# Aerospace A question on thrust

1. Dec 6, 2004

### Morphose

I am playing around here at home and was thinking about how thrust works. This is assuming we are in space...

Let's say I had a mass of 10 metric tons (including the engine) and an engine that produces 100kg of thrust (I don't know if you would refer to it in seconds, minutes, or hours). But how fast would this mass be traveling in say 1, 6, 12, 24,and 48 hrs.?

Any claculations you could give me to help explain it would be great.

I don't know how this works, but I figured if anyone could explain it to me, somebody here could.

2. Dec 6, 2004

### ceptimus

Really, you should measure the thrust in Newtons. You can only convert from mass (kg) to Newtons if you already know the acceleration. On Earth we are used to the acceleration due to gravity $$g \approx 9.81 ms^{-2}$$

OK, for your example, let's say the thrust is 1000 Newtons (this is pretty much what I think you meant, since if g were 10 instead of 9.81, then 100kg would be 1000 N on Earth.

The formula that gives speed after a certain time is

$$v = u + at$$

where u is the starting speed (zero if we start from rest) t is the time in seconds and a is the acceleration.

You can get the acceleration from the mass and the thrust by the formula

$$a = F/m$$

where F is the force (the thrust) and m is the mass.

You have a force of 1000 N and a mass of 10,000 kg, so a = 0.1 m/s/s

An hour of this (3600 seconds) would give a speed increase of

$$v = 3600 * 0.1 = 360 m/s \approx 805 mph$$

Last edited: Dec 6, 2004
3. Dec 6, 2004

### Morphose

ok, thanks...

I'll have to write this stuff down and see if I can wrap my noggin around it. But you have helped out, so thank you for your time. I'll have to reply to see if I can understand it properly.

4. Dec 6, 2004

### Morphose

ok, I think I have it, but let me try here...

If I have an engine that has a force of 1500N (150kg), and a Space vessel with a mass of 250 Tonnes (250,000kg), traveling for 48 hrs. (172,800 seconds). Then it would look like this;

a = F / m
a = 1500 / 250,000
a = 0.006

v = u + at
v = 0 + 172,800 * 0.006 = 1037 m/s or about 2318 mph

is that about right? do I have an unsteady handle on it?

5. Dec 6, 2004

### ceptimus

Looks good to me. I make it 2319.3 mph, but that's just an insignificant rounding error.

With real space rockets, at least the current generation of chemical fuelled ones, the math is made more complex by the fact that as the fuel is used up, the mass of the rocket gets gradually less. So you have a roughly constant force from the rocket motor, but this produces an increasing acceleration as the rocket gets lighter.

But if your rocket were fuelled by some very efficient power source, maybe a nuclear reactor powering an ion drive, then the mass of the rocket would change relatively little, so the math you did here would be pretty accurate.

Last edited: Dec 6, 2004
6. Dec 7, 2004

### Morphose

Thanks you. That part is done, now how about distance? Do you or anybody else know how to calculate distance being traveled as speed increases?

We'll assume that we are using an Ion Propulsion system with limitless amount of fuel to use (or so much as to make no difference for a long time). But let's say this;

Force = 1,000 N
Mass = 160,000 kg
Time = 86,400 sec. 24hrs.
Starting speed = 0

How far would this vessel have trraveled in the afore-mentioned 24 hr span?

7. Dec 7, 2004

### Morphose

Or more precisely, what claculation would I use to find the distance traveled? I have an Excel sheet open to help me with these calculations, so as long as I can input into that to use as a reference, I'm cool

8. Dec 7, 2004

### Staff: Mentor

I'm guessing you haven't had physics or calculus yet, so the easy way is to average the speed (add the starting speed to the end speed and divide by 2) and multiply by time. d=v*t

If you have had calculus, you can calculate distance directly from acceleration by integrating the velocity equation. If you're not ready for that yet, don't worry about it...

9. Dec 7, 2004

### Morphose

No, I haven't had Calculus or Physics, they never offered it when I went to school, and I can't get approval to attend here at the college (requisites I can't pass). I am very interested in physics and things of that nature, but unfortunately, interest is not a major in college... =(

So, in order to calculate the distance traveled from the previous information;
a = F/m
a = 1000/160,000
a = 0.00625

v = u + at
v = 0 + 0.00625*86,400
v = 540 m/s

d = v*t
d = (540/2)86,400
d = 23328000 m
d = 23328 km

Is this correct? Do I have a proper grasp of it?

10. Dec 7, 2004

### ceptimus

Yes that's right.

The most useful formulas linking constant acceleration with time, velocity and distance are:

$$v = u + at$$

$$s = ut + \frac{1}{2}at^2$$

$$v^2 = u^2 + 2as$$

in all of these:

v is the final velocity in metres/second

u is the starting velocity in metres/second

s is the distance traveled in metres

a is the acceleration in metres/second/second

t is the time taken in seconds

so for your question we can use

$$s = ut + \frac{1}{2}at^2$$

$$s = 0 \times 86400 + \frac{1}{2} \times \frac {1000}{160000} \times 86400^2 = 23328000 \ metres$$

Actually the formulas work whatever units you use, so you could use miles, hours, and miles per hour if you wanted (and miles/hour/hour for the acceleration). But it's better to stick to metres, kilograms and seconds normally. That way, when you start calculating forces, power required, and energy use etc., all the units work together nicely, without having to remember lots of conversion constants (like how many killowatt hours of energy it takes to accelerate one pound by one mile an hour). If you stick to S.I. units as they are called, then all the conversion factors are exactly 1.

Last edited: Dec 7, 2004
11. Dec 8, 2004

### Morphose

Alright, I understand this so far. But what is the calculation to find the time to travel a specified distance?

12. Dec 8, 2004

### ceptimus

You have to rearrange one of the above formulae to give you the time you want.

Note that the first formula doesn't have an 's' in it, so it's no use to you. The third one doesn't have a 't' so to link time and distance you need this one:

$$s = ut + \frac{1}{2}at^2$$

rearranging, we get

$$\frac{1}{2}at^2 + ut -s = 0$$

this is a quadratic, so it can be solved using the quadratic formula to give:

$$t = \frac{-u \pm \sqrt{u^2 + 2as}}{a}$$

Note that the $$\pm$$ (plus or minus) means that you will usually get two answers to this equation. One will generally be negative and you can ignore it. To get two positive answers, the acceleration has to be the opposite direction to the initial velocity (that is a deceleration). Then both answers are right as the object will pass through the given distance twice, once slowing down, and again going the other way.

Last edited: Dec 8, 2004
13. Dec 8, 2004

### ceptimus

If you are starting from zero speed, u = 0, so the formula simplifies to:

$$t = \frac{\sqrt{2as}}{a}$$