How would I approach this problem? What formula would I use to answer each part? Thank you in advance for any guidance...
As part of the launch of the space shuttle, solid-rocket boosters (SRBs) first take it to an altitude of 45.0 km while they accelerate it to a speed of 4973 km/hr. The SRBs then separate from the shuttle and fall back to earth. Assume that the acceleration due to gravity is the same throughout their fall, and neglect air resistance.
1) How fast would the SRBs be moving (in km/hr) when they reached the earth's surface if they were traveling upward when they released the space shuttle?
2) How long (in mins.) after they began to fall would it take the SRBs to reach the ground?
3) While the space shuttle is being taken up, starting from rest, what is its acceleration (and the acceleration of the SRBs), in m/s^2?
4) While the space shuttle is being taken up, starting from rest, what is its acceleration (and the acceleration of the SRBs), in g's?
5) How long does it take to reach the 45 km altitude?
The following might provide you with some insight into what you seek.
The nature of the Space Shuttle's acceleration during its ascent to orbit is somewhat difficult to define in that it varies continuously throughout the flight. For the same reason, it is also very misleading to quote an average acceleration. We know that acceleration is the rate of change of velocity with time, expressed mathematically,
a = (V2 - V1)/t, with a = acceleration in ft/sec/sec, V2 = final velocity in ft/sec, and V1 = initial velocity. You have no doubt seen the familiar equation from physics, F = ma where F = force in pounds, m = mass(weight in pounds divided by the acceleration due to gravity in ft/sec/sec), and a = acceleration. Solving this for a yields a = F/m. We therefore have two means by which we might compute the shuttle's acceleration.
The first equation offers a quick and simplistic means of computing the "average" acceleration of the shuttle over its 8 min.-38sec. powered flight. Knowing V2, the final velocity of 25,725 fps, V1, the initial velocity of zero, and the time = 518 seconds, we can solve for a, such that a = [V2 - V1]/t = [25725 - 0]fps/518sec = 49.66 ft/sec/sec or ~34 MPH/sec. "average" acceleration over the entire flight. It must be stressed that this computed acceleration is an "average" acceleration only and, as such, does not reflect the true acceleration profile of the vehicle throughout its flight. In fact, the shuttle actually does have this specific acceleration at two points in its trajectory.
The use of the term "average" acceleration can often be misleading, for two primary reasons: 1) The average acceleration does not reflect the actual variations of thrust and weight throughout the flight and 2) it implies a straight line uniform variation of both velocity and acceleration throughout the flight, which we know is untrue. Lets see what a more realistic view of the acceleration profile might look like.
In general, for a rocket moving under the influence of a constant thrust, the rocket mass is continuously decreasing as the propellants are consumed and, therefore, the rocket's acceleration is continuously increasing throughout the flight. Simultaneously, the forward acceleration is constantly being reduced by the pull of gravity on the rocket in proportion to the cosine of the flight path angle as measured between the direction of flight and the local horizontal. The effect of gravity on the rocket's vertical acceleration gets smaller with increasing altitude due to the continuous pitching over of the rocket as it approaches its target altitude. Ultimately, as the rocket reaches the required burnout altitude, velocity, and flight direction, gravity has no impact on the forward acceleration of the rocket.
Now, for our shuttle, the picture is complicated a bit more. First off, the shuttle Solid Rocket Boosters (SRB's) are designed to provide a varying thrust profile. A typical solid rocket motor ignites at maximum thrust, maintains this thrust for some time period, and then ramps down to a final burnout thrust of perhaps 70-80 percent of its initial thrust. The shuttle SRB thrust profile is roughly as follows: The initial, single motor, vacuum thrust at ignition is ~3,300,000 pounds. This continues for about 50 seconds, at which point the thrust drops to ~2,310,000 pounds over the next 5 seconds or so. The reason for the drop off is to assist in limiting the g level on the vehicle to 3g's and to reduce the dynamic pressure on the vehicle. From this point on, the thrust declines almost linearly, to a value of ~1,342,000 pounds at ~112 seconds burntime and then drops off to zero over the next 11+ seconds. Now, another subtle factor that modifies the initial thrust is the reduction in thrust caused by atmospheric pressure at liftoff. (Thrust = At(Pc - Pa)Cf where At = the throat (chamber exit) area of the motor, Pc = chamber pressure, Pa = ambient pressure, and Cf = exhaust nozzle coefficient which reflects the increase in thrust caused by the motor nozzle.) So in reality, our initial sea level ignition thrust is
~3,050,000 pounds.
Our picture is further complicated by the fact that the SSME engines on the orbiter also vary their thrust throughout the flight. The SSME vacuum design rated thrust (DRT) is 470,000 pounds. It has the ability to throttle this thrust down to 65% of DRT, or 305,500 pounds. (This throttle down is used as the shuttle passes through the maximum dynamic pressure region shortly after liftoff and again, toward the end of the orbiter's flight to restrict the g level on the orbiter to 3 g's.) It also has the capability of throttling the thrust up to a maximum of 109% of DRT or 512,300 pounds. As with the SRB's, the sea level DRT of the engine is reduced by atmospheric pressure to 375,000 pounds. The actual thrust at liftoff is 104% of DRT at sea level or 390,000 pounds.
Knowing the thrust variations of the SSME's and the SRB's and the corresponding propellant consumption rates, it is now possible to visualize how much variation there actually is in the acceleration profile of the shuttle during its ascent to orbit. Lets see what we can derive.
A first order computation can be derived as follows. The initial liftoff acceleration may be accurately defined as a = 7,270,000 lbs/[4,468,000 lbs/32.2fps^2] = 52.4 ft/sec/sec minus the full 32.2 ft/sec/sec decelerating effect of gravity at its maximum, yielding a = ~20.2 ft/sec/sec. or ~13.8 MPH/sec. Similarly, its acceleration just before orbiter burnout is a = 916,500lbs/[303,920 lbs/32.2fps^2] = 97 ft/sec/sec or ~66 MPH/sec. Now, using just these two extremes of the flight, we can derive an apparent "average" acceleration over the flight of (20 + 97)/2 = ~58.5 ft/sec/sec, considerably different, but how believable. Since we only used the accelerations at the two extremes of the flight, the averaging does not take into account the "designed in" variations that take place in
between. So at best, this number is suspect.
A somewhat more accurate method of deriving a "phase averaged" acceleration is to determine, and combine, the time averaged accelerations over each major phase of the flight. For instance, the net acceleration at liftoff is ~20 fps^2 as determined above. Just before SRB burnout, the forward acceleration due to thrust is ~70 fps^2, minus an approximate 19 fps^2 component of gravity at that point, for a net forward acceleration of 51 fps^2. Thus the average acceleration of this 120 second SRB action phase is ~35.5 fps^2. After SRB jettisoning, we have an ~30 fps^2 forward acceleration due to thrust, minus the same gravity component of 19 fps^2, for a net forward acceleration of ~11 fps^2. At orbiter burnout, we have the same ~97 fps^2 as defined earlier. Thus the average acceleration of the final 398 second orbiter burn phase is ~54 fps^2. Now time averaging these boundry accelerations over the time periods during which they act, [(35.5(120)) + (54(398))]/518, results in an overall average acceleration of ~49.7 fps^2, exactly what we derived in the first place. Now lets see if by using these averages we can arrive at our final velocity of 25,725 fps. For the first 120 seconds we achieve a velocity increase of 35.5 fps^2 x 120 sec. = ~4260 fps. For the final 398 seconds we add a velocity increase of 54 x 398 = ~21,492 fps for a grand total of 25,752 fps. What do you know, pretty close to the required and actual burnout velocity.
Another way of deriving time averaged numbers is as follows. The average of the initial shuttle liftoff acceleration and the SRB burnout acceleration, due to thrust alone, is (52 + 70)/2 = 61 fps^2. Subtracting from this the time averaged gravity component of ~25.5 fps^2 yields an average SRB phase forward acceleration of 35.5 fps^2. The average of the post SRB acceleration and the orbiter burnout acceleration, due to thrust alone, is
(30 + 97)/2 = 63.5 fps^2. Subtracting the time averaged gravity component of ~9.5 fps^2 yields an average post SRB phase forward acceleration of ~54 fps^2. Not all that surprisingly, the same numbers we derived in the previous method.
Clearly, the finer we break down the time periods, the more accurate "average" acceleration calculations become. Obviously if one wishes to know, and understand, the actual acceleration profile of the shuttle during ascent, it will be necessary to examine in very fine detail, the actual weight and thrust variations vs time for the entire 518 seconds of flight. In conclusion, it must be emphasized, that the quoting of average accelerations for steadily decreasing masses under the influence of a constantly varying forces, is dangerous, often misleading, and should be done with great caution.
Ref: 1--Aerofax Datagraph # 5, Rockwell International Space Shuttle by Dennis Jenkins,Aerofax,Inc.,Arlington,Texas,1989.
2--The Space Shuttle Operators Manual by K.M. Joels and G.P. Kennedy, Ballantine Books, NY, 1988.
3--Miscellaneous reports, specifications, and personal computations.
To approach this problem, you can use the equations of motion and the principles of free fall. Let's break down each part of the problem and the formulas to use:
1) To determine the final speed of the SRBs when they reach the Earth's surface, you can use the equation of motion for uniform acceleration: v^2 = u^2 + 2as, where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the distance traveled. In this case, the SRBs are traveling upward, so the initial velocity would be negative (since we consider upward as the positive direction).
2) To calculate the time it takes for the SRBs to reach the ground, you can use the formula s = ut + (1/2)at^2, where s is the distance traveled, u is the initial velocity, t is the time, and a is the acceleration. You'll need to rearrange the equation to solve for time.
3) To calculate the acceleration of the SRBs (and the space shuttle) while they are being taken up, you can use the equation a = (v - u) / t, where v is the final velocity, u is the initial velocity, and t is the time taken. In this case, the initial velocity is 0 m/s since it starts from rest.
4) To express the acceleration in g's, you need to divide the acceleration by the acceleration due to gravity on Earth. 1 g is equal to 9.8 m/s^2, so dividing the acceleration by 9.8 will give you the acceleration in g's.
5) To find the time it takes to reach the 45 km altitude, you can use the kinematic equation s = ut + (1/2)at^2, where s is the distance traveled, u is the initial velocity, t is the time, and a is the acceleration. In this case, the space shuttle starts from rest, so the initial velocity is 0 m/s.
Now that you understand the formulas to use for each part, you can plug in the given values and calculate the answers.