Momentum in a space shuttle and thrust

[fogvajarash]

Homework Statement

The space shuttle, with an initial mass M = 2.41 x 10⁶kg, is launched from the surface of the Earth with an initial net acceleration a = 26.1 m/s^2 . The rate of fuel consumption is R = 6.90 x 10³kg/s. The shuttle reaches outer space with a velocity v₀= 4632m/s, and a mass of M₀= 1.45x10⁶kg. How much fuel must be burned after this time to reach a velocity v_f = 5343 m/s

Homework Equations

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The Attempt at a Solution

I've tried first setting up the exhaust velocity of the gases. As we are given the net acceleration, we can say that the net force F is: F = F_t - W (where F_t is thrust force, and it is given by F_t = Rv_ex where v_ex is the exhaust velocity). By clearing this equation, I found that the exhaust velocity was v_ex = 1.254 x 10₄m/s.

For the other part, we will have that momentum is conserved in the following way:

m₀v₀ = m_gv_ex + (m₀ - m_g)v_f

Where m_g and v_ex is the mass of the gas expelled and exhaust velocity. By clearing this equation, I found out that we had the following equation:

(m₀v₀ - m₀v_f)/(v_ex - v_f) = m_g

Keep in mind that v_ex is negative. I found that the final mass of the gases was 5.74 x 10⁴kg, however, it's wrong. Can someone help me in finding what i did wrong in my procedure? Thank you.

 

[mukundpa]

Whether the exhaust velocity is relative to ground or space or relative to space shuttle?
Is the velocity of gases relative to space remains constant during the increase in velocity form Vi to Vf ?

 

[fogvajarash]

Whether the exhaust velocity is relative to ground or space or relative to space shuttle?
Is the velocity of gases relative to space remains constant during the increase in velocity form Vi to Vf ?

I'm sorry? I don't get what you are saying. Shouldn't the velocity of the gases be reepect to the ground? And in the exercise the exhaustion velocity is assumed to be constant (stated).

 

[gneill]

Exhaust velocity is relative to the spacecraft . As such, the speed of exhausted material in some inertial frame of reference will change as the ship speed changes. The mass of the ship also changes over time as the fuel is burned and ejected. This complicates the analysis by momentum conservation.

Another approach to the problem might be to consider the Rocket Equation for the change in velocity. The details of mass change and velocity exhaust speed have already been "solved" in that equation.

 

[fogvajarash]

Exhaust velocity is relative to the spacecraft . As such, the speed of exhausted material in some inertial frame of reference will change as the ship speed changes. The mass of the ship also changes over time as the fuel is burned and ejected. This complicates the analysis by momentum conservation.

Another approach to the problem might be to consider the Rocket Equation for the change in velocity. The details of mass change and velocity exhaust speed have already been "solved" in that equation.

Thanks gneill! I finally got it using the equation Δv = v_exln(m₀/m). The answer was 7.99 x 10⁴kg, however is it possible to get to this answer without any calculus or external formulas? (the rocket equation was derived using calculus)

 

[gneill]

Thanks gneill! I finally got it using the equation Δv = v_exln(m₀/m). The answer was 7.99 x 10⁴kg, however is it possible to get to this answer without any calculus or external formulas? (the rocket equation was derived using calculus)

The acceleration won't be constant due to the changing mass, so the usual kinetic motion formulas won't apply. When parameters are changing, calculus is the way to go.

 

[Adib Karimi]

Excuse me why should Vex be relatived to nozzle or ground?

 

[gneill]

Excuse me why should Vex be relatived to nozzle or ground?

Exhaust velocity is the speed at which the burnt fuel exits the nozzle of the rocket. Clearly it must be associated with the device where the fuel is being burned (combustion chamber), which is tied to the rocket.

 

[Adib Karimi]

Exhaust velocity is the speed at which the burnt fuel exits the nozzle of the rocket. Clearly it must be associated with the device where the fuel is being burned (combustion chamber), which is tied to the rocket.

I'm confused about relative velocity, could u please explain it? and relate it to Vex to Vnozzle?

 

[gneill]

I'm confused about relative velocity, could u please explain it? and relate it to Vex to Vnozzle?

"Relative" means with respect to. Something that is measured with respect to some particular point of reference. The point of reference might be the ground, or a moving vehicle, or any other frame of reference you might choose.

Suppose there are two cars moving in the same direction along a straight road. Vehicle one has a speed of 50 kph while the other, vehicle two, has a speed of 60 kph with respect to the road surface. The relative speed of vehicle two with respect to vehicle one is 10 kph.

 

[Adib Karimi]

"Relative" means with respect to. Something that is measured with respect to some particular point of reference. The point of reference might be the ground, or a moving vehictle, or any other frame of reference you might choose.

Suppose there are two cars moving in the same direction along a straight road. Vehicle one has a speed of 50 kph while the other, vehicle two, has a speed of 60 kph with respect to the road surface. The relative speed of vehicle two with respect to vehicle one is 10 kph.

Thank you sir for your knowledge. would u please say what's the direction of Vex, and Vex in relative to Vnozzle why is (V-Vex)?
please make me understand why plane uses more fuel during taking off and landing, actually first one is evident but I can't just understand it meanwhile landing!

 

[gneill]

Thank you sir for your knowledge. would u please say what's the direction of Vex, and Vex in relative to Vnozzle why is (V-Vex)?

The exhaust is shooting out of the back of the rocket. If the rocket is assumed to be moving in a positive direction with speed V according to some observer, and the exhaust is departing in the negative direction with speed Vex with respect to the rocket, then the observer will see the exhaust moving with velocity V - Vex from his point of view.

please make me understand why plane uses more fuel during taking off and landing, actually first one is evident but I can't just understand it meanwhile landing!

Please start your own separate threads for new questions. Use the Post New Thread button at the top right of the thread list in the forum.

 

[Adib Karimi]

The exhaust is shooting out of the back of the rocket. If the rocket is assumed to be moving in a positive direction with speed V according to some observer, and the exhaust is departing in the negative direction with speed Vex with respect to the rocket, then the observer will see the exhaust moving with velocity V - Vex from his point of view.

Please start your own separate threads for new questions. Use the Post New Thread button at the top right of the thread list in the forum.

If Vex has negative direction respect to shuttle so V is positive then observer will see exhaust with (V-Vex) it means that we assume rocket is not moving at that time when exhaust moving with relative velosity?!

 

[Adib Karimi]

Why planes use more fuel when they are taking off or landing?
How planes land and how they face with drug forces?

 

[gneill]

If Vex has negative direction respect to shuttle so V is positive then observer will see exhaust with (V-Vex) it means that we assume rocket is not moving at that time when exhaust moving with relative velosity?!

Imagine that there are two observers. Observer A is located on the ground, watching the rocket move at velocity V (according to him). Observer B is located on the shuttle and therefore moving along with it. Observer B watches the exhaust shoot out the engine nozzle with speed Vex according to him. According to B the exhaust has velocity -Vex in his frame of reference. Observer A still sees the shuttle (and observer B along with it!) moving at velocity V. Observer A will see the exhaust moving at V - Vex. If you want to could say: Vex_A = V - Vex, that is, the velocity of the exhaust gas at that instant in the reference frame of observer A is V - Vex.

 

[gneill]

Why planes use more fuel when they are taking off or landing?
How planes land and how they face with drug forces?

Start a NEW thread. Go to the main thread list for Introductory Physics and click on Post New Thread.

 

[Adib Karimi]

Please just imagine two cars x and y are moving at the opposite direction Vx=60 km/h then Vy=-50 km/h . Velocity of x respect to y is 110 km/h and velocity of y respect to x is -110 km/h it means that Vy_x=Vy-Vx=-50-60=-110 OK ?! so I think for observer B Vex equals Vex-V
The other problem I have with this is that when there is a mass of -dm velocity is v+dv so momentum of shuttle at t+dt is Pf=(v+dv)*(m+dm)+(Vex-(v+dv))*(-dm) Pi=m*v
but I don't know why in the equation there is no dv in momentum of exhaust ?!

 

[gneill]

Please just imagine two cars x and y are moving at the opposite direction Vx=60 km/h then Vy=-50 km/h . Velocity of x respect to y is 110 km/h and velocity of y respect to x is -110 km/h it means that Vy_x=Vy-Vx=-50-60=-110 OK ?! so I think for observer B Vex equals Vex-V

If Observer B is aboard the shuttle, then according to him V = 0; The shuttle is not moving with respect to himself, and he sees the exhaust leaving with velocity -Vex.

The other problem I have with this is that when there is a mass of -dm velocity is v+dv so momentum of shuttle at t+dt is Pf=(v+dv)*(m+dm)+(Vex-(v+dv))*(-dm) Pi=m*v
but I don't know why in the equation there is no dv in momentum of exhaust ?!

If the calculation is being done in the frame of reference of the rocket then the exhaust velocity is a constant.