Rocket Equation &

Multistaging
PROFESSOR CHRIS CHATWIN
LECTURE FOR SATELLITE AND SPACE SYSTEMS MSC
UNIVERSITY OF SUSSEX
SCHOOL OF ENGINEERING & INFORMATICS 25TH APRIL 2017
Orbital Mechanics & the Escape
Velocity
 The motion of a space craft is that of a body with a certain
momentum in a gravitational field
 The spacecraft moves under the combines effects of its momentum
and the gravitational attraction towards the centre of the earth
 For a circular orbit V = wr (1)
F = mrw2 = centripetal acceleration (2)
And this is balanced by the gravitational
attraction
F = mMG / r2 (3)
𝐺𝑀 1/2 𝐺𝑀 1/2
𝑤= 𝑜𝑟 𝑉= (4)
𝑟3 𝑟
Orbital Velocity V
 𝐺 = 6.67 × 10−11 N𝑚2 𝑘𝑔2 Gravitational constant
 M = 5.97× 1024 kg Mass of the Earth
 𝑟0 = 6.371× 106 𝑚
 Hence V = 7900 m/s
 For a non-circular (eccentric) orbit
1 𝐺𝑀𝑚2
 = 1 + 𝜀 𝑐𝑜𝑠𝜃 (5)
𝑟 ℎ2
 ℎ = 𝑚𝑟𝑉 angular momentum (6)
ℎ2
 𝜀= −1 eccentricity (7)
𝐺𝑀𝑚2 𝑟0
1
𝐺𝑀 2
 𝑉= 𝑟
1 + 𝜀 𝑐𝑜𝑠𝜃 (8)
Orbit type

 For ε = 0 circular orbit

 For ε = 1 parabolic orbit
 For ε > 1 hyperbolic orbit (interplanetary fly-by)
Escape Velocity
 For ε = 1 a parabolic orbit, the orbit ceases to be closed and the space
vehicle will not return
 In this case at r0 using equation (7)
ℎ2
 1= −1 but ℎ = 𝑚𝑟𝑣 (9)
𝐺𝑀𝑚2 𝑟0
1
𝑚2 𝑉02 𝑟02 2𝐺𝑀 2
 2= therefore 𝑉0 = (10)
𝐺𝑀𝑚2 𝑟0 𝑟0

 𝐺 = 6.67 × 10−11 N𝑚2 𝑘𝑔2 Gravitational constant

 M = 5.975× 1024 kg Mass of the Earth
 𝑟0 = 6.371× 106 𝑚 Mean earth radius
1
2𝐺𝑀 2
 𝑉0 = 𝑟0
Escape velocity (11)
 𝑉0 = 11,185 m/s (~ 25,000 mph)
 7,900 m/s to achieve a circular orbit (~18,000 mph)
Newton’s 3rd Law & the Rocket
Equation
 N3: “to every action there is an equal and opposite reaction”
 A rocket is a device that propels itself by emitting a jet of matter.
 The momentum carried away results in a force acting to accelerate the rocket
in a direction opposite to that of the jet
 Like a balloon expelling its gas and providing thrust

 A rocket is different to a gun because a bullet is given all its energy at the
beginning of its flight. The energy of the bullet then decreases with time due to
the losses against air friction.
 A cannon shell or a bullet is a projectile
 A rocket is a vehicle
Rocket Equation

𝑑𝑚
 Thrust 𝐹 = −𝑉𝑒 negative because the mass of the rocket
𝑑𝑡
decreases with time (14)
 The acceleration of the rocket under this force is given by Newton’s 2nd
Law
𝑑𝑉
 𝐹 = 𝑚 𝑑𝑡 (15)
𝑑𝑉 1 𝑑𝑀
 Therefore = − 𝑀 𝑉𝑒 (16)
𝑑𝑡 𝑑𝑡
𝑑𝑀
 𝑑𝑉 = −𝑉𝑒 𝑀 (17)
 Integrate between limits of zero
and V, for a change in mass M0 to M
gives the result
Rocket Equation

𝑉 𝑀 𝑑𝑀

0
𝑑𝑉 = −𝑉𝑒 𝑀0 𝑀
(18)
𝑀
 𝑉= −𝑉𝑒 𝑙𝑜𝑔𝑒 (19)
𝑀0
𝑀
 𝑉= 𝑉𝑒 𝑙𝑜𝑔𝑒 𝑀0 (20)
 This is Tsiolkovsk’s Rocket Equation
 The rocket equation shows that the
final speed depends upon only two
numbers
• The final mass ratio
• The exhaust velocity
It does not depend on the thrust; size of engine; time of burn
Exit velocity depends on the fuel

𝑚
 Gunpowder 𝑉𝑒 ≈ 2000
𝑠
𝑚
 Liquid fuel 𝑉𝑒 ≈ 4500
𝑠
𝑉𝑒ℎ𝑖𝑐𝑙𝑒 𝑚𝑎𝑠𝑠 + 𝑝𝑟𝑜𝑝𝑒𝑙𝑙𝑎𝑛𝑡 𝑚𝑎𝑠𝑠 𝑀0
 Mass ratio = = (21)
𝑉𝑒ℎ𝑖𝑐𝑙𝑒 𝑚𝑎𝑠𝑠 𝑀
𝑀0
 = 20 𝑖𝑚𝑝𝑙𝑖𝑒𝑠 95% 𝑜𝑓 𝑡ℎ𝑒 𝑖𝑛𝑖𝑡𝑖𝑎𝑙 𝑚𝑎𝑠𝑠 𝑖𝑠 𝑓𝑢𝑒𝑙 (22)
𝑀𝑉
A rocket can travel faster than its
exhaust speed
 A rocket can travel faster than its exhaust speed Ve
 This appears to be counter intuitive if we think of the exhaust as
pushing against something. But this is not the case
 All the action and reaction takes place inside the rocket where an
accelerating force is being developed against the walls of the
combustion chamber and the inside of the nozzle
 A rocket will exceed its exhaust speed when
𝑀0
 𝑙𝑜𝑔𝑒 =1 (25)
𝑀
𝑀0
 ie = e = 2.718 (26)
𝑀
Diminishing returns

 It is clear that increasing the mass ratio, that is: increasing the mass
of fuel leads to diminishing returns
 For Ve = 1000 m/s, V ⇒ 3000 m/s
 A higher mass ratio will produce a higher velocity but only with a diminishing
return
 To escape the earth’s gravitational field a velocity of around 11km/s is
required.
 This can only be achieved with a high exhaust velocity and a large mass ratio
Gravity loss

 Our main result neglects the so called “gravity loss” that is the work
done against gravity. If this were included
𝑀0
 𝑉 = 𝑉𝑒 𝑙𝑜𝑔𝑒 −𝑔𝑡 (27)
𝑀
𝑀0 𝑀0 𝑀
 𝑉 = 𝑉𝑒 𝑙𝑜𝑔𝑒 −𝑔 1− (28)
𝑀 𝑚 𝑀0
𝑀0 𝑀
 𝑔 1− can account for 1200 m/s
𝑚 𝑀0
Multi stage Rockets

 As previously demonstrated a velocity of about 11 km/s is required

to achieve escape from earth’s gravity
 A velocity of 8 km/s is required to achieve a circular orbit
 For a single stage rocket, with modern fuel Ve ~ 4 km/s
 This implies a mass ratio of:
 ~ 16 to achieve escape velocity 15/16th fuel ~94%
 ~ 7.4 to achieve orbit 6.4/ 7.4 fuel ~ 86%
 Although the latter is currently possible, the former, ie escape
velocity can only be achieved by multi stage rockets
Multi-stage rockets

 For a single stage rocket

𝑀0 𝑀𝑠 +𝑀𝐹 +𝑀𝑃
 𝑅0 = =
𝑀 𝑀𝑆 +𝑀𝑃

 MF = fuel mass
 MP = payload mass
 MS = structural mass (depends on design – engines, pumps, fuel
tanks, control systems)
 In general, we mayexpect the structural mass is kept to a minimum
and is a constant proportion of the fuel mass for stages using the
same fuel.
Multi-stage rockets
Two-stage rocket
 This rocket is divided into two stages
 The first rocket stage is ignited and burns until all its fuel is exhausted, this
gives the whole stack a velocity defined by the rocket equation, with
the mass ratio of:
𝑀0 𝑀𝑠 +𝑀𝐹 +𝑀𝑃 𝐵𝑒𝑓𝑜𝑟𝑒
 𝑅1 = = 𝑀
𝑀 𝑀𝑆 + 𝐹 +𝑀𝑃 𝐴𝑓𝑡𝑒𝑟
2
 The first stage burns out, is dropped off and the 2nd stage is ignited. It
then gains additional velocity defined again by the rocket equation
with mass ratio
1 1
𝑀𝑠 + 𝑀𝐹 +𝑀𝑃 𝐵𝑒𝑓𝑜𝑟𝑒
 𝑅2 = 2
1
2
𝑀 +𝑀𝑃 𝐴𝑓𝑡𝑒𝑟
2 𝑆
 The second stage begins its burn with the payload, half the structural
mass and half the fuel mass and ends with half the structural mass and
the payload
 The final velocity is the sum of the two velocity increments
Single stage and two stage rocket
compared
 So to compare the performance of single stage and two stage
rockets we need to calculate:
 𝑉0 = 𝑉𝑒 𝑙𝑜𝑔𝑒 𝑅0
 𝑉 = 𝑉𝑒 𝑙𝑜𝑔𝑒 𝑅1 + 𝑉𝑒 𝑙𝑜𝑔𝑒 𝑅2
 For example:
 Total mass of 100 tonnes; Payload of 1tonne
 Ve = 2.7 × 103 m/s
 𝑀𝑆 = 10% 𝑜𝑓 𝑓𝑢𝑒𝑙 𝑚𝑎𝑠𝑠
 Therefore MF= 90 tonnes; MS= 9 tonnes; MP = 1 tonne
Single stage and two stage rocket
compared
9+90+1
 𝑉0 = 2700 𝑙𝑜𝑔𝑒 = 6217 m/s single stage final velocity
9+1

 Now divide the rocket into two smaller ones, each with half the fuel
and the structural mass shared equally
9+90+1
 𝑉1 = 2700 𝑙𝑜𝑔𝑒 = 1614 m/s
9+45+1
4.5+45+1
 𝑉2 = 2700 𝑙𝑜𝑔𝑒 = 5986 m/s
4.5+1

 Total velocity increment = 𝑉1 + 𝑉2 = 7600 m/s

Three stage rocket

90+9+1
 𝑅1 = = 1.4286; 𝑉1 = 2700 𝑙𝑜𝑔𝑒 𝑅1 = 963 𝑚/𝑠
60+9+1
60+6+1
 𝑅2 = = 1.8108; 𝑉2 = 2700 𝑙𝑜𝑔𝑒 𝑅2 = 1603 𝑚/𝑠
30+6+1
30+3+1
 𝑅3 = = 8.5; 𝑉3 = 2700 𝑙𝑜𝑔𝑒 𝑅3 = 5778 𝑚/𝑠
3+1
 Total velocity increment = 𝑉1 + 𝑉2 + 𝑉3 = 963+1603+5778 = 8344 m/s
Multi stage rocket summary
comparison
 Single stage Velocity increment = 6217 m/s

 Two stage Velocity increment = 7600 m/s

 Three stage Velocity increment = 8344 m/s

End