Forces - Space Shuttle Takeoff Calculations

[Taylan]

Homework Statement

[/B]
The total mass of a space shuttle and its launch vehicle is
M=2000t.

a) What must be the minimum size of the thrust force, to make the rocket move?

b) The actual thrust of the rocket is F=30MN. What is its acceleration in the beginning?

c) Assume that a mass of a 1000t of fuel is burnt by the rocket in 120 seconds and the velocity of the gas that is coming out is 3.6x10^3 m/s. What velocity will the rocket have after 1000t of fuel is consumed if the acceleration is assumend to be constant?

Homework Equations

F= ma
F(Thrust)= ( [change in] m / t).v
m1v1 = m2v2

The Attempt at a Solution

a) 2x10^6(9.81) = 19.62MN

b) so the thrust is said to be 30 MN
F= ma --> a= F/m
m= 2x10^6
F= [30x10^6 - 2x10^6(9.81)] / 2x10^6 = 5.19 m/s2.

c) for c ı am confused because I just used to conservation of momentum to calculate the velocity that will be reached by the rocket but without using the time (120seconds) anywhere in my calculations;

0 = (-3.6x10^3)(1000t) + (2000t-1000t)(v2)
v2= 3.6x10^3 m/s

 

[willem2]

b) so the thrust is said to be 30(2000t).

No it was said to be 30 Mega Newtons.

For point c, the rocket exhaust after the rocket gained some speed, won't have the same speed as the rocket exhaust at t=0 and v=0, so this
conservation of momentum calculation won't work.
Since the acceleration is apparently constant (very unrealistic), you can calculate the speed of the rocket at any given t. The only thing you will need to know is how long the rocket can keep that up.

 

[gneill]

a) no units for mass is given in the question so I assumed it to be kg.

The mass is said to be 2000t, the "t" being shorthand for a metric tonne (1000 kg).

 

[Taylan]

Thank you for correcting my confusions about a and b. I am still unclear about c tho. Can you give me some other tips pls?

 

[gneill]

See if you can find the thrust (force) using the given information.

 

[Taylan]

See if you can find the thrust (force) using the given information.

Yes I used F=(mass/t)v to find the thrust which gives out 30NM which is also stated in part b actually. But I wasn't able to link this with the velocity

 

[gneill]

Yes I used F=(mass/t)v to find the thrust which gives out 30NM which is also stated in part b actually. But I wasn't able to link this with the velocity

Which of Newton's laws might tell you how the velocity is changing?

 

[Taylan]

Which of Newton's laws might tell you how the velocity is changing?

Newton's first law says velocity will change if an external force is applied

 

[gneill]

Newton's first law says velocity will change if an external force is applied

Yes. How will velocity change when a force is applied? (which other of Newton's laws tells you this?)

 

[Taylan]

Yes. How will velocity change when a force is applied? (which other of Newton's laws tells you this?)

the second law says F=ma

 

[gneill]

the second law says F=ma

Right. So which of those variables do you have information about?

 

[Taylan]

Right. So which of those variables do you have information about?

So I can get to know the net force by net force = thrust - mg. and mass is known. this way ı found the acceleration which is asked in b. So ı know f,m and a.
but still not velocity?

 

[gneill]

How is velocity related to acceleration?

 

[Taylan]

How is velocity related to acceleration?

ah so am I using u=0m/s a= 5.19m/s2 (from b) and t=120s ? a= (v-u)/t. But I already found the acceleration in b and they gave me all that info again in c which ı can use to find thrust and acceleration once again. and thrust was actually also given. that confused me I think. do you think the part c is weirdly formulated with all those extra info?

 

[gneill]

do you think the part c is weirdly formulated with all those extra info?

Perhaps, but the point of part (c) may have been largely to get you find out about how to obtain the thrust.

Edit:

Strictly speaking you're using the info from part (c). They could have used different numbers there (but happened not to).

 

[Taylan]

Perhaps, but the point of part (c) may have been largely to get you find out about how to obtain the thrust.

Edit:

Strictly speaking you're using the info from part (c). They could have used different numbers there (but happened not to).

I see. Thanks a lot for your help! :)

 

[haruspex]

do you think the part c is weirdly formulated with all those extra info?

It is worse than that - it is contradictory.
There is enough information to calculate the initial acceleration, as you did in part b, and to find the acceleration after 120 seconds. Since the latter is double the initial acceleration, which is to be taken as the constant value?

Are you sure it doesn't say to assume the rate of change of acceleration is constant? Or, even more reasonably, that the thrust should be assumed constant?

 

[Taylan]

Yeah it says acceleration is constant, not the rate of change of acceleration :O

 

[haruspex]

Yeah it says acceleration is constant, not the rate of change of acceleration :O

To highlight the peculiarity of that, assuming the initial acceleration lasts for 120 seconds gives a speed of 624m/s; taking the thrust to be constant gives 2500m/s.

 

[Taylan]

To highlight the peculiarity of that, assuming the initial acceleration lasts for 120 seconds gives a speed of 624m/s; taking the thrust to be constant gives 2500m/s.

and how do you calculate the 2500m/s ?

 

[haruspex]

and how do you calculate the 2500m/s ?

Actually I blundered... let me recalculate that...

It's 1300m/s or so.
The standard rocket equation is logarithmic.
Do you know how to write down and solve differential equations?

 

[Taylan]

No. But i came across this '' Tsiolkovsky rocket equation '' . Would that be useful?

 

[haruspex]

No. But i came across this '' Tsiolkovsky rocket equation '' . Would that be useful?

Yes, that is the standard equation.