). The pad can support a rocket before that threshold.
one vehicle
upper stage
Thrust = force (N). Acceleration = velocity change per second (m/s²). Velocity = motion state (m/s); it accumulates over time.
spent stage
Single-stage: empty tanks stay attached
Serial staging: lower stage burns, separates; upper continues
Burnout → discard → next stage accelerates
Propellant depleted
spent tanks + engines
inert mass discarded
less inert load
Ideal rocket equation · one stage’s mass ratio
Mass ratio m₀/mf = 4. Dropping spent dry mass leaves less inert load for the next stage, improving its mass ratio (all else equal).
Serial stages fire one after another; strap-on boosters are parallel helpers jettisoned when spent. Staging adds hardware, separation events and failure modes; discarded parts are not automatically recovered. Ideal example omits gravity, drag, structure and staging losses—no launch prediction.
Spent stage kept attached
base + payload
old dry mass
next propellant
ratio = (base + dry + propellant) / (base + dry)
Spent stage jettisoned
same propellant
ratio = (base + propellant) / base → larger (same next stage)
Takeaway: jettisoning spent stages matters because the next burn no longer accelerates their empty structure.
What mass ratio 4 means · idealized one-stage example
Before burn: m₀ = 4 units
1
After burn: mf = 1 unit remains
3 propellant units expelled
So m₀/mf = 4/1 = 4. In a real vehicle, “remaining” includes payload, tanks, engines, and residual propellant—not just payload.
Schematic: dropping old dry mass reduces the next stage’s carried base mass, so its fixed propellant gives a larger mass ratio.
From 🧩 staging: less spent mass to accelerate; guidance now shapes the flight path.
altitude ↑
velocity increasingly horizontal
powered: thrust follows attitude
A small initial tilt lets gravity bend the path; the guidance system gradually pitches over instead of holding the nose straight up.
downrange distance →
Earth surface (schematic)
After burnout, tanks and engines are inert mass. Separation leaves the next stage to accelerate without that load.
Spent hardware is discarded—not automatically recovered.
Altitude is height; orbit also needs enough sideways speed. This powered curve is still under thrust—not yet the free-fall orbit.
smaller mass remains to accelerate
The powered rocket curves over; only after engine cutoff does the satellite coast in free fall. Schematic, not to scale.
r = Rₑ + h
For a circular orbit, gravity supplies the inward (centripetal) acceleration:
v tangent
Cancel m; multiply by r: v² = GM/r → v = √(GM/r)
G: gravitational constant; M: Earth’s mass; GM: standard gravitational parameter; r: distance from Earth’s center. m is the satellite mass.
Earth
gravity inward
Orbit is not hovering: the satellite is always falling toward Earth, but its sideways speed carries it forward as Earth curves away.
Rₑ ≈ 6,370 km; h ≈ 200 km, so r ≈ 6,570 km = 6.57 × 10⁶ m. GM ≈ 3.986 × 10¹⁴ m³/s².
v = √[(3.986 × 10¹⁴)/(6.57 × 10⁶)] ≈ 7,790 m/s ≈ 7.8 km/s.
Ideal period: T = 2πr/v ≈ 5.3 × 10³ s ≈ 88 min. Circular-orbit assumptions only.
At a fixed radius, altitude alone is not enough. A tangential speed below circular speed gives a bound ellipse (which may intersect Earth); above circular but below escape gives an ellipse; at escape speed or more, the path is unbound.
Escape is different: at this radius, ideal escape speed = √2 × circular speed ≈ 11.0 km/s; orbital speed is not “escaping.”
The 7.8 km/s result is an ideal circular speed at that radius—not the full launch requirement. Real ascent must account for atmospheric drag, Earth’s rotation, gravity and steering losses, target inclination, and detailed guidance. A real insertion orbit may be elliptical rather than circular.
Takeaway: altitude chooses the orbital radius; sideways speed makes the fall curve around Earth. Guidance and thrust build that speed during ascent.
Next questions: How can a rocket steer in space after engine cutoff? How do launch windows and orbital inclination shape a mission?
Speed at one radius · ideal two-body, tangential case
0
v₍c₎ ≈ 7.8 km/s
v₍esc₎ ≈ 11.0 km/s
Below v₍c₎ ellipse; may hit Earth
At v₍c₎ circular orbit
v₍c₎ < v < v₍esc₎ bound ellipse
≥ v₍esc₎ escape
How a gravity turn builds sideways speed
Small initial tilt starts the turn
Gravity curves velocity; thrust follows attitude
Mostly horizontal speed; cutoff → free-fall orbit
Blue arrows: velocity direction · grey arrows: gravity · gold arrows: flight progression (schematic)
Guidance turns velocity toward horizontal
speed v
vₓ = v cos θ vᵧ = v sin θ
θ from horizontal
vertical, vᵧ
During the turn, θ decreases as the flight direction becomes more horizontal, so the horizontal share vₓ grows. Guidance steers powered ascent; after cutoff, gravity—not continued upward thrust—curves the orbit.
horizontal, vₓ