# Thrust and acceleration of a launch vehicle

1. Mar 2, 2014

### coggo8

1. The problem statement, all variables and given/known data

Here is the data I have been given;
mass = 3.4*10^6 kg (at lift off)
mass of first stage propellants = 2.0*10^6
propellant consumption rate = 1.4*10^4 kg/s
g = 9.8 m/s/s
Find the thrust of the engines at lift-off.

2. Relevant equations

a = T-mg/m

3. The attempt at a solution

So mg = (3.4*10^6)*9.8 = 3.332*10^7
m = 3.4*10^6

∴a = T-3.332*10^7/3.4*10^6

I have two unknowns here, I thought there might be a way of finding either acceleration or thrust using mass, propellant mass and propellant consumption?
I have searched through my HSC texts and haven't been able to find an equation where I can find 'a' or 'T' without needing one to find the other.

2. Mar 2, 2014

### Staff: Mentor

Hi coggo8. The question seems not as complicated as the data might at first lead you to believe.

The rocket won't lift off unless thrust > weight.

Too easy. (unless you haven't represented the question correctly)

3. Mar 2, 2014

### coggo8

I see what you mean, so I could answer this question quite simply with; T>3.4*10^6?

However I have just read part (b) of the question which says 'If that thrust is sustained, find the time over which the first stage operates and acceleration of the vehicle at that time.'

So, I can find how long the first stage will operate using this method;
Mass of propellant/Consumption rate of propellant
= 2.0*10^6/1.4*10^4
= 142.85 seconds = 2 minutes 23 seconds (approx.)

Acceleration of the vehicle at that time? As we are assuming thrust>weight perhaps I could use a number slightly greater than the weight but still close, 3.41*10^6?
As you can see I am still stuck with two unknowns :(

4. Mar 2, 2014

### Staff: Mentor

What units would that quantity have?

Methinks there is a lift-off acceleration specified in the textbook, which you have failed to disclose.

Perhaps if you knew the height the rocket must attain before the fuel runs out, you could work backwards and determine the necessary acceleration?

Last edited: Mar 2, 2014
5. Mar 2, 2014

### D H

Staff Emeritus
Or even more likely, there is an exhaust velocity which coggo8 has failed to disclose.

6. Mar 2, 2014

### coggo8

Nope, no lift-off acceleration (otherwise this question would be simple), and it would be strange for them to expect me to use a random number greater than the weight because if I were to do that, I would have to use an infinitely close number?! Especially considering this is an assignment and that would be very ambiguous..

7. Mar 2, 2014

### coggo8

I have considered all of the above :(
Surely either acceleration or thrust could be calculated using fuel consumption rate, fuel weight and mass of the rocket?

8. Mar 2, 2014

### D H

Staff Emeritus
Surely not. Knowing how fast the exhaust leaves the rocket is crucial.

Please provide the relevant information. You have left out something very important because you think it is irrelevant. That missing information is very relevant.

9. Mar 2, 2014

### Staff: Mentor

I asked what units this would have. You replied m s-1

Prove it.

10. Mar 2, 2014

### coggo8

I have not left any information out. Here is the statement.

"An Apollo launch vehicle has a total mass of 3.4*10^6 kg at lift-off. The first stage contains 2.0*10^6 kg of propellant (fuel and oxygen) which is consumed at a rate of 1.4*10^4 kg/s."

I provided all of this information at the beginning of the thread.
Mass of vehicle = 3.4*10^6 kg (at lift-off)
Mass of propellants = 2.0*10^6 kg
Consumption rate of propellants = 1.4*10^4 kg/s

11. Mar 2, 2014

### Staff: Mentor

Perhaps a google search would tell you the altitude to be reached before the first stage drops off? Or some useful detail like that...

12. Mar 2, 2014

### coggo8

Thrust is measured in Newtons (N). Thrust is a force, that is why we measure it in newtons.
Don't ask why I said ms-1 earlier.

Also T>3.4*10^6 would be incorrect. Because the downward force of the vehicle is actually (3.4*10^6)*9.8 = 3.332*10^7 N
∴ T>3.332*10^7 N

13. Mar 2, 2014

### coggo8

Also Part (c) asks me to find the speed of the exhaust gases.

14. Mar 3, 2014

### D H

Staff Emeritus
There is something missing. One cannot compute the acceleration or thrust of a rocket knowing only the fuel consumption rate, fuel weight, and mass of the rocket.

Think of it this way: You are sitting on a flatbed railroad car with a pile of uniformly-sized rocks, with the car in turn sitting on a frictionless track. The car will accelerate when you throw rocks at some rate from the car. The acceleration increases if you instead use some device that ejects rocks at a very high speed, even if you keep the number of rocks ejected per minute the same as the rate at which you threw them.

15. Mar 4, 2014

### coggo8

I fully understand the concepts involved. I am merely trying to think of alternative ways to solve this problem.

I will find help with this elsewhere. You D H, as a 'mentor', are very condescending. A mentor is supposed to stimulate thought and be suggestive. I don't find this "Please provide the relevant information. You have left out something very important because you think it is irrelevant. That missing information is very relevant." at all thought-stimulating, suggestive, supportive, helpful, intuitive, enlightening or open-minded. I find it condescending and arrogant. I cannot fathom how a person of your calibre (engineering degree) is unable to suggest any form of alternative to put this problem remotely close to a solution.

I DO however thank NascentOxygen for sharing your suggestions and even still when you could not find a solution, suggesting an alternative to find more information.

16. Mar 4, 2014

### Staff: Mentor

I'm certain that when you ask your teacher how this question is to be answered, s/he will realize there's a missing detail. Please come back and let us know what it is. You can't be expected to solve a problem where an essential piece of information has been left out, albeit inadvertently.

In the event of this being an exam question, you may still be eligible for full marks if you were to identify those details which are needed to be able to fully answer the question.

17. Mar 5, 2014

### D H

Staff Emeritus
That's the mass of the vehicle, not the weight. NascentOxygen suggested using weight.

That's the weight of the vehicle, and that is a good lower limit on the thrust. From this, you should be able to calculate a lower limit on the exhaust velocity.

18. Mar 5, 2014

### BvU

I love this. Nasty of you to bite the helping hand. As you notice, DH is magnanimous enough to understand your outcry is just an utterance of frustration.

With only one google apollo launch thrust you find back your T > 3.33 x 107 N and the one or two bits extra info the missing of which caused all this fuzz :
•Stage I was called the S-IC. It had five rocket engines that used liquid oxygen and kerosene as fuel. Each engine produced 1.5 million pounds (6,675,000 newtons) of thrust. Together, the engines could generate 7.5 million pounds (33,375,000 newtons) of thrust. Think of thrust as the strength of a rocket engine. This thrust pushed the entire vehicle assembly more than 36 miles (57.9 km) vertically at a speed of 9,030 feet (2,752 m) per second (fps). At that point, the S-IC's engines shut off. Explosive bolts connecting S-IC to the rest of the Saturn V vehicle detonated, jettisoning stage I into the Atlantic Ocean.
Even if it was your teacher that accidentally left one of these bits, you owe DH an apology. You're smart enough to understand that getting angry is one thing, making up another. And that running away is not a solution.

Oh, and the launch vehicle is a Saturn missile. Apollo is the spacecraft. Now that is definitely something to pass on to teacher...

19. Mar 5, 2014

### BvU

But perhaps teacher isn't doing so badly: At launch time the acceleration of the missile isn't all that big (it's a stately thing to see such a launch) and the speed isn't either. In fact we don't even need T > m0g: T=m0g will make the thing take off just as well thanks to dm/dt<0. Let's stick to the = sign.

We do know that F = dp/dt = dm/dt v + m dv/dt, right? For rocket + exhaust gas v = vrocket - vexhaust. (using scalars, up=positive)

F = -mg and if we ignore dv/dt and let vrocket- vexhaust = -ve, we have the exhaust gas speed wrt the rocket. Thrust is constant (given), so assume ve constant as well.

By the time the fuel is almost gone, we still have F = dp/dt = dm/dt (-ve) + m dv/dt = -mg.
No extra info needed to get dv/dt !

20. Mar 5, 2014

### Staff: Mentor

That's good thinking, BvU! It involves a differential equation. We'd better first ask is coggo8 au fait with first-order differential equations?

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