Rocketry: The Gateway To Understanding Space Technology

Few people living today are aware the Space Age began on October 4, 1957 with the launch of the Russian Sputnik satellite. (See newspaper clipping). Sputnik was a resounding event, and had the impact of a train collision on the American educational system and citizens'  consciousness. As the attached graphic shows, Sputnik was a 184 lb. satellite that orbited the Earth every 96 minutes at an altitude of about 900 km (600 miles).

I was 11 at the time, and had much more interest earlier on that particular day (a Friday) in how my Milwaukee Braves would fare in Game 3 (the next day, Saturday) of the World Series against the New York Yankees - than in any space exploits.

Some days later, we heard the first radio 'beeps' from Sputnik. The signal had been picked up by an RCA receiving station at Riverhead, New York and relayed to the NBC studios in Manhattan, when most of us alive then in the U.S. heard it over the Huntley-Brinkley Report.

How did the nation, including  politicos, react? (This was during the Eisenhower administration which alas, for most of the population today constitutes ancient history!) According to Paul Dickson, author of Sputnik: The Shock of the Century, p. 23:

"Polls taken within days of the launch showed that Americans were concerned - so concerned that almost every person surveyed was willing to see the national debt limit raised and forgo a proposed tax cut in order to get the United States moving in space"

The Sputnik moment triggered a space competition that would ultimately see the United States reaching the Moon before the Russians. It disclosed a collective willingness to sacrifice financially, via raising the debt limit,  and rescinding a proposed tax cut to achieve it.  And with very good reasons. By the time of Sputnik's launch in October, 1957, the Russians were producing some 66,000 engineers a year compared to the United States' 22,000. In addition, the key subjects of higher math and physics were almost nowhere to be found in the U.S. secondary school curriculum - nor were there the teachers to teach them. All this had to be factored into the coming expense to get the U.S. on a competitive par with the Soviets. Teacher education and training alone came to over $1 billion by the time of the Apollo 11 lunar landing.

By the early 1960s, physics, astronomy, math as well as the hobby of model rocketry had become a part of many students' lives. Those who might have been mildly interested in psychology or medicine turned instead to mathematics, physics, and rocket engineering. And the memory of Sputnik became the driving impetus leading many of us to want to build our own rockets - with their own payloads.
Space phenomena such as auroras, asteroids, cosmic rays and solar flares - while important -  also often provoke the desire to learn more about space technology. After all, getting a space telescope in orbit above Earth is the optimum way to observe celestial objects, only surpassed by  sending space probes to actually land on asteroids to take samples, e.g.

http://www.esa.int/Our_Activities/Space_Engineering_Technology/Asteroid_Impact_Mission/The_art_of_landing_on_an_asteroid

Thus the study of space often itself begins for many with the study of space vehicles and also construction of rockets. If one then begins by building and launching simple rockets - he or she goes a long way to becoming informed about the physics of space flight overall, as well as energized to learn more about advanced space propulsion systems.

My own rocketry exploits ran in parallel to my astronomy devotion. At the same time I was building my own refracting telescopes to observe celestial phenomena from M13 (globular cluster)  in Hercules to M 8 (Lagoon Nebula) in Sagittarius, I was also constructing my own model rockets to launch publicly - e.g. at Mgsr. Edward Pace High (where the entire school would be let out to watch a launch).  Below is shown a typical 2- stage design I'd use, with dimensions.


My single stage rockets were up to 16-18" in length and generally contained a payload section with parachute, within which a lizard, frog or cricket was often placed - on a comfortable wad of cotton to withstand the g-forces. A tiny side panel was cut out with plastic glued over it to allow a kind of small window.

Rocketeering, of course, also required a thorough study of the related physics principles. What is it that causes that single stage rocket to be thrust upwards? And can one compute the altitude from certain basic parameters? The first question can be answered with respect to the diagram below, and a model rocket launched by Colorado high school rocketeers::

As indicated in the diagram, the rocket’s motion changes when a fraction of its mass (D m) is released in the form of ejected gases. Since the gases acquired their own momentum, the rocket receives a compensating momentum in the opposite direction.  Therefore, the rocket is accelerated as a result of a push from the gases. In free space, or a vacuum, the entire system works independent of the presence of any opposing medium.

Assume at some time t, the momentum of the rocket plus fuel is: (M + D m)v, then at some later time: (t + D t), the rocket ejects some fraction of mass D m, so the rocket’s velocity must increase to (v + D v). By appealing to Newton’s 3^rd law via an application of conservation of momentum,  we may write:

Total initial momentum of the rocket system = Total final momentum of the system

Then we get:

(M + D m)v = M(v + D v) + D m(v – v’)

Where v’ is the velocity with which the fuel is ejected relative to the rocket. The equation can then be simplified to yield:  Mdv = v dm, which may be integrated, viz.:

M ò_v1^v2  dv = v ò_m1^m2  dm

Or, letting m2 = M _f, m1 = M _i and v2 = v _f, v1 = v_i^:

v _f – v_i^:   = v’ ln [M _i / M _f]

Where the left side shows the difference between the final and initial velocities, M _i refers to the initial rocket mass (fuel plus rocket proper) and M _f refers to the final rocket mass with fuel expended.  (In general, for most rockets,    M _i  >> M _f ).

To see how this works, say a model rocket is launched by an amateur group in central Colorado (see image  at top)  of initial  total mass 1.0 kg. They used a Zn S (zinc sulfide) solid fuel engine for which the exhaust gases attained a velocity of 100 ms ^-1 relative to the rocket for 3 seconds.  After this interval, the rocket mass decreased to 0.05 kg.   We can then find the rocket’s acceleration and estimate the altitude assuming zero air drag and a near –vertical launch angle.

We have: M _i = 1.0 kg, M _f  = 0.05 kg

 Therefore, the difference between initial and final velocities is:

v _f – v_i^:    =  v’ ln [M _i  / M _f]

v _f – v_i^:     = (100 m/s) ln [1.0 kg/ 0.05kg]

v _f – v_i^:    = (100 m/s) ln(20) = (100 m/s) (3) = 300 m/s

The altitude can be estimated by using the kinematic eqn.

s = ½ at²

where s is the vertical displacement for an acceleration a, over time t.

s = ½ (300 ms^-2) (3)²  

s = 450 m  or  1 485 ft.

For the rocket itself, the key parameter is usually the thrust, on which the rocket's velocity will depend. The graphic below gives an idea how the average thrust of a given rocket engine is obtained:

Total Impulse, as indicated above, is extremely critical and a property of the solid rocket engine one uses - each of which has a specified value in model rocketry with units in Newton-seconds or pound -seconds. The 16 0z. in the factor on the extreme right is because this is generally regarded as the upper limit for the model rocket. Above this and there is too high a danger of instability - mainly that the design will not allow the center of gravity to be ahead of the center of pressure as it needs to

be.

Let's say the model rocketeer wishes to compute the velocity v2 from the equation above, using the units (feet-pounds) as indicated.  Let the total impulse of the engine be 10 pound-seconds, and the burn time of the engine be 2 seconds. Then the force F or thrust is:

(10 lb-sec)/ 2 sec  x (16 oz/ 1 lb) = 80 oz.

Then the velocity v2 = (80 oz/ 10 oz   - 1)  32 ft/ sec/sec (2 sec)

Assuming the average weight of the rocket at lift off is 10 oz.

v2 = (8 - 1) (64 ft/ sec)  =  7( 64 ft/ sec)=  448 ft/ sec or about 135 m/sec

Which is a reasonable value.

Many model rocket enthusiasts  of course, go on to full amateur rocketry which entails the construction of large metallic tube rockets capable of going thousands of feet in altitude. Different safety rules apply to these "full fledged" rockets, and their design and construction is also much more complex because now instead of buying a ready -made engine as in the case of the model rocket, you are designing your own. This means you need to get the design of the rocket engine nozzle very precise.  A typical design layout is shown below:

The computation for the effective thrust coefficient C_F shown at the bottom is related to the physical specs including the atmospheric pressure,  P_a    , the chamber pressure,  P_e  , the ratio of specific heats (C_p /C_v )  for combustion products k .

The combustion chamber cross-sectional area (A_e / A _t)  is the ratio of nozzle exhaust area to throat area) given by:


(A_e / A _t)   =  (M _t / M _c)  [(1 + M _c ² (k -1 )/2/ 1 + M _t ² (k -1 )/2] ^k+1/2(k+1) 


Where M _t is the Mach number of the gases in the throat, and M _c is the Mach number at the end of the cylindrical section.

Generally, the diameter of the nozzle throat needs to be about one third the diameter of the combustion chamber, while the angle of the converging section of the nozzle needs to be approximately 30 degrees, and the angle for the diverging section 15 degrees. The failure to properly design the nozzle is probably responsible for most amateur rocket misfires.


Of course, those of us doing any rocketry back in the early 60s didn't have access to computers or even scientific calculators like the spoiled students today. Nope, we used the one reliable instrument we had, the slide rule. For me, the good ol' Mannheim type as shown below:

Today, most of these instruments are relegated to mathematical displays in certain museums, but I still have mine and even check it out every now and then, computing a tangent, cube root or ...a rocket's thrust, velocity.

In more than a few ways, today's space exploits and technological developments - including many citizens' continued interest in space- was incepted by the launch of Sputnik - and more critically, the proactive response to it.

For those who'd like to learn a lot more, I provide the link to MIT's Astrodynamics course below:

https://ocw.mit.edu/courses/aeronautics-and-astronautics/16-346-astrodynamics-fall-2008/

Enjoy!

 

Coronal Mass Ejections and Collisionless Shocks (2) - With Problems

In the previous instalment I noted that the dissipation associated with collisionless plasma can be attributed to Landau damping and this may be the most common form to excite a shock associated with CMEs. But even more germane is the two stream instability which can be thought of as the inverse of Landau damping, where a greater number of particles that move slower than the wave phase velocity v_ph (as compared with those that move faster), leads to an energy transfer from the wave to the particles.  In the case of the two stream instability, when an electron stream is injected to the plasma, the particle's velocity distribution function has a "bump" on its tail:


In the region where the slope is positive (¶f(v)  / ¶ v > 0) there is a greaterI number of faster than slower particles so a greater amount of energy is transferred from particles to associated (e.g. Alfven) waves.  Since f _eb contains more fast than slow particles a wave is excited, and there is inverse Landau damping such that plasma oscillations with v_ph in the positive gradient region are unstable. We expect  v_ph  to occur where the gradient:(¶f(v)/ ¶ v) ê_{wr/ k}  = maximum. Instability can result provided: 

w_e  ²   >   k²v²  or  êkv ê  <   w_e  ^.

Resonant electrons (at v_ph  »  w_e/ k)   are the first to be affected by the local wave-particle interactions and have distributions altered by the wave electric field, E1, such that the total energy balance:

S E_tot = ½ E1 _w + E1 _k    =  ½[ ½ e_o | E1|²]  +  ½ r | ṽ|²         

For which the wave energy (E1 _w) is balanced against the kinetic (particle) energy (E1 _k ). Thus, for S E_tot = const. then as ṽ decreases, ½ e_o | E1 |²]   increases. In Landau damping the exact opposite occurs, so as the decrement associated with ¶f(v)/ ¶ v decreases, the wave amplitude (and  E1 _w) decreases, particle kinetic energy increases- i.e. energy is fed to the particles at expense of waves.

For reference, the longest known radio signatures for coronal shocks are the Type II radio bursts. These are narrow band (Dn/n ~ 0.1)  ) radio emissions excited at the local plasma frequency for which the drift rate :

dn/dt ~ ( - 0.2 MHz/s)

Where the negative sign indicates a drift toward lower frequencies. This is as the shock propagates outwards through the corona. Derived measurements disclose a radial velocity on the order of 1000 km/s. In 60 per cent of cases there is a clear harmonic structure with a frequency of occurrence of once every 100 hours. The Type II bursts are accompanied by weak polarization. The original excitation is usually by a fast mode MHD shock though the mechanisms previously considered, i.e. via Landau damping and two stream instability, must also be considered.

If we know the ambient free electron density associated with the event, it is possible to estimate the pure plasma  frequency (n)  using:

1)      n   =   9000  Ö N    or    2) n   =   9  Ö N   

      

In the above (1) is used to get MHz if the units for N are in c.g.s. units, and (2) for S.I. units. Bear in mind:w_e   =   2p n.

It is also useful to note that radio waves cannot propagate through a medium in which the plasma frequency  is greater than the radio frequency.

Problems:

1)For a hydrogen plasma, a lab generated shock of Mach number M= 80  with the temperature at T =   10 ⁵    K is found to have an ion gyroradius of 0.22 cm.  If    Ω _e  =  1.9 x 10 ⁷  /s  what would be the magnitude of the associated magnetic field (B)?

2)Find the associated plasma frequency and use it to obtain the Debye length and plasma parameter if n is a value typical of the solar corona  (10 ¹⁶ /m³). Find also the frequency n of the associated radio emission.

3)If the shock speed is u =  10 ⁷  cm/ sec, estimate the wave number k _w.  

4)Hence, or otherwise, use the equation below to obtain the ion velocity  v _o   if the  frequency w  = 10 ¹⁰ / s

(Note: m is the electron mass, and M =  m_i, the ion mass.)

5)Based on all the above information, estimate the maximum magnetic field B _m using the shock equation.

6) The frequency of a critical layer as a function of electron density is given by:

n   =  e/2p  Ö( N/ ε_o m _e )

Show that this can be reduced to the simplified form:

n   =   9  Ö N

(Hint: ε_o  =  8.85 x  10 ^-12  farad m^-1 )