A rocket and its gas exhaust velocity

In summary: The mass of the massive object influences the trajectory of the object, as the initial trajectory of the object (circumference centred in the massive object) is changed by the mass of the massive object.
  • #1
Guillem_dlc
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Homework Statement
Along the present problem you may assume that there is no air friction, that the ignition processes of the different rocket phases are instantaneous, and that fuel capsules have negligible mass.

a) Consider a 1-phase rocket such that its gas exhaust velocity is [itex] u [/itex], its initial mass (payload plus fuel) is [itex] m_0 [/itex], and its final mass (payload) is [itex] m_f [/itex]. Assuming that the rocket is vertically launched from the Earth's surface, and that the Earth is at rest, determine the amount of fuel required for the rocket to reach a maximum height [itex] h [/itex].
Relevant Equations
Energy conservation
This isn't right, is it?
[tex] -\dfrac{GM}{R}+\dfrac12 v^2=-\dfrac{GM}{R+h}[/tex]
[tex] v=\sqrt{\dfrac{GM}{R}}\left( 1-\sqrt{\dfrac{R}{R+h}}\right) [/tex]
He's doing energy conservation. The mechanical energy at the Earth's surface is equal to the energy when the speed is [itex]0[/itex].
 
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  • #2
Then how do you determine the amount of fuel required to v ?
 
  • #3
Guillem_dlc said:
This isn't right, is it?
[tex] -\dfrac{GM}{R}+\dfrac12 v^2=-\dfrac{GM}{R+h}[/tex]
[tex] v=\sqrt{\dfrac{GM}{R}}\left( 1-\sqrt{\dfrac{R}{R+h}}\right) [/tex]
Don't worry about the rocket part, it's just this step
 
  • #4
anuttarasammyak said:
Then how do you determine the amount of fuel required to v ?
You can find the speed with
[tex] \Delta V=u\cdot \ln \left( \dfrac{m_0}{m_f}\right) [/tex]
It's Tsiolkovski's equation.
 
  • #5
Guillem_dlc said:
He's doing energy conservation. The mechanical energy at the Earth's surface is equal to the energy when the speed is [itex]0[/itex].
Yes. In order for this step to become meaningful, one has to reason a bit about the problem. What pattern of burn will give you the best performance?

Do you burn steadily from surface to final altitude?
Do you hover for half an hour on your rockets and then burn hard and fast?
Do you burn halfway up, then turn off the rockets and coast?
Do you burn as hard and as fast as you can immediately and then coast the rest of the way?

Two heuristics help make that decision easy. First, the longer your flight takes, the more time the Earth's gravity has to decrease your upward velocity. Second, any excess velocity you have when you get to the goal is wasted.
 
  • #6
Tsiolkovski's equation is derived from the law of conservation of momentum in case of no external forces. I wonder if it stands with gravity force working.

I agree with v of OP as the initial speed of projectile popped up with no self propulsion. So the case is the rocket gets speed v by instantaneous burning of all the fuel at launching, like a cannon or a bullet.
 
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  • #7
I believe the question being asked by @Guillem_dlc is purely in regard to this algebraic step:
Guillem_dlc said:
[tex] -\dfrac{GM}{R}+\dfrac12 v^2=-\dfrac{GM}{R+h}[/tex]
[tex] v=\sqrt{\dfrac{GM}{R}}\left( 1-\sqrt{\dfrac{R}{R+h}}\right) [/tex]
And I agree, that is quite wrong. It should yield
##v=\sqrt{\frac{2GMh}{R(R+h)}}##.

I did wonder if it is a defensible approximation, but it does not appear to be.
 
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  • #8
anuttarasammyak said:
I agree with v of OP as the initial speed of projectile popped up with no self propulsion. So the case is the rocket gets speed v by instantaneous burning of all the fuel at launching, like a cannon or a bullet.
Yes, in this problem it is considered to burn all the fuel instantly at first and continue until gravity cancels out all the initial speed.
 
  • #9
haruspex said:
I did wonder if it is a defensible approximation, but it does not appear to be.
I also think that, but I think it might indicate the speed variation.

Otherwise, there's no meaning in the minus sign.

If you do the velocity variation considering circular trajectories I think that's correct. Since the speed: [tex] V=\sqrt{\dfrac{GM}{R}} [/tex]

Although not on top it shouldn't have speed
 
  • #10
Guillem_dlc said:
Yes, in this problem it is considered to burn all the fuel instantly at first and continue until gravity cancels out all the initial speed.
OK. So you should consider how much fuel of relative speed u should go out instantly so that the rocket body gets velocity v, not by the post #4.
 
  • #11
I think the algebra is wrong there (or there is a typo), it should have been
$$v=\sqrt\frac{2GM}{R}\sqrt{1-\frac{R}{R+h}}$$ which yields the same expression as that of post #7

And i don't know if we can use the approximation $$\sqrt{1-\frac{R}{R+h}}\approx \frac{1}{\sqrt{2}}\left (1-\sqrt\frac{R}{R+h}\right )$$ but that's the only way to justify the book's formula.
 
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  • #12
anuttarasammyak said:
OK. So you should consider how much fuel of relative speed u should go out instantly so that the rocket body gets velocity v, not by the post #4.
Yes
 
  • #13
Delta2 said:
which yields the same expression as that of post #7
And here you substitute the velocity for the expression of [itex] u\cdot \ln \left( \dfrac{m_o}{m_f}\right) [/itex] and isolate the mass
 
  • #14
"I) Transfer from initial (circular) to transfer (elliptical) orbit.
- The initial trajectory of [itex] m [/itex] is a circumference centred in the massive object [itex] M [/itex] of radius [itex] r=r_p [/itex]. The velocity of [itex] m [/itex] is constant, and can be easily obtained as follows:
[tex] e=-\dfrac{GM}{2r_p}=\dfrac12 v_p^2-\dfrac{GM}{r_p} [/tex]
[tex] v_p=\sqrt{\dfrac{GM}{r_p}}" [/tex]
Perhaps it is related to this as the [itex]2[/itex] seems to disappear, and the rocket always makes an elliptical orbit when it leaves the Earth.
 
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  • #15
Guillem_dlc said:
"I) Transfer from initial (circular) to transfer (elliptical) orbit.
- The initial trajectory of [itex] m [/itex] is a circumference centred in the massive object [itex] M [/itex] of radius [itex] r=r_p [/itex]. The velocity of [itex] m [/itex] is constant, and can be easily obtained as follows:
[tex] e=-\dfrac{GM}{2r_p}=\dfrac12 v_p^2-\dfrac{GM}{r_p} [/tex]
[tex] v_p=\sqrt{\dfrac{GM}{r_p}}" [/tex]
Perhaps it is related to this as the [itex]2[/itex] seems to disappear, and the rocket always makes an elliptical orbit when it leaves the Earth.
Possibly, but the problem statement does not begin with a circular ballistic orbit. It begins from the surface of the Earth.
 
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  • #16
jbriggs444 said:
Possibly, but the problem statement does not begin with a circular ballistic orbit. It begins from the surface of the Earth.
Right. Well, it must be mistaken.
 

1. What is a rocket's gas exhaust velocity?

A rocket's gas exhaust velocity is the speed at which the gas is expelled from the rocket's engine. This velocity is typically measured in meters per second (m/s) or kilometers per second (km/s).

2. How does a rocket's gas exhaust velocity affect its performance?

A rocket's gas exhaust velocity is a crucial factor in determining its performance. A higher exhaust velocity means the rocket can achieve greater speeds and travel further distances. It also allows for a more efficient use of fuel, resulting in longer flight times.

3. What factors affect a rocket's gas exhaust velocity?

The gas exhaust velocity of a rocket is affected by several factors, including the type of fuel used, the design of the engine, and the ambient pressure and temperature of the environment. The efficiency of the rocket's nozzle also plays a significant role in determining the exhaust velocity.

4. How is a rocket's gas exhaust velocity calculated?

The gas exhaust velocity of a rocket can be calculated using the ideal rocket equation, which takes into account the mass of the rocket, the mass of the fuel, and the change in velocity. This equation is typically used in conjunction with other performance metrics to determine the overall effectiveness of a rocket.

5. Can a rocket's gas exhaust velocity be increased?

Yes, a rocket's gas exhaust velocity can be increased through various means, such as using more efficient fuels, improving the design of the engine and nozzle, and optimizing the ambient conditions. However, increasing the exhaust velocity often comes at the cost of added complexity and expense.

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