The space shuttle has a weight 2.00 x 10^7 N. The shuttle is accelerated upward at a rate of 5.50 m/s^2. What is the mass of the shuttle? What is the thrust (force) of the combined engines? What is the magnitude of the net force applied to the space shuttle?
thrust= mg+ ma
solve for mass from weight=mg
solve for thrust with the top equation.
Net thrust=ma
The following might be of some help to you.
1/23/02
What is the final burnout velocity of the Space Shuttle?
The final burnout velocity of the Space Shuttle, or any rocket for that matter, is a function of the altitude at burnout and the orbit that the rocket is being injected into. For a purely circular orbit the burnout velocity is one value. For an elliptical orbit, with the burnout altitude being the low point in the orbit, the burnout velocity is higher. For instance, if the Space Shuttle Orbiter(SSO) wants to be injected into a 100 nautical mile high circular orbit, it must attain a final velocity at burnout of 25,568.5 feet per second or 17,429 MPH. with its velocity vector(direction) perpendicular to a line from the SSO to the center of the earth. If, on the other hand, it wants to be injected into an elliptical orbit with a low altitude of 100nm and a maximum altitude of 250nm, its burnout velocity must be 26, 067 fps. or17,769 MPH, again with its velocity direction as described before.(There are other ways to reach the final conditions but I won't confuse you with them here) Taking it one step further, if it wants to be injected into a 250 nm high circular orbit, its burnout velocity must be 25,044 fps or 17,072 MPH. You might have noticed here that the higher you go, the smaller the velocity required to maintain the orbit. This holds true for all space vehicles no matter what the altitude. One might conclude here that why not fly all satellites at higher altitudes as the final velocity required is less. The flaw in that reasoning is that the rocket has to expend that much more energy to push the spacecraft up to the higher altitude in the first place, fighting gravity all the way, such that it really takes more energy.
To give you a brief idea of what a rather typical Space Shuttle launch might look like, let me give you the following rough numbers representing a mission to a 250 nm altitude. Just be aware that these numbers are general, vary with each mission, and may be different from todays real numbers due to improvements in the Space Shuttle. They are from some information I developed while working on the Space Shuttle many years ago.
The Gross Liftoff weight of the Shuttle for a specific mission carrying a 25,000 pound payload in the Orbiter is about 4,457,667 pounds. The liftoff thrust is approximately 7,245,000 pounds, (probably more today with the improved solid rockets), with 375,000 pounds coming from each of the Orbiter's engines and 3,060,000 pounds coming from each of the solid rocket boosters. 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 2.06 minutes, at nominally 163,000 feet and 25 nm downrange, the Solid Rocket Boosters burnout and are jettisoned, 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 ~450,000 pounds each. At approximately 8.34 minutes, 57nm altitude and 761 nm downrange, the main Orbiter engines cutoff at a velocity of approximately 25,725 fps(17,536 MPH). At about 8.52 minutes, the External Tank is jettisoned and the Orbiter is on its own now. At a suitable 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 will now remain in this orbit with an orbital period of ~90 minutes, until the deorbit maneuvers are executed for reentry.
The acceleration of the SpaceShuttle is a bit more difficult to define as it varies constantly throughout the flight. You might have heard of 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 in ft/sec/sec. Solving this for a yields a = F/m.
Now, in a rocket with a constant thrust, the mass is continuously decreasing as the propellant is consumed and therefore the acceleration is increasing continuously during the flight. Simultaneously, the acceleration due to thrust is constantly being reduced by the pull of gravity on the rocket and, as the rocket pitches over with increasing altitude to acquire the required burnout attitude, the effect of gravity on its acceleration is reduced. Ultimately, the rocket reaches the desired altitude, required velocity, and necessary direction of flight to sustain its position at a constant altitude.
An extremely simplistic view of the average acceleration of the Shuttle over its 8.34 minute flight can be calculated from the familiar equation of motion Vf = Vo + at where Vf is the final velocity, Vo is the initial velocity, a is the acceleration, and t is time. Solving this equation for a, we get a = [Vf - Vo]/t = [25725 - 0]fps/500sec = 51.45 ft/sec/sec or ~35 MPH/sec. average acceleration over the entire flight. Its initial liftoff acceleration is a = 7,245,000 lbs/[4,457,667 lbs/32.2fps^2] = 52.33 ft/sec/sec minus the full 32.2 ft/sec/sec decelerating effect of gravity at its maximum, yielding a = 20.13 ft/sec/sec. or 13.72 MPH/sec. Similarly, its acceleration just before burnout is a = 1,410,000 lbs/[260000 lbs/32.2fps^2] = 174.6 ft/sec/sec or 119 MPH/sec. Be careful as these are only averages. You must remember that acceleration during the early portion of the flight is affected the most by gravity and at the end of the flight, the effect of gravity on acceleration is almost zero.
I hope this has given a little understanding of the Space Shuttle acceleration picture.
Thanks but that just made me more confused.
To find the mass of the shuttle, we can use Newton's second law of motion, which states that force is equal to the mass of an object multiplied by its acceleration. In this case, the force is the weight of the shuttle, given as 2.00 x 10^7 N, and the acceleration is 5.50 m/s^2.
To find the mass, we rearrange the equation:
Force = mass x acceleration
mass = Force / acceleration
mass = 2.00 x 10^7 N / 5.50 m/s^2
mass ≈ 3.64 x 10^6 kg
Therefore, the mass of the space shuttle is approximately 3.64 x 10^6 kg.
To find the thrust (force) of the combined engines, we need to account for the opposite direction of the force applied by the engines, which is downward. In this case, the thrust will oppose the weight of the shuttle.
Therefore, the thrust force of the combined engines is equal to the weight of the shuttle:
Thrust force = 2.00 x 10^7 N
Finally, to find the magnitude of the net force applied to the space shuttle, we need to consider all the forces acting on it. In this case, there are only two forces: the thrust force of the engines and the weight of the shuttle.
Net force = Thrust force - Weight
Net force = 2.00 x 10^7 N - 2.00 x 10^7 N
Net force = 0 N
The magnitude of the net force applied to the space shuttle is zero. This means that the forces are balanced, resulting in no acceleration in any direction.