Rocket Launch: Upthrust Increasing?

  • Thread starter Tommy1995
  • Start date
  • Tags
    Launch
In summary, there is an initial upthrust provided by the combustion of the engine during the launch of a rocket, but the thrust remains constant even as the rocket's mass decreases due to fuel consumption. This results in a gradual increase in acceleration until the rocket reaches a constant acceleration rate. The efficiency of the rocket also increases as the weight decreases, resulting in even greater fuel savings. The main engines and solid rocket boosters are not constantly at full thrust during launch, and the thrust is actively adjusted to maintain a constant acceleration rate. A simplified model of the rocket's motion can be described by the equation m \dot{v} = v_{ex} \dot{m} –mg.
  • #1
Tommy1995
39
0
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?
 
Physics news on Phys.org
  • #2
Tommy1995 said:
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?

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
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
Drakkith said:
Once the thrust is going fully, the rocket lifts off with about the same force it will have throughout the flight.

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 doesn't that mean the upthrust force increases ? hence force increases throughout the flight?
 
  • #5
Tommy1995 said:
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 doesn't that mean the upthrust force increases ? hence force increases throughout the flight?

No, the force stays the same, but as mass decreases acceleration increases.
 
  • #6
The decrease in weight also results in additional acceleration due to the increase in net force available to accelerate the rocket.
 
  • #7
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
phinds said:
No, it means that the thrust provided is only slightly more than the force of gravity holding the rocket down.
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
(found via a previous thread, by google)
The thrust remains constant but the weight of the fuel decreases so the rocket picks up speed.
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
russ_watters said:
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.

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
russ_watters said:
The thrust is not necessarily constant. On the space shuttle, for example, the main engines are throttled twice after launch.
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.

I'm not clear on if the SRBs are constant thrust.
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.

Srbthrust2.jpg
 
Last edited:
  • #11
"were" :frown:
 
  • #12
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:
[tex]m \dot{v} = v_{ex} \dot{m} –mg[/tex]
Where m and v are the time-dependent mass and velocity of the rocket. And [itex]v_{ex}[/itex] 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 [itex]v_{ex} \rho A[/itex] (Where A is simply the area of the exhaust, and [itex]\rho[/itex] is the density of the fuel). We can use this in the equation from before, to get:
[tex] m \dot{v } = {v_{ex}}^2 A \rho -mg[/tex]
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, [itex]v_{ex}[/itex] 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:
[tex]\dot{a} = - \frac{{v_{ex}}^3 \rho^2 A^2}{m^2}[/tex]
So the rate of change of acceleration of the rocket will be positive (since [itex]v_{ex}[/itex] 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 [itex]F=m \dot{v}[/itex] In which case,
[tex] \dot{F} = - \dot{m} g [/tex]
Or we could define [itex]F= \dot{P}[/itex] (where P is the momentum of the rocket), so then we get:
[tex] \dot{F} = - \dot{m} (g- \dot{v}) [/tex]
And since I have been assuming that the rocket is still close to the earth, [itex]\dot{v}[/itex] 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
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
Tommy1995 said:
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? :/
This is the whole basis for multi stage rocket systems: don't carry any more mass than you need to.
 
  • #15
Tommy1995 said:
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? :/
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
BruceW said:
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).

It would be really handy to re-fuel half way up. :wink:
 
  • #17
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
A really long mains lead, perhaps.
 
  • #19
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
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:
  • #21
read said:
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?

That depends on what you mean by efficiency. In terms of minimal energy usage, you want a very high massflow, with a relatively small exhaust velocity. You run out of reaction mass really quickly this way though, so it isn't really ideal. If you instead want to minimize your usage of reaction mass (which is typically the limiting factor), you want the highest possible exhaust velocity at all times.
 
  • #22
cjl said:
That depends on what you mean by efficiency. In terms of minimal energy usage, you want a very high massflow, with a relatively small exhaust velocity. You run out of reaction mass really quickly this way though, so it isn't really ideal. If you instead want to minimize your usage of reaction mass (which is typically the limiting factor), you want the highest possible exhaust velocity at all times.

I mean that I want to maximally use the kinetic energy of the gases to accelerate my rocket. So, at the moment we do not care how much energy has been used to produce the mass flow.

What strategy of velocity of emitted gases V_g(m) should I use to minimize the factor:

K(m)/(K(m)+Q(m0-m))

where K(m) is kinetic energy of the rocket when its mass is 'm', Q(m0-m) total kinetic energy of the emitted gases. m0 is initial mass when the rocket has zero speed.
 
  • #23
read said:
I mean that I want to maximally use the kinetic energy of the gases to accelerate my rocket. So, at the moment we do not care how much energy has been used to produce the mass flow.

What strategy of velocity of emitted gases V_g(m) should I use to minimize the factor:

K(m)/(K(m)+Q(m0-m))

where K(m) is kinetic energy of the rocket when its mass is 'm', Q(m0-m) total kinetic energy of the emitted gases. m0 is initial mass when the rocket has zero speed.

Sorry "to maximize the factor:"
 
  • #24
Actually, that is a really good question. Using the same old calculation (without gravity) gives:
[tex]v_{ex} \dot{m}=m \dot{v} [/tex]
(where [itex]v_{ex}[/itex] is the velocity of the propellant with respect to the rocket, i.e. the exhaust velocity). Then, doing a similar calculation, to find the rate of change of total kinetic energy of the system, I get:
[tex]\frac{dQ}{dt}=- \frac{1}{2} \dot{m} {v_{ex}}^2 [/tex]
(where the left hand side of the equation is the rate of change of the total kinetic energy in the system. So you can relate this to the rate at which chemical energy is being used up, since change in kinetic energy is due to a change in chemical energy). Note: when the rate of change of mass of the rocket is negative, the rate of change of total kinetic energy of the system is positive, as we would expect. Also, we know that the kinetic energy of the rocket is given by:
[tex]T=\frac{1}{2}mv^2[/tex]
(Where I am using the symbol T to mean the KE of the rocket). Also, the rate of change of the kinetic energy of the rocket is:
[tex]\frac{dT}{dt}=\dot{m}v( \frac{1}{2} v+ v_{ex}) [/tex]
So now, using our two equations for rate of change of rocket KE with time, and rate of change of total KE with time, we get:
[tex]\frac{dT}{dQ}= - {(\frac{v}{v_{ex}})}^2 - \frac{v}{v_{ex}} [/tex]
Awesome. This equation is telling us the rate of change of KE of the rocket, with respect to change in the total KE of the system (i.e. this equation is the 'instantaneous' efficiency). Now, we want to find what value [itex]v_{ex}[/itex] must take, for this equation to be maximised, so if we take the partial derivative with respect to [itex]v_{ex}[/itex], then we get [itex]v_{ex}=-2v[/itex] and if this condition is fulfilled, then the instantaneous efficiency is at its maximum value of 1/2

So, after all that, we find that to get max efficiency for the rocket, we require [itex]v_{ex}=-2v[/itex] so this means that the exhaust velocity must always be at twice the velocity of the rocket (and in opposite direction, of course). So, equivalently, this means we require that the actual velocity of the propellant must always be equal and opposite to the velocity of the rocket (as viewed by an inertial reference frame).

So, I agree with your book, that initially the propellant must come out slowly, and gradually begin to come out more quickly. But I find that the efficiency is 1/2, not 1. In fact, an efficiency of 1 does not make sense, because this would imply that the total kinetic energy of all the expelled propellant is zero. But this is clearly not possible.

Also, I should say, this is for an ideal rocket, and a real rocket is probably much more complicated. But it is nice to work out that the max theoretical efficiency is 1/2, assuming I've done the calculations right ;)
 
  • #25
mrspeedybob said:
Would it make any significant difference to launch a rocket with some initial velocity?

We do this already. Rockets are launched easterly from as close to the equator as practical to provide initial velocity due to Earth's rotation.

For example the launches from Merritt Island Florida, at 28.5 degrees north latitude, yields a "free" initial speed of 1,472Km/h before the rocket even leaves the pad.
 
  • #26
Also, the thrust from a "fixed" thrust rocket increases a bit as it gets above the atmosphere. The exhaust doesn't have to fight atmospheric pressure in getting out of the nozzle.
 
  • #27
BruceW said:
Actually, that is a really good question. Using the same old calculation (without gravity) gives:
[tex]v_{ex} \dot{m}=m \dot{v} [/tex]
(where [itex]v_{ex}[/itex] is the velocity of the propellant with respect to the rocket, i.e. the exhaust velocity). Then, doing a similar calculation, to find the rate of change of total kinetic energy of the system, I get:
[tex]\frac{dQ}{dt}=- \frac{1}{2} \dot{m} {v_{ex}}^2 [/tex]
(where the left hand side of the equation is the rate of change of the total kinetic energy in the system. So you can relate this to the rate at which chemical energy is being used up, since change in kinetic energy is due to a change in chemical energy). Note: when the rate of change of mass of the rocket is negative, the rate of change of total kinetic energy of the system is positive, as we would expect. Also, we know that the kinetic energy of the rocket is given by:
[tex]T=\frac{1}{2}mv^2[/tex]
(Where I am using the symbol T to mean the KE of the rocket). Also, the rate of change of the kinetic energy of the rocket is:
[tex]\frac{dT}{dt}=\dot{m}v( \frac{1}{2} v+ v_{ex}) [/tex]
So now, using our two equations for rate of change of rocket KE with time, and rate of change of total KE with time, we get:
[tex]\frac{dT}{dQ}= - {(\frac{v}{v_{ex}})}^2 - \frac{v}{v_{ex}} [/tex]
Awesome. This equation is telling us the rate of change of KE of the rocket, with respect to change in the total KE of the system (i.e. this equation is the 'instantaneous' efficiency). Now, we want to find what value [itex]v_{ex}[/itex] must take, for this equation to be maximised, so if we take the partial derivative with respect to [itex]v_{ex}[/itex], then we get [itex]v_{ex}=-2v[/itex] and if this condition is fulfilled, then the instantaneous efficiency is at its maximum value of 1/2

So, after all that, we find that to get max efficiency for the rocket, we require [itex]v_{ex}=-2v[/itex] so this means that the exhaust velocity must always be at twice the velocity of the rocket (and in opposite direction, of course). So, equivalently, this means we require that the actual velocity of the propellant must always be equal and opposite to the velocity of the rocket (as viewed by an inertial reference frame).

So, I agree with your book, that initially the propellant must come out slowly, and gradually begin to come out more quickly. But I find that the efficiency is 1/2, not 1. In fact, an efficiency of 1 does not make sense, because this would imply that the total kinetic energy of all the expelled propellant is zero. But this is clearly not possible.

Also, I should say, this is for an ideal rocket, and a real rocket is probably much more complicated. But it is nice to work out that the max theoretical efficiency is 1/2, assuming I've done the calculations right ;)

Thank you for calculations. I have also done similar thing, but I calculated Q' and T' differently. I use the frame of reference where the the rocket has speed V. Then,

Q'= -m'(V_ex-V)^2/2
T'=((mV^2)/2)'=mVV'

For T' we take derivative only for the speed, not mass, because we are interested only in final gain in the kinetic energy of the rocket with final mass m.

Since mV'=-m'V_ex, we have T'=-m'*V*V_ex

Then it is easy to find:
efficiency of power (or energy rate) T'/(T'+Q')=2(V_ex/V)/(1+(V_ex/V)^2), i.e. the maximum 100% is achieved at V_ex=V.

If we now assume V_ex=V and solve the equation of motion

mV'=-m'V

we will get

V(m) = v0*m0/m +const.

If const=0 than this would mean that the rocket initially had speed v0 and the engine has been already working so gases had speed v0 with respect to the rocket and we will then calculate the efficiency to 100%. Simply because in the frame where we do the calculation the emitted gases always has velocity==zero, according to our assumption V_ex=-V.

But this looks extremely odd because we do the calculation with respect to some frame where the rocket has already got initial velocity v0... If we choose v0=0, which is fair, meaning that we start with gas velocity 0 and rocket speed 0, than we will not accelerate.

Initially one should consider a rocket that is not accelerating but moving inertially, then we switch on engine. The easiest is to calculate in center of momentum reference frame, i.e. when the rocket has initial speed zero. In this case I cannot get 100% of efficiency.
 
  • #28
read said:
Q'= -m'(V_ex-V)^2/2
T'=((mV^2)/2)'=mVV'
We get different answers for these two expressions. The way I worked out these two, is this way: Suppose our rocket at one instant has mass m, moving at speed v, then a short time later, the mass and velocity of our ship have both changed by some small amounts, to [itex]v+ \Delta v[/itex] and [itex]m+ \Delta m[/itex] So by conservation of mass, we will need to have a small amount of propellant ejected, with mass equal to [itex]- \Delta m[/itex] and we define it to be moving at speed [itex]v_{ex}+v[/itex] (i.e. the exhaust velocity is the relative velocity of the propellant with respect to the rocket). Therefore, using conservation of momentum, (i.e. total momentum before = total momentum after), we get the equation:
[tex]mv = (m+ \Delta m)(v+ \Delta v) - \Delta m(v_{ex} +v)[/tex]
Now, cancelling some terms, and dividing by the small time interval [itex]\Delta t[/itex], we get:
[tex]m \frac{\Delta v}{\Delta t} + \frac{\Delta m \Delta v}{\Delta t} = v_{ex} \frac{\Delta m}{\Delta t} [/tex]
Now, take the limit that these small changes are infinitesimally small changes, then the second term on the left hand side disappears (because it is much smaller than the other terms), so we get left with:
[tex]m \frac{dv}{dt} = v_{ex} \frac{dm}{dt} [/tex]
hooray! Now, it is possible to find things like the rate of change of the KE of the rocket and the rate of change of total KE, by using a very similar line of reasoning as above. Right, so to find out what will be the small change in KE of the rocket, we simply use the KE of rocket after, minus the KE of the rocket before, so we get:
[tex]\Delta T = \frac{1}{2} (m+ \Delta m)(v+ \Delta v )^2 - \frac{1}{2}mv^2 [/tex]
And straight away, I am going to ignore any terms which have a product of more than one 'delta', since these terms are going to be much smaller, when we take our limit of very small changes, so therefore, we get:
[tex]\Delta T = mv \Delta v + \frac{1}{2} v^2 \Delta m [/tex]
So, now, again dividing by [itex]\Delta t[/itex] and taking the limit of very small change, we get:
[tex]\frac{dT}{dt}=mv \frac{dv}{dt} + \frac{1}{2} v^2 \frac{dm}{dt} [/tex]
And now we simply use the relation [itex]\dot{m} v_{ex} = m \dot{v} [/itex] with the first term of our equation for T, to get the nicer expression:
[tex]\frac{dT}{dt} = \frac{dm}{dt} v (v_{ex} + \frac{1}{2} v ) [/tex]
hooray. And now, to find the change in KE of the system, we use the KE of the rocket after, and plus the KE of the propellant, and minus the KE of the rocket before (since we want to find the total change in KE of the system). So we get:
[tex]\Delta Q = \Delta T - \frac{1}{2} \Delta m (v_{ex} + v)^2 [/tex]
Note: we are adding the KE of the propellant, but since the mass of the propellant is [itex]- \Delta m [/itex], we get a minus sign. So again, discarding any terms where there is a product of delta's, we have:
[tex]\Delta Q = mv \Delta v - v_{ex} v \Delta{m} - \frac{1}{2} v_{ex}^2 \Delta m [/tex]
So, divide by delta t and take limit of small changes:
[tex]\dot{Q} = mv \dot{v} - \dot{m} v_{ex} v - \frac{1}{2} v_{ex}^2 \dot{m} [/tex]
But luckily, using our equation [itex]\dot{m} v_{ex} = m \dot{v} [/itex] the first two terms on the right-hand side cancel, so we end up with:
[tex]\dot{Q} = - \frac{1}{2} \dot{m} v_{ex}^2 [/tex]
 

1. How does upthrust affect a rocket launch?

Upthrust, also known as buoyancy force, is the upward force exerted by a fluid on an object immersed in it. In the case of a rocket launch, as the rocket moves upwards through the air, it experiences an increasing amount of upthrust. This helps to counteract the force of gravity and allows the rocket to continue its ascent.

2. What factors can cause an increase in upthrust during a rocket launch?

The main factor that affects upthrust during a rocket launch is the speed of the rocket. As the rocket moves faster, it displaces more air, causing an increase in upthrust. Other factors that can contribute to an increase in upthrust include the shape and size of the rocket, as well as the density of the air around it.

3. How is upthrust measured during a rocket launch?

Upthrust is typically measured using force sensors or load cells that are attached to the rocket. These sensors can detect the amount of force exerted on the rocket due to upthrust and provide real-time data on its changes throughout the launch.

4. Can an increase in upthrust cause problems during a rocket launch?

In most cases, an increase in upthrust is beneficial for a rocket launch as it helps to lift the rocket off the ground and keep it stable during its ascent. However, if the upthrust becomes too strong, it can cause the rocket to deviate from its intended flight path and potentially lead to a failed launch. Therefore, it is important to carefully calculate and monitor the upthrust during a rocket launch.

5. How can scientists and engineers control and manipulate upthrust during a rocket launch?

Scientists and engineers can control upthrust during a rocket launch by adjusting the shape and size of the rocket, as well as the amount of propellant used. They can also use fins and other aerodynamic features to help control the direction and strength of upthrust. Advanced simulation and testing techniques are also used to predict and optimize upthrust during a rocket launch.

Similar threads

  • Other Physics Topics
Replies
32
Views
4K
  • Aerospace Engineering
Replies
31
Views
3K
Replies
4
Views
2K
Replies
31
Views
3K
  • Aerospace Engineering
Replies
5
Views
1K
  • Aerospace Engineering
Replies
24
Views
4K
  • Aerospace Engineering
7
Replies
237
Views
12K
  • Classical Physics
Replies
3
Views
686
  • Other Physics Topics
Replies
14
Views
2K
  • Classical Physics
Replies
1
Views
719
Back
Top