Why is the thrust equation same under gravitational force?

[f3sicA_A]

Homework Statement

Why is the thrust equation of a rocket the same irrespective of whether a gravitational force acts on the rocket or not (assuming no air resistance)?

Relevant Equations

$$f_t=-\dot{m}v_\mathrm{ex}$$

The homework statement isn't exactly as is mentioned above. The actual problem statement is as follows:

A rocket (initial mass $m_0$) needs to use its engines to hover stationary, just above the ground. (a) If it can afford to burn no more than a mass $\lambda m_0$ of its fuel, for how long can it hover?

This is problem 3.8 from John R. Taylor's Classical Mechanics; however, my question is not related to the main problem itself but one particular aspect of it. Now, in the same textbook (John R. Taylor's Classical Mechanics), the equation for the thrust of a rocket has been derived as follows:

Let $m$ be the mass of the rocket at a time $t$, that is, the momentum of the rocket at this point is $P(t)=mv$. At time $t+\mathrm{dt}$, let the mass of the rocket be $m+\mathrm{d}m$ (where $\mathrm{d}m$ is negative), that is, the total momentum of the rocket and the fuel is $P(t+\mathrm{d}t)=(m+\mathrm{d}m)(v+\mathrm{d}v)-\mathrm{d}mv_\mathrm{ex}$. Now, assuming there is no external force on the rocket, we can write $\mathrm{d}P$ as:

$$\mathrm{d}P=(m+\mathrm{d}m)(v+\mathrm{dv})-mv-\mathrm{d}mv_\mathrm{ex}$$

However, since there is no external force acting on the rocket, $\mathrm{d}P=0$; therefore, we get:

$$m\mathrm{d}v=-\mathrm{d}mv_\mathrm{ex}$$

Dividing both sides by $\mathrm{d}t$ (the book uses the differential of a quantity as a finite term and just mentions that it is justified):

$$m\dot{v}=-\dot{m}v_\mathrm{ex}$$

Here, $-\dot{m}v_\mathrm{ex}$ has been defined as the magnitude of the thrust force. However, the important point to note here is that this formular has been derived assuming no external force (including gravity). Now, in the original question, the following hint has been given on how to solve the problem:

Write down the condition that the thrust just balance the force of gravity. You can integrate the resulting equation by separating the variables $t$ and $m$. Take $v_\mathrm{ex}$ to be constant.

That is, it wants me to balance out the thrust and gravitational force using the above derived formula for thrust (which has been derived in the case of no external force). How are we allowed to do that given that in this case there is an explicitly acting gravitational force?

 

[andrewkirk]

How long it can hover depends on the downward speed $v$ of the gaseous fuel it ejects, which will depend on the usable energy density relating to the ejected fuel. The problem needs to specify that density in order to allow the reader to find a solution.
Given such a density $A$, we can write the energy conservation equation:
$$A\ dm = \frac12 v^2\, dm$$
where the left side shows the energy consumed in period dt and the right side shows the kinetic energy of the ejected gas.
You can use that to work out $v$.
Then you need to analyse impulse and momentum to write a differential equation for $m$, which you can solve. You only need to worry about the momentum of the ejected gas, as the momentum of the rocket remains zero.

 

[f3sicA_A]

How long it can hover depends on the downward speed $v$ of the gaseous fuel it ejects, which will depend on the usable energy density relating to the ejected fuel. The problem needs to specify that density in order to allow the reader to find a solution.

My apologies, I should have mentioned that the problem does not provide a numerical value for the exhaust speed of the gaseous fuel the rocket ejects; however, it requires us to assume that the value is a constant $v_\mathrm{ex}$.

Then you need to analyse impulse and momentum to write a differential equation for $m$, which you can solve.

I do not see the need for bringing in impulse when momentum alone suffices (from which we can find the thrust force), and using the thrust force, we can balance the forces as such:

$$T=mg$$

Where $T$ is the thrust force. However, my main question still remains as to what the magnitude of this thrust force would be. As shown from the derivation in the original problem, the magnitude of the thrust force is $T=-\dot{m}\cdot v_\mathrm{ex}$; however, this formula is derived assuming there is no gravitational force acting on the rocket (which is clearly not the case here) but the problem requires me to use the balance equation $-\dot{m}\cdot v_\mathrm{ex}=mg$, upon solving which, I get the following hover time:

$$t=-\frac{v_\mathrm{ex}}{g}\ln{(1-\lambda)}$$

 

[jbriggs444]

Where $T$ is the thrust force. However, my main question still remains as to what the magnitude of this thrust force would be. As shown from the derivation in the original problem, the magnitude of the thrust force is $T=-\dot{m}\cdot v_\mathrm{ex}$; however, this formula is derived assuming there is no gravitational force acting on the rocket (which is clearly not the case here) but the problem requires me to use the balance equation $-\dot{m}\cdot v_\mathrm{ex}=mg$, upon solving which, I get the following hover time:

$$t=-\frac{v_\mathrm{ex}}{g}\ln{(1-\lambda)}$$

I do not understand your concern. If the rocket is hovering with a fixed $v_\mathrm{ex}$ then clearly $\dot{m}$ is being continuously adjusted to produce that result. The balance equation is what you get when you do a force balance on the hovering rocket.

 

[f3sicA_A]

I do not understand your concern. If the rocket is hovering with a fixed $v_\mathrm{ex}$ then clearly $\dot{m}$ is being continuously adjusted to produce that result. The balance equation is what you get when you do a force balance on the hovering rocket.

Right, my question is simply that how do we know that the magnitude of the thrust force is given by $T=-\dot{m}v_\mathrm{ex}$. As far as my understanding goes, this equation for the magnitude of thrust force is derived by assuming that there is no gravitational force acting on the rocket (as shown in my derivation in the original question), how do we know that the magnitude of the thrust force remains the same irrespective of whether there are external forces such as gravitational force acting on the rocket or not?

 

[haruspex]

As far as my understanding goes, this equation for the magnitude of thrust force is derived by assuming that there is no gravitational force acting on the rocket

No, the thrust is purely the force provided by the rocket engine. It does not depend on what other forces may be present. The net force is thrust+gravity+ whatever.

 

[Filip Larsen]

however, this formula is derived assuming there is no gravitational force acting on the rocket (which is clearly not the case here)

The idea, I guess, is to derive the thrust force relationship assuming no external forces and then apply this thrust in situations where there are other forces, like gravity. You can also add gravity already to derivation of the thrust and arrive at the same result, thus concluding that while the rocket acceleration clearly is different with and without gravity the magnitude of the thrust itself is independent of this.

 

[f3sicA_A]

The idea, I guess, is to derive the thrust force relationship assuming no external forces and then apply this thrust in situations where there are other forces, like gravity. You can also add gravity already to derivation of the thrust and arrive at the same result, thus concluding that while the rocket acceleration clearly is different with and without gravity the magnitude of the thrust itself is independent of this.

I see, I will try to derive the thrust equation again whilst considering the force of gravitation. Thank you for your response!

 

[kuruman]

Your question is like saying "I saw in my textbook that the net force on a mass hanging from a string is $T-mg.~$ I understand that the weight of an object in free fall is $mg.$ How are we allowed call the weight $mg$ given that in this case there is an explicitly acting force of tension?"

 

[f3sicA_A]

Your question is like saying "I saw in my textbook that the net force on a mass hanging from a string is $T-mg.~$ I understand that the weight of an object in free fall is $mg.$ How are we allowed call the weight $mg$ given that in this case there is an explicitly acting force of tension?"

Whilst your analogy now makes sense, that is only because I have been able to make sense of the math which proves that irrespective of whether gravitational force acts on the rocket or not, which I initially couldn't, and hence this thread. However, now that I have worked through the problem as follows:

$$P(t)=mv$$
$$P(t+\mathrm{d}t)=(m+\mathrm{d}m)(v+\mathrm{d}v)-\mathrm{d}m(v-v_\mathrm{ex})$$
$$\implies\mathrm{d}P=m\mathrm{d}v+v_\mathrm{ex}\mathrm{d}m$$
$$\implies-mg\mathrm{d}t=m\mathrm{d}v+v_\mathrm{ex}\mathrm{d}m$$
$$\implies m\dot{v}=-v_\mathrm{ex}\dot{m}-mg$$

Where $-v_\mathrm{ex}\dot{m}$ still remains to be the thrust, clearly unchanged irrespective of whether or not there is a gravitational force acting on the rocket. Again, as I mentioned earlier, I wasn't able to see this line of reasoning up until now; however, thanks to @Filip Larsen suggesting that I should try to work it out with gravitational force involved, I realized where I was going wrong.

 

[kuruman]

Whilst your analogy now makes sense, that is only because I have been able to make sense of the math $\dots$

I am trying to convey the idea that you shouldn't need to redo the math for the thrust with gravity included because of superposition. When you write the net force, $$\mathbf{F}_{\text{Net}}=\sum_i\mathbf{F}_i$$superposition allows you to do the math for each force separately and then add all the forces together as vectors to find the net force. This means that in this case, when you turn "on" gravity, all you have to do is tack on the weight of the rocket acting in an opposite direction to the thrust.