Variable Mass Systems and the Rocket Equation

(Topic)

Variable Mass Systems and the Rocket Equation

The particle-system equations developed earlier assumed a fixed set of particles.

A rocket does not satisfy that assumption.

As propellant is expelled, mass crosses the boundary of the object we call “the rocket.” The rocket mass decreases, and the escaping material carries momentum with it.

The central lesson is

|-----------------------------------------------------------------------------------------| |For an open variable mass system, momentum flux across the boundary must be included. | ------------------------------------------------------------------------------------------

(1)

For one stream of transferred mass, a useful Newtonian equation is

|--------------------------| | dv dm | |m ---= Fext + (u − v )---,| ---dt------------------dt---

(2)

where

m(t) is the instantaneous mass of the chosen body,
v is the body’s velocity in an inertial frame,
u is the inertial-frame velocity of the material crossing the boundary,
dm∕dt is the signed rate of change of the body’s mass.

For a rocket,

dm ----< 0. dt

(3)

If exhaust leaves backward with speed u_e relative to the rocket, the thrust magnitude is

|------(------)---| | dm-- | T = ue − dt . | -------------------

(4)

With no external force, integration gives the ideal rocket equation

|----------(----)--| Δv = u ln m0- .| | e mf | --------------------

(5)

1 Why the fixed-mass equation needs modification

For a fixed set of particles,

dP- Fext = dt .

(6)

If the selected system is the rocket body alone, however, particles continually leave the system.

Writing

P = mv

(7)

and then expanding

-d(mv ) = m dv-+ v dm-- dt dt dt

(8)

does not by itself produce the correct rocket equation.

The missing physics is the momentum carried across the system boundary by the exhaust.

Figure 1. A variable mass body is an open system. Material crossing the boundary transports momentum, so mass flow must appear explicitly in the momentum balance.

2 Closed material system versus open rocket system

There are two useful ways to analyze rocket motion.

2.1 Closed material system

Choose a system containing the rocket and all propellant that will later be expelled.

No mass crosses this enlarged system boundary.

Ordinary total momentum conservation applies directly when external impulse is negligible.

2.2 Open rocket system

Choose only the instantaneous rocket and its remaining propellant.

Mass crosses the boundary through the exhaust.

The momentum carried by that mass must be included as a flux term.

Both viewpoints produce the same physics when used consistently.

3 One-stream variable mass momentum balance

Consider a body with instantaneous mass m and velocity v.

During a short time dt, its mass changes by dm.

The material transferred across the boundary has inertial velocity u.

First-order momentum bookkeeping gives

|----------------------------| m--dv-=-Fext-dt +-(u-−-v-)dm.-

(9)

Divide by dt:

|--------------------------| |m dv-= Fext + (u − v )dm-.| ---dt------------------dt---

(10)

Define the stream velocity relative to the body:

-------------- u = u − v.| --rel----------

(11)

Then

|----------------------| |m dv-= Fext + ureldm-.| ---dt--------------dt---

(12)

Figure 3. The variable mass equation separates ordinary external force from momentum transfer associated with material crossing the system boundary.

4 Sign convention for a rocket

Let e_x point in the desired forward direction.

Suppose exhaust leaves backward relative to the rocket at speed u_e > 0:

|u---=-−-u-e--.| ---rel-----e-x--|

(13)

The rocket is losing mass:

|--------| |dm | |----< 0.| -dt-------

(14)

Therefore

ureldm-- dt

(15)

points forward.

Define the positive propellant mass-flow rate

|----------------| | dm | |˙mp = − ---> 0.| ---------dt-------

(16)

Then the thrust vector is

|--------------| |T = uem˙p ex. | ---------------

(17)

Its magnitude is

|----------| T = uem˙p. | ------------

(18)

5 Differential derivation of rocket thrust

Consider a one-dimensional rocket.

At time t, the rocket has mass m and velocity v.

During dt, the rocket ejects positive mass

dμ > 0.

(19)

The remaining rocket mass is

m − d μ,

(20)

and its new velocity is

v + dv.

(21)

If exhaust leaves backward at speed u_e relative to the rocket, its inertial velocity to first order is

v − ue.

(22)

Ignoring external impulse for the moment,

pi = mv.

(23)

Final momentum is

pf = (m − d μ)(v + dv) + dμ(v − ue).

(24)

Expand and discard the second-order product dμdv:

p_f	= mv + mdv − v dμ + v dμ − u_e dμ	(25)
	= mv + mdv − u_e dμ.	(26)

Momentum conservation gives

mv = mv + m dv − uedμ.

(27)

Therefore

|--------------| |m dv = uedμ. | ---------------

(28)

Since

dμ = − dm,

(29)

we obtain

|----------------| |m dv = − ue dm. | -----------------

(30)

Figure 2. Differential rocket derivation. The expelled propellant carries backward momentum, producing a forward change in rocket momentum.

6 Rocket equation with external force

Include an external force F_ext along the line of motion.

The differential equation becomes

|------------------------| m dv = − uedm + Fextdt.| --------------------------

(31)

Dividing by dt,

|----------------------| | dv dm | m dt-= − ue-dt-+ Fext.| ------------------------

(32)

Because

dm − ----= m˙p, dt

(33)

this is

|----------------| | dv- | m dt = T + Fext,| ------------------

(34)

with

T = uem˙p.

(35)

The mass is time dependent even when thrust is constant.

Figure 4. Thrust equals effective exhaust speed times positive propellant mass-flow rate: T = u_eṁ_p.

7 Ideal Tsiolkovsky rocket equation

Assume:

one-dimensional motion,
no external force during the idealized burn,
constant effective exhaust speed u_e,
Newtonian mechanics.

Then

m dv = − ue dm.

(36)

Separate variables:

dm-- dv = − ue m .

(37)

Integrate from initial mass m₀ and velocity v₀ to final mass m_f and velocity v_f:

∫ vf ∫ mf dm dv = − ue ----. v0 m0 m

(38)

Therefore

v_f − v₀	= −u_e_m₀^m_f	(39)
	= −u_e ln .	(40)

Thus

|----------(----)--| | m0- | Δv = ue ln mf .| --------------------

(41)

8 Mass ratio

Define the mass ratio

|---------| | m0- | R = mf . | -----------

(42)

Then

|--------------| -Δv--=-ue-ln-R.-|

(43)

Solving for mass ratio,

|------------| -R-=--eΔv∕ue.|

(44)

The exponential dependence is one of the central design constraints of rocketry.

Figure 5. Ideal rocket delta-v grows logarithmically with mass ratio. Equivalently, required mass ratio grows exponentially with required delta-v.

9 Example 1: ideal delta-v

A rocket has

m₀	= 12000 kg,	(45)
m_f	= 4000 kg,	(46)

and effective exhaust speed

ue = 3000m ∕s.

(47)

The mass ratio is

R = 3.

(48)

Therefore

Δv	= 3000 ln 3	(49)
	= 3296 m∕s.	(50)

Thus

|----------------| |Δv ≈ 3.30 km ∕s.| ------------------

(51)

10 Example 2: required mass ratio

Suppose

ue = 3500 m ∕s

(52)

and required ideal delta-v is

Δv = 7000 m ∕s.

(53)

Then

7000∕3500 2 R = e = e = 7.39.

(54)

Therefore

|m-----------| |--0 = 7.39. | -mf----------|

(55)

Only

1 --= 0.135 R

(56)

of the initial mass remains after the modeled propellant expenditure.

11 Propellant fraction

Define

m = m − m . p 0 f

(57)

The propellant fraction relative to initial mass is

|------------------| | mp 1 | |fp = --- = 1 − --.| ------m0--------R---

(58)

For

R = 7.39,

(59)

we obtain

|-----------| fp-≈-0.865.--

(60)

Thus about 86.5% of the initial mass is expelled in this idealized single-stage example.

12 Specific impulse

Rocket performance is often expressed using specific impulse:

|---------| I = ue, | |sp g0 | -----------

(61)

where

g = 9.80665 m∕s2 0

(62)

is standard gravity.

Therefore

|----------| ue-=--g0Isp.-

(63)

The thrust equation becomes

|-------------| T--=-m˙pg0Isp.--

(64)

The ideal rocket equation can be written

|-------------(----)---| |Δv = g0Ispln m0- .| | mf | -----------------------

(65)

13 Example 3: thrust from specific impulse

An engine has

Isp = 320 s

(66)

and propellant flow rate

m˙p = 25kg ∕s.

(67)

The effective exhaust speed is

u_e	= g₀I_sp	(68)
	= (9.80665)(320)	(69)
	= 3138 m∕s.	(70)

The thrust is

T	= ṁ_pu_e	(71)
	= (25)(3138)	(72)
	= 7.85 × 10⁴ N.	(73)

Thus

|------------| T--≈-78.5kN.--

(74)

14 Constant propellant mass-flow rate

m˙p = constant,

(75)

then

m (t) = m0 − m˙pt.

(76)

The ideal velocity change from time 0 to t is

|------------------------------| | ( m0 ) | v (t) − v0 = ueln ---------- .| -------------------m0-−-m˙pt-----

(77)

The burn time to reach final mass m_f is

|----m---−-m---| tb = --0-----f.| --------m˙p------

(78)

15 Acceleration during a constant-thrust burn

For no external force and constant thrust,

m dv-= T. dt

(79)

Therefore

|------------| | --T-- | a (t) = m (t).| --------------

(80)

As propellant is consumed, m(t) decreases.

Thus acceleration increases even if thrust remains constant.

16 Vertical flight with constant gravity

Take upward as positive.

Neglect drag and assume constant gravitational acceleration g.

Then

m dv-= T − mg. dt

(81)

Using

dm T = − ue---, dt

(82)

we obtain

dv = − uedm--− g dt. m

(83)

Integrating over the burn,

|--------------(-----)-------| | m0- | |vf − v0 = ueln m − gtb.| -------------------f---------

(84)

The quantity

|---| gtb--

(85)

is the gravity loss in this simplified vertical constant-g model.

Figure 6. In simplified vertical flight, ideal rocket delta-v is reduced by gravity acting throughout the burn.

17 Example 4: vertical burn with gravity

Suppose an ideal rocket would produce

Δvideal = 3000m ∕s.

(86)

The burn lasts

tb = 80s.

(87)

Neglect drag and use

2 g = 9.81m ∕s .

(88)

The gravity loss is

gtb = (9.81)(80) = 785m ∕s.

(89)

Therefore

|--------------------| -Δvactual ≈-2215-m-∕s.

(90)

This is a deliberately simplified vertical model.

18 Drag and other external forces

In general,

m dv = − u dm + F dt. e ext

(91)

Fext = − mg − D

(92)

for vertical upward flight with drag magnitude D, then

dm-- D- dv = − ue m − gdt − m dt.

(93)

Thus, schematically,

|---------------------------------| Δv--=-Δvideal −-Δvgravity-−-Δvdrag.-

(94)

The ideal rocket equation provides propulsive delta-v. External forces determine how much of that capability appears as actual vehicle velocity change.

19 Staging

A single-stage vehicle must carry payload, structure, tanks, engine, and all remaining propellant.

Discarding empty structure allows a later stage to begin with a smaller inert mass burden.

For ideal sequential stages,

|-------------------(-----)--| | ∑ m0,j | |Δvtotal = ue,j ln ----- .| -----------j----------mf,j----

(95)

Each stage has its own effective exhaust speed and mass ratio.

Figure 7. Staging discards inert structure so later propulsion does not have to accelerate hardware that is no longer useful. Ideal stage delta-v values add.

20 Example 5: two ideal stages

Stage 1 provides

u_e,1	= 3000 m∕s,	(96)
R₁	= 3.0.	(97)

Stage 2 provides

u_e,2	= 3400 m∕s,	(98)
R₂	= 2.5.	(99)

Then

Δv₁	= 3000 ln 3 = 3296 m∕s,	(100)
Δv₂	= 3400 ln 2.5 = 3115 m∕s.	(101)

Therefore

|-------------------| Δvtotal = 6411m ∕s. | ---------------------

(102)

21 Mass accretion: a different variable mass problem

Variable mass is not unique to rockets.

Suppose a cart of mass m and velocity v collects incoming material moving at inertial velocity u.

Now

dm--> 0. dt

(103)

The one-stream equation is

|--------------------------| | dv dm | |m ---= Fext + (u − v)---.| ---dt------------------dt--

(104)

If the incoming material is initially at rest in the laboratory,

u = 0,

(105)

and no external horizontal force acts:

m dv- = − vdm-. dt dt

(106)

Thus

d-(mv ) = 0. dt

(107)

The cart slows as it gains stationary mass.

Figure 8. A cart that captures stationary material gains mass and slows. The incoming material brings its own momentum into the open-system balance.

22 Example 6: cart collecting stationary mass

A cart has initial mass

m = 100kg 0

(108)

and speed

v0 = 6.0m ∕s.

(109)

It collects stationary material until its mass becomes

mf = 150 kg.

(110)

With no external horizontal force,

m0v0 = mf vf.

(111)

Therefore

|------------------------| | 100 | vf = ---(6.0) = 4.0m ∕s.| ------150-----------------

(112)

Kinetic energy decreases because capture is inelastic.

23 Multiple mass streams

If several independent streams cross the boundary, the momentum-flux contributions add.

Schematically,

|--------------------------| | dv- ∑ | m dt = Fext + urel,km˙k, | ----------------k-----------

(113)

provided each signed mass-flow rate and relative stream velocity are defined consistently.

The sign convention must be stated explicitly before using the equation.

24 Effective exhaust velocity

The simple derivation treats exhaust as a stream with relative speed u_e.

Real rocket thrust can also include a pressure contribution at the nozzle exit.

A more detailed propulsion model uses

T = m˙pve + (pe − pa )Ae,

(114)

where v_e is nozzle-exit exhaust speed relative to the vehicle, p_e is exit pressure, p_a is ambient pressure, and A_e is exit area.

An effective exhaust velocity can then be defined by

|----------| | T | |ue = --- .| ------m˙p---

(115)

25 What the ideal rocket equation does not include

The ideal equation

Δv = ue ln(m0 ∕mf )

(116)

does not by itself include:

gravitational loss,
aerodynamic drag,
steering loss,
pressure variation unless absorbed into effective u_e,
changing exhaust performance,
structural mass constraints,
relativistic effects.

It is an integrated momentum relation for idealized propulsion, not a complete trajectory model.

26 Why the logarithm appears

The differential relation

dm dv = − ue---- m

(117)

contains the factor

1 -- . m

(118)

Integrating 1∕m produces

lnm.

(119)

The logarithm is therefore a direct consequence of thrust acting on a continuously decreasing mass.

27 Energy is not the easiest route

A rocket carries internal energy in its propellant.

That energy becomes rocket kinetic energy, exhaust kinetic energy, thermal energy, pressure work, and other losses.

Because the exhaust retains substantial kinetic energy, simply equating propellant energy to rocket kinetic energy does not yield the rocket equation.

Momentum balance is the natural starting point.

28 Center of mass of rocket plus exhaust

If the full rocket-plus-exhaust material system experiences no external force,

Ptotal = constant.

(120)

Therefore the center-of-mass velocity of the complete material system remains constant.

The rocket accelerates forward because the exhaust acquires backward momentum.

29 Common mistakes

Applying F = d(mv)∕dt to the rocket body alone without a momentum-flux term.
Forgetting that the rocket has dm∕dt < 0.
Using exhaust speed in the laboratory frame when the rocket equation requires exhaust velocity relative to the rocket.
Losing the minus sign in mdv = −u_e dm.
Using Δv = u_e(m₀∕m_f) instead of the logarithmic relation.
Confusing thrust with impulse.
Confusing specific impulse in seconds with exhaust velocity in m∕s.
Forgetting the factor g₀ in u_e = g₀I_sp.
Adding gravity loss to ideal delta-v instead of subtracting it in the simple vertical model.
Assuming constant thrust means constant acceleration.
Treating a variable mass open system as if it were a fixed set of particles.
Mixing signed mass-flow rate dm∕dt with positive propellant flow rate ṁ_p without stating the convention.
Assuming the ideal rocket equation is a complete launch trajectory model.
Forgetting that staging mass ratios must be defined for each stage using the mass actually carried by that stage.

30 Practice exercises

Derive
$dv dm m ---= Fext + (u − v )---. dt dt$ (121)
Starting from differential momentum conservation, derive
$m dv = − uedm$ (122)

for a one-dimensional rocket with no external force.
Integrate the differential rocket equation to obtain
$Δv = ueln(m0 ∕mf ).$ (123)
A rocket has u_e = 2800 m∕s and R = 4. Find its ideal delta-v.
What mass ratio is required for Δv = 6000 m∕s with u_e = 3200 m∕s?
Find the propellant fraction corresponding to R = 5.
An engine has I_sp = 300 s and propellant flow rate 40 kg∕s. Find effective exhaust speed and thrust.
A rocket burns 5000 kg of propellant at constant rate 50 kg∕s. Find the burn time.
A constant-thrust rocket has initial mass 10000 kg and final mass 5000 kg. By what factor does its acceleration change if external forces are neglected?
Derive the simplified vertical result
$Δv = ue ln(m0 ∕mf ) − gtb.$ (124)
A vertical burn has ideal delta-v 2500 m∕s and lasts 60 s. Estimate the gravity loss at constant g.
A cart of mass 80 kg moving at 5 m∕s collects 20 kg of stationary material. Find its final speed.
Explain physically why a rocket can accelerate in empty space.
Show that the center-of-mass velocity of rocket plus exhaust remains constant when external force is zero.
For two ideal rocket stages, derive
$Δv = ∑ u ln(m ∕m ). total e,j 0,j f,j j$ (125)

31 Summary

For one stream crossing an open system boundary,

|--------------------------| | dv- dm--| |m dt = Fext + (u − v )dt .| ----------------------------

(126)

For a rocket with effective exhaust speed u_e and positive propellant flow rate

dm ˙mp = − ---, dt

(127)

the thrust magnitude is

|----------| T = uem˙p. | ------------

(128)

With no external force,

|----------------| |m dv = − ue dm. | -----------------

(129)

Integration gives

|------------------| | ( m0 ) | Δv = ue ln --- .| -------------mf-----

(130)

Using specific impulse,

|----------| ue-=--g0Isp.-

(131)

For simplified vertical flight with constant gravity and no drag,

|------------------------------| -Δvactual =-ue-ln-(m0-∕mf-)-−-gtb.

(132)

The next article returns to fixed-mass particle systems and develops angular momentum.

References

[1] K. E. Tsiolkovsky, Exploration of Cosmic Space by Means of Reaction Devices, 1903.

[2] G. P. Sutton and O. Biblarz, Rocket Propulsion Elements, 9th ed., Wiley, 2017.

[3] J. R. Taylor, Classical Mechanics, University Science Books, 2005.

[4] D. Kleppner and R. Kolenkow, An Introduction to Mechanics, 2nd ed., Cambridge University Press, 2014.

[5] OpenStax, University Physics, Volume 1, Rice University, 2016.

"Variable Mass Systems and the Rocket Equation" is owned by bloftin.
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Variable Mass Systems and the Rocket Equation