The space shuttle has a mass of 2.0 × 106 kg. At lift-off the engines generate an upward
force of 30 × 106 N.
a. What is the weight of the shuttle?
b. What is the acceleration of the shuttle when launched?
c. The average acceleration of the shuttle during its 10 minute launch is 13 m/s2. What
velocity does it attain?
What is the acceleration of the Space Shuttle during its ascent to orbit?
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 answer these questions, we can use Newton's second law of motion, which states that the force acting on an object is equal to the mass of the object multiplied by its acceleration (F = ma).
a. To find the weight of the shuttle, we need to find the force due to gravity acting on it. Weight is defined as the force experienced by an object due to gravity. So, we can calculate the weight using the equation:
Weight = mass × acceleration due to gravity
Acceleration due to gravity is approximately 9.8 m/s^2. Therefore, the weight of the shuttle is:
Weight = 2.0 × 10^6 kg × 9.8 m/s^2 = 19.6 × 10^6 N
b. The upward force generated by the engines is equal to the net force acting on the shuttle, as there are no other forces stated in the problem. So, we can use the equation:
Net force = mass × acceleration
Given that the net force is 30 × 10^6 N and the mass is 2.0 × 10^6 kg, we can rearrange the equation to solve for acceleration:
acceleration = Net force / mass
acceleration = (30 × 10^6 N) / (2.0 × 10^6 kg)
acceleration = 15 m/s^2
Therefore, the acceleration of the shuttle during launch is 15 m/s^2.
c. We can use the following equations of linear motion to find the final velocity of the shuttle when its average acceleration is 13 m/s^2:
v^2 = u^2 + 2as
where v is the final velocity, u is the initial velocity (which is usually assumed to be 0 in these types of problems), a is the average acceleration, and s is the displacement or distance traveled.
Since the average acceleration is given as 13 m/s^2 and the displacement is not given, we cannot directly solve for the final velocity. However, we can use the fact that the shuttle's initial velocity is zero.
v^2 = 0^2 + 2 × 13 m/s^2 × s
Simplifying the equation, we have:
v^2 = 26s m^2/s^2
Taking the square root of both sides, we get:
v = sqrt(26s) m/s
Therefore, the velocity attained by the shuttle depends on the distance it has traveled during its 10-minute launch. Without knowing the displacement, we cannot determine the exact velocity attained by the shuttle.
To answer these questions, we need to apply Newton's second law of motion, which states that the force acting on an object is equal to the mass of the object multiplied by its acceleration (F = m * a). We will also use the formula for weight, which is the force of gravity acting on an object given by the formula W = m * g, where W is the weight, m is the mass, and g is the acceleration due to gravity.
a. To find the weight of the shuttle, we can use the formula W = m * g. Weight is the force of gravity acting on an object, so we need to multiply the mass of the shuttle (2.0 × 10^6 kg) by the acceleration due to gravity (g = 9.8 m/s^2):
W = (2.0 × 10^6 kg) * (9.8 m/s^2)
W = 19.6 × 10^6 N
Therefore, the weight of the shuttle is 19.6 × 10^6 N.
b. To find the acceleration of the shuttle when launched, we can use Newton's second law of motion, F = m * a. In this case, the force acting on the shuttle is the upward force generated by the engines (30 × 10^6 N), and the mass of the shuttle is given (2.0 × 10^6 kg). Solving for acceleration:
F = m * a
30 × 10^6 N = (2.0 × 10^6 kg) * a
a = (30 × 10^6 N) / (2.0 × 10^6 kg)
a = 15 m/s^2
Therefore, the acceleration of the shuttle when launched is 15 m/s^2.
c. To find the velocity that the shuttle attains during its 10 minute launch, we can use the average acceleration and the time taken. We can use the formula v = u + a * t, where v is the final velocity, u is the initial velocity (which we assume to be zero in this case), a is the average acceleration, and t is the time:
v = u + a * t
v = 0 + (13 m/s^2) * (10 min * 60 s/min)
v = 0 + (13 m/s^2) * (600 s)
v = 0 + 7800 m/s
Therefore, the velocity that the shuttle attains during its 10-minute launch is 7800 m/s.
See how far you gan get using
Weight = M*g and
F = M*a
You need to make mor of an effort to get help here.