The rocket equation
Tonight's idea
A rocket is a machine for throwing mass backwards. At some instant it has mass \(m\) and speed \(v\). In a short time it burns a small mass \(dm\) of fuel and throws it out of the back at speed \(v_e\) relative to the rocket (the exhaust speed). Ignore gravity and air for now. Nothing pushes from outside, so momentum is conserved. Before: \(mv\). After: a lighter, faster rocket plus exhaust moving at \(v - v_e\):
\[mv = (m - dm)(v + dv) + dm\,(v - v_e).\]
Expand the right side: \(mv + m\,dv - v\,dm - dm\,dv + v\,dm - v_e\,dm\). The \(mv\) and \(v\,dm\) terms cancel, and \(dm\,dv\), two small things multiplied, is negligible. What survives is
\(m\,dv = v_e\,dm\), so \(dv = v_e\,\dfrac{dm}{m}\).
Each gram of fuel gives the same push, shared by whatever mass remains, and that mass keeps shrinking. Add up the pushes as the mass falls from \(m_0\) (full) to \(m_1\) (empty):
\[\Delta v = v_e \int_{m_1}^{m_0} \frac{dm}{m} = v_e \ln\frac{m_0}{m_1}.\]
This is the Tsiolkovsky rocket equation, written down in 1903, more than fifty years before anything reached orbit. The logarithm is the cruel part: to double \(\Delta v\) you must square the mass ratio.
Watch one: \(v_e = 3\) km/s, mass ratio 10
Kerosene burning in oxygen gives roughly 3 km/s. Build a rocket that is 90% fuel by mass, so \(m_0/m_1 = 10\):
\[\Delta v = 3 \ln 10 = 3 \times 2.303 = 6.9 \text{ km/s}.\]
Low orbit needs 7.8 km/s, and fighting gravity and air on the way up costs about another 1.6, so call it 9.4 km/s. The ratio required is \(e^{9.4/3} = e^{3.13} \approx 23\): fuel must be \(1 - 1/23 = 95.6\%\) of the launch mass. The remaining 4.4% must be tanks, engines, guidance and payload, and tanks and engines alone typically weigh more than that. Exercise 4 shows the way out.
Your turn
- A rocket has \(v_e = 2.5\) km/s and launch mass four times its empty mass. Find \(\Delta v\). What fraction of the launch mass was fuel?
- What mass ratio gives \(\Delta v = v_e\)? What fraction of the launch mass is fuel? Repeat for \(\Delta v = 2v_e\). Twice the speed needs how many times the fuel fraction?
- Hydrogen burning in oxygen gives \(v_e \approx 4.4\) km/s. What mass ratio delivers the 9.4 km/s of the example, and what fuel fraction? Compare with kerosene's 95.6%. Why do engineers care so much about exhaust speed?
- Staging. A 100 t rocket with \(v_e = 3\) km/s burns 60 t of fuel, drops its 10 t empty first stage, then the 30 t second stage burns 20 t more, leaving 10 t at the end. (a) Find \(\Delta v\) by adding the two burns. (b) A single-stage rocket with the same 80 t of fuel must carry all 10 t of structure to the end: 100 t down to 20 t. Find its \(\Delta v\). (c) Where did the difference come from?
Stretch: gravity bites during the burn
Burning straight up, gravity subtracts \(g\,dt\) from each \(dv\), so \(\Delta v = v_e\ln(m_0/m_1) - gT\) for a burn lasting \(T\) seconds. The example rocket (ratio 10, \(v_e = 3\)) burns for 150 s. How much is lost? Why do rockets tip towards the horizontal soon after launch, and why do engines burn as hard as the structure allows?
Answers
- \(\Delta v = 2.5 \ln 4 = 2.5 \times 1.386 = 3.5\) km/s. Empty mass is a quarter of launch mass, so fuel was three quarters: 75%.
- \(\ln(m_0/m_1) = 1\) means \(m_0/m_1 = e = 2.72\), fuel fraction \(1 - 1/e = 63\%\). For \(2v_e\): ratio \(e^2 = 7.39\), fuel fraction \(1 - e^{-2} = 86\%\). Twice the speed needs only 1.37 times the fuel fraction, but the fraction cannot pass 100%, so the empty rocket is squeezed from 37% of launch mass down to 14%. The squeeze on what is left is the real cost.
- \(e^{9.4/4.4} = e^{2.14} = 8.5\), fuel fraction \(1 - 1/8.5 = 88\%\). That leaves 12% for structure and payload instead of 4.4%, nearly three times as much. Exhaust speed sits inside the exponent, so a 47% gain in \(v_e\) cuts the required ratio from 23 to 8.5. (Hydrogen's drawback is that it is so light it needs huge tanks.)
- (a) First burn \(3\ln(100/40) = 3 \times 0.916 = 2.75\) km/s; second \(3\ln(30/10) = 3 \times 1.099 = 3.30\) km/s; total 6.04 km/s. (b) \(3\ln(100/20) = 3\ln 5 = 4.83\) km/s. (c) Staging gained 1.2 km/s from the same fuel because the second burn did not have to accelerate 10 t of empty tank. Fuel spent moving dead weight is fuel wasted; dropping the weight early is the only escape from the logarithm.
Stretch. \(gT = 9.81 \times 150 = 1{,}470\) m/s, about 1.5 km/s, so \(\Delta v = 6.9 - 1.5 = 5.4\) km/s. The loss is \(gT\), proportional to burn time and nothing else, which is why the thrust is as high as the structure allows: finish the burn fast. Once the rocket is moving horizontally, gravity pulls sideways to its motion and no longer subtracts from its speed, so rockets pitch over as soon as they are clear of the thickest air.