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f3sicA_A

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- 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:

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,

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

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:

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?

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?