The space shuttle has a total mass of 2.0 x 10^6 kg. At liftoff, the engines exert a total thrust of 35 x 10^6 N. Calculate the shuttle's weight. Calculate the shuttle's acceleration at liftoff. If the acceleration averages 13 m/s/s over the first 10 minutes, what velocity does it attain?
The weight is 2.0x106 x 2.205 = 4,410,000 lbs. (actually 4,458,000 with a 25,000 lb. payload)
Liftoff thrust is 35x10^6N .102 x 2.205 = 7,871,850 lbs. (actually 7,245,000 lbs.)
Net liftoff acceleration
a = 7,871,850(32.2)/4,410,000 - 32.2 = 25.27fps^2.
Velocity attained after 10 minutes at an average acceleration of 13x3.281 = 42.653fps is 42.653(10)60 = 25,592 fps. = ~17,450 mph.
The following might be of some interest to you.
1/23/02
<< How much thrust at liftoff does the space shuttle produce?>>
The Space Shuttle liftoff thrust is a total of 7,245,000 pounds. This is made up of 3,060,000 pounds from each of two solid rocket boosters (SRB's) and 375,000 pounds from each of the three SSME (SPace Shuttle Main Engines). As you have probably seen in launches on TV, the SSME's ignite first and when Launch Control is certain that they are performing to specification values, the two SRB's are ignited and the hold down clamps released. The SRB's burn for about two minutes after which they are separated from the SSET (Space Shuttle External Tank). The SSME's continue to burn for an additional 6 minutes-34 seconds for a total of 8min-34 sec of burn time.
The liftoff weight of the Shuttle with a 25,000 pound payload on board is ~ 4, 458,000 pounds. This gives the vehicle a liftoff thrust to weight ratio of 1.625 and an initial acceleration of 20.13 feet per second or 13.72 MPH. At ~6.5 seconds after liftoff, the Shuttle has cleared the tower. At ~7.3 seconds it initiates its pitchover maneuver to acquire its proper flight path. At ~59 seconds it passes through the point of maximum dynamic pressure on the vehicle. At 2.00 minutes, at nominally 163,000 feet and 25 nm downrange, the Solid Rocket Boosters burnout and are jettisoned at 2.1 minutes, and the Orbiter and External Tank continue their flight with the thrust from the three Orbiter engines which have now reached their maximum thrust of ~470,000 pounds each. At approximately 8 minutes-34 seconds, 57nm altitude and 761 nm downrange, the main Orbiter engines cutoff at a velocity of approximately 25,725 fps(17,536 MPH). At about 8 minutes-52 seconds, the External Tank is jettisoned and the Orbiter is on its own now. At the appropriate time, the Orbiter fires its Orbital Maneuvering Engines to increase the velocity of the Orbiter by 342 fps which places the Orbiter in an elliptical transfer orbit to its 250 nm altitude destination. Approximately 45 minutes later, the Orbiter reaches its 250 nm final altitude with a velocity of 24,706 fps where the Orbital Maneuvering Engines fire again to increase the vehicle velocity by 338 fps to achieve the required circular orbital velocity of 25,044 fps(17,072 MPH). The Orbiter now remains in this orbit with an orbital period of ~90 minutes, until the deorbit maneuvers are executed for reentry.
Some of the performance numbers I have quoted you are from data that is several years old. Recent performance enhancements of the SRB's and the SSME's may have altered the numbers quoted but they are close enough to give you a broad idea as to the Shuttle's performance.
I hope this gives you a clearer picture of the Space Shuttle liftoff and launch profile.
References:
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 calculate the space shuttle's weight, we can use the equation:
Weight = mass x gravitational acceleration
where the gravitational acceleration is approximately 9.8 m/s^2 (on Earth).
Weight = (2.0 x 10^6 kg) x 9.8 m/s^2
Weight = 19.6 x 10^6 N
Therefore, the shuttle's weight is 19.6 x 10^6 N.
To calculate the shuttle's acceleration at liftoff, we can use Newton's second law of motion, which states that force is equal to mass times acceleration:
Force = mass x acceleration
Rearranging the equation, we can solve for acceleration:
acceleration = Force / mass
acceleration = (35 x 10^6 N) / (2.0 x 10^6 kg)
acceleration = 17.5 m/s^2
Therefore, the shuttle's acceleration at liftoff is 17.5 m/s^2.
To calculate the velocity the shuttle attains, we can use the equation:
velocity = initial velocity + (acceleration x time)
In this case, the initial velocity is assumed to be 0 m/s since the shuttle is starting from rest.
velocity = 0 m/s + (13 m/s^2 x 10 minutes x 60 seconds/minute)
velocity = 0 m/s + (13 m/s^2 x 600 seconds)
velocity = 0 m/s + 7800 m/s
velocity = 7800 m/s
Therefore, the shuttle attains a velocity of 7800 m/s over the first 10 minutes.
To calculate the shuttle's weight, we can use the equation:
Weight = mass × gravitational acceleration
where the mass of the shuttle is given as 2.0 x 10^6 kg and the gravitational acceleration is approximately 9.8 m/s^2.
Weight = (2.0 x 10^6 kg) × (9.8 m/s^2) = 19.6 x 10^6 N
Therefore, the weight of the space shuttle is 19.6 x 10^6 N.
To calculate the shuttle's acceleration at liftoff, we can use Newton's second law of motion:
Force = mass × acceleration
The total force exerted by the engines is given as 35 x 10^6 N, and we already know the mass of the shuttle as 2.0 x 10^6 kg.
35 x 10^6 N = (2.0 x 10^6 kg) × acceleration
Dividing both sides of the equation by 2.0 x 10^6 kg, we can find the acceleration:
acceleration = (35 x 10^6 N) / (2.0 x 10^6 kg) = 17.5 m/s^2
Therefore, the acceleration of the space shuttle at liftoff is 17.5 m/s^2.
To calculate the velocity attained by the shuttle, we can use the equation:
Final velocity = Initial velocity + (acceleration × time)
Given that the acceleration is 13 m/s^2, and the time is given as 10 minutes, we need to convert the time to seconds. There are 60 seconds in one minute.
10 minutes = 10 × 60 seconds = 600 seconds
Plugging these values into the equation, we get:
Final velocity = 0 + (13 m/s^2 × 600 s) = 7800 m/s
Therefore, the velocity attained by the space shuttle is 7800 m/s.