# Momentum in a space shuttle and thrust

1. Nov 3, 2013

### fogvajarash

1. The problem statement, all variables and given/known data
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 vf = 5343 m/s

2. Relevant equations
-

3. 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 = Ft - W (where Ft is thrust force, and it is given by Ft = Rvex where vex is the exhaust velocity). By clearing this equation, I found that the exhaust velocity was vex = 1.254 x 104m/s.

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

m0v0 = mgvex + (m0 - mg)vf

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

(m0v0 - m0vf)/(vex - vf) = mg

Keep in mind that vex 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.

2. Nov 3, 2013

### 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 ?

3. Nov 3, 2013

### fogvajarash

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).

4. Nov 3, 2013

### Staff: Mentor

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.

5. Nov 3, 2013

### fogvajarash

Thanks gneill! I finally got it using the equation Δv = vexln(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)

6. Nov 3, 2013

### Staff: Mentor

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.

7. Apr 28, 2015

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

8. Apr 28, 2015

### Staff: Mentor

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.

9. Apr 28, 2015

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

10. Apr 28, 2015

### Staff: Mentor

"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.

11. Apr 29, 2015

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!

12. Apr 29, 2015

### Staff: Mentor

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.

13. Apr 30, 2015

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?!!

14. Apr 30, 2015

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

15. Apr 30, 2015

### Staff: Mentor

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: VexA = V - Vex, that is, the velocity of the exhaust gas at that instant in the reference frame of observer A is V - Vex.

16. Apr 30, 2015

### Staff: Mentor

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

17. May 1, 2015

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 ?!!

18. May 1, 2015

### Staff: Mentor

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.

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

Draft saved Draft deleted

Similar Discussions: Momentum in a space shuttle and thrust
{"extraction_info": {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.8833810687065125, "perplexity": 1168.9377933124363}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2017-34/segments/1502886103270.12/warc/CC-MAIN-20170817111816-20170817131816-00506.warc.gz"}