# A rocket's launch

1. Dec 18, 2012

### Tommy1995

When a rocket is launched there is an upthrust which is provided by the combustion of its engine. However, during the launch a rocket initially remains still then gradually lifts off with increasing speed. Does this mean that the upthrust of the rocket increases gradually over the initial stages of launch?

2. Dec 18, 2012

### phinds

No, it means that the thrust provided is only slightly more than the force of gravity holding the rocket down. The thrust remains constant but the weight of the fuel decreases so the rocket picks up speed.

3. Dec 18, 2012

### Staff: Mentor

Once the thrust is going fully, the rocket lifts off with about the same force it will have throughout the flight. (But not always) However acceleration to any speed takes time. And since you can't really see acceleration, but velocity, it may seem like it just sits there doing nothing. Plus the really big rockets like the space shuttle are actually accelerating quite fast, it's just that they are so big that it looks like they don't really move for the first few seconds.

4. Dec 19, 2012

### Tommy1995

So upthrust would be a force "F" and this force will be going against the weight force of the rocket "mg" however as the mass of the rocket decreases so too does the wieight force so doesnt that mean the upthrust force increases ? hence force increases throughout the flight?

5. Dec 19, 2012

### Staff: Mentor

No, the force stays the same, but as mass decreases acceleration increases.

6. Dec 19, 2012

### Staff: Mentor

The decrease in weight also results in additional acceleration due to the increase in net force available to accelerate the rocket.

7. Dec 19, 2012

### sophiecentaur

Initially, the efficiency of the rocket is very low - the energy supplied by the fuel is vast but the rate of increase of Gravitational Potential Energy is very low (it climbs slowly at first). An alternative way of starting it off would save a lot of fuel and involve lighter fuel tanks - hence the interest in horizontal take off (flying) launch platforms.

8. Dec 19, 2012

### Staff: Mentor

I'm not sure if it is "slightly", but according to this, the acceleration of the space shuttle at takeoff is about 2/3g.
http://www.aerospaceweb.org/question/spacecraft/q0183.shtml
That's a little confusingly worded and simplistic, so let me clarify:

The thrust is not necessarily constant. On the space shuttle, for example, the main engines are throttled twice after launch. I'm not clear on if the SRBs are constant thrust.

As I mentioned above, the change in weight should not be overlooked. For the space shuttle, for example, a halving of the weight is almost as big a factor as the halving of the mass initially. Using the numbers listed above and assuming constant thrust, the acceleration in terms of the fractional increase is:

a=f/m = 1.75f/(0.5m) = 3.5

So if acceleration starts at .67g, after the mass is cut in half, the new acceleration is 2.4g.

9. Dec 19, 2012

### sophiecentaur

Interesting. That had never struck me before. A double whammy! It must also mean that there is an even greater saving in fuel because the efficiency (GPE gained per unit of fuel) will also go up.

10. Dec 19, 2012

### D H

Staff Emeritus
That's correct. The engines were throttled down to about 64% before max-Q (maximum dynamic pressure), then back up to full, and then once acceleration reached 3g, the engines were continuously throttled down so as to keep acceleration at a constant 3g.

They weren't. They also had reduced thrust at max-Q, about 1/3 less than the thrust right after launch. The SRBs weren't actively throttleable. The variable thrust was instead attained passively by designing the grain geometry so that the surface area of the flame front gradually decreased after launch and then increased again after about 50 seconds into the launch. After about 80 seconds thrust slowly started dropping again as propellant started reaching depletion.

Last edited: Dec 19, 2012
11. Dec 19, 2012

### Staff: Mentor

"were"

12. Dec 19, 2012

### BruceW

Hey all, I was interested in this thread, so I did a basic derivation of the motion of the rocket (and I introduce a lot of assumptions), so I could get a feel for some of the physics behind it. It pretty much predicts the same as what you would guess anyway. It is a very simplistic model, so don't take it too seriously, but I think it captures some of the principles.

If we choose a reference frame fixed to the earth, neglecting the rotation of the earth, and using non-relativistic mechanics, we get:
$$m \dot{v} = v_{ex} \dot{m} –mg$$
Where m and v are the time-dependent mass and velocity of the rocket. And $v_{ex}$ is the velocity of the ejected fuel with respect to the rocket (i.e. the exhaust velocity).

We can go even further by saying that since the rate of fuel lost is $v_{ex} \rho A$ (Where A is simply the area of the exhaust, and $\rho$ is the density of the fuel). We can use this in the equation from before, to get:
$$m \dot{v } = {v_{ex}}^2 A \rho -mg$$
I am defining g to be positive, v to be positive (i.e. up is positive, and the rocket is always moving up). From now on in this post, I am going to use the approximation that the rocket is moving slowly, and near to the earth’s surface and that the fuel consumption is roughly constant (i.e. the exhaust velocity and fuel density are constant). I know this is not always going to be true, but I just wanted to think about the simple case to begin with. So therefore, $v_{ex}$ is going to be negative and g will be roughly 9.81. So what does this tell us about the acceleration of the rocket (under the assumptions I have stated)? It tells us:
$$\dot{a} = - \frac{{v_{ex}}^3 \rho^2 A^2}{m^2}$$
So the rate of change of acceleration of the rocket will be positive (since $v_{ex}$ is negative). So the acceleration of the rocket will increase because it starts to lose mass. (which is what you’d expect).

How about the force on the rocket? Well it depends on the definition of ‘force’ you want to use. We could define $F=m \dot{v}$ In which case,
$$\dot{F} = - \dot{m} g$$
Or we could define $F= \dot{P}$ (where P is the momentum of the rocket), so then we get:
$$\dot{F} = - \dot{m} (g- \dot{v})$$
And since I have been assuming that the rocket is still close to the earth, $\dot{v}$ will be less than g.

So using either of the two definitions of ‘force’ on the rocket, this force will be increasing with time. This is in the simplest case, when fuel use is constant, and the rocket is still moving slow close to surface of the earth. So when gravity begins to decrease, this will further increase the acceleration of the rocket. (More than what my model predicts). As Sophiecentaur said, a double whammy.

13. Dec 20, 2012

### Tommy1995

Interesting, I was thinking the net force upwards would increase over time even if the force provided by the fuel is constant because.

Fnet = Upthrust + (-Mg) where upwards is positive

Now after like 50seconds or so, assuming the mass is now halved

Fnet' = Upthrust + (-Mg/2)

∴Fnet'>Fnet

Furthermore, as the Fnet after 50s increases I was thinking that there would be also a double whammy increase in the acceleration because from F=ma, when considering the rocket, there would not only be a greater Fnet there would also be a lesser mass of the rocket. Would this be true? :/

14. Dec 20, 2012

### sophiecentaur

This is the whole basis for multi stage rocket systems: don't carry any more mass than you need to.

15. Dec 20, 2012

### BruceW

Yes, it will become easier for the rocket to accelerate because of 2 reasons: 1) the rocket is carrying less fuel, so there is less mass for gravity to act on. 2) the rocket has less mass, so a smaller force will be able to accelerate it by the same amount.

I see this double whammy as something which is very annoying for rocket-makers. Because to get a rocket into space, you need to put fuel into your rocket. But unfortunately, this makes the mass increase, so you will get less acceleration for the same force. And on top of that, when you put more fuel into your rocket, gravity will have a stronger hold on your rocket. So in the end, you have to put a lot of fuel into your rocket, most of which gets used up just to get your rocket into space (mucho expensive).

16. Dec 20, 2012

### sophiecentaur

It would be really handy to re-fuel half way up.

17. Dec 20, 2012

### BruceW

that's true. I am eagerly awaiting a Terry Pratchett style nanotube that reaches up into space. That would be able to re-fuel the rocket!

18. Dec 20, 2012

### sophiecentaur

A really long mains lead, perhaps.

19. Dec 23, 2012

### mrspeedybob

Would it make any significant difference to launch a rocket with some initial velocity?

For example suppose I created vertical launch tube which extended 240 meters below ground. My rocket is accelerated by electromagnets at 3g for 4 seconds so that by the time it is at ground level and the engines are started it is already going 120 m/s. Would this be any more or less beneficial then launching from a platform 720 meters above ground level?

On the one hand the launch tube would seem to result in 4 seconds less engine burn time to reach orbital velocity which should save fuel. On the other hand, the kinetic energy added by the launch tube would be equal to the gravitational potential energy added by the platform so they should be equivalent.

20. Jan 9, 2013

Hello BruceW,

Might be you can help?

Regarding efficiency of using fuel: K/K+Q, where Q=total kinetic energy of the emitted gases, K=final kinetic energy of rocket.

Sometime ago I read in a popular book of Gerard[/PLAIN] [Broken] t'Hooft "Playing with Planets" where he says that efficiency can be near 100% if the velocities of emitted gas gradually increases during the flight. I guess he knows what he says...

However I cannot get this by calculations. Assume for simplicity no gravity.

Differentiating momentum equality we have (similar to your equation):
m' Vg = m Vr',
where m - mass Vg a is the velocity of the gas relative to the rocket, Vr' is acceleration of rocket.

If we assume |Vg|=|Vr|, than it looks like that all gas power is transformed to the rocket kinetic power. For every moment the gas kinetic energy is zero in the frame where the rocket has speed Vr. Solving the above diff equation one can get the simplest formula for the Vr:

Vr(m)=v0*m0/m

m0 initial mass, v0 initial speed. Using this formula one can calculate again the efficiency to 100%, but the formula looks odd because we stay in the reference frame where the rocket has initial velocity v0. If i modify it to be "correct":

Vr(m)=v0*m/m0-v0

than simple calc gives the efficiency 1-m/m0.

For the constant gas speed velocity v0 we have Vr(m)=v0*ln(alpha), alpha=m0/m,
and efficiency ln(alpha)^2/(alpha-1), which is worse than previous.

If I am a captain of rocket and I want to use my fuel in a most efficient way, which time (or mass) dependence of gas velocity I should choose?

Last edited by a moderator: May 6, 2017