Answer: Energy of Rocket in Uniform Gravity Field

In summary, The change in total energy of the rocket during its motion in a uniform gravitational field can be found by solving the equation ##\frac{dE}{dt}=\frac{dm}{dt}(gz+\frac{v^2}{2}-uv)## where ##u## - is the velocity of burned propellant relative to the rocket. The problem is that the condition of the problem is not correct. For solutions need to know either the ##u##or acceleration of the rocket ##\ddot z## .
  • #1
sergiokapone
302
17

Homework Statement


How does change the total energy of the rocket during its motion inin a uniform gravitational field?

2. The attempt at a solution
My idea is to write the law of conservation of energy systems - "rocket- gases" ##\frac{dE_{total}}{dt}=0##
But get very cumbersome terms , the physical sense , I can not understand.
Besides, it is not clear how to express the change of potential energy of emitted gases .
 
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  • #2
sergiokapone said:
How does change the total energy of the rocket during its motion inin a uniform gravitational field?

I'm sorry, what does this mean?
 
  • #3
##E=\frac{mv^2}{2}+mgz##
 
  • #4
I found
##\frac{dE}{dt}=\frac{dm}{dt}(gz+\frac{v^2}{2}-uv)##
 
  • #5
Ok, my problem is:
Find mass vs time for the rocket, which is a rise in a uniform gravitational field , if the value ##-u^2 \dot m ## is constant ?
where ##u## - is the velocity of burned propellant relative to the rocket.
I use
##m \ddot z=-mg-u\dot m##
But I have no idea how to find ##m(t)##.
 
  • #6
sergiokapone said:
I found
##\frac{dE}{dt}=\frac{dm}{dt}(gz+\frac{v^2}{2}-uv)##
Why are you treating velocity as constant? I'm pretty sure the only "thing" with constant velocity is the ejected mass from the rocket.
sergiokapone said:
Ok, my problem is:
##m \ddot z=-mg-u\dot m##
So the force that the ejected matter exerts on the rocket is in the same direction as the force that the Earth exerts on the rocket?
 
  • #7
*comment retracted*
 
  • #8
BiGyElLoWhAt said:
So the force that the ejected matter exerts on the rocket is in the same direction as the force that the Earth exerts on the rocket?
Of course not, because ##-u\dot m## is positive value, because ##\dot m## - is negative.

BiGyElLoWhAt said:
Why are you treating velocity as constant?
Velocity ##v(t)## is unknovn.
 
  • #9
sergiokapone said:
Ok, my problem is:
Find mass vs time for the rocket, which is a rise in a uniform gravitational field , if the value ##-u^2 \dot m ## is constant ?
where ##u## - is the velocity of burned propellant relative to the rocket.
I use
##m \ddot z=-mg-u\dot m##
But I have no idea how to find ##m(t)##.
If ##u^2\dot{m}## is constant what do you know about m? Is there any reason why u should vary?
Also, from your energy equation, you have a product on the RHS, you have to treat it as such.
 
  • #10
It seems to me that the condition of the problem is not correct. For solutions need to know either the ##u##or acceleration of the rocket ##\ddot z## .
 
  • #11
sergiokapone said:
It seems to me that the condition of the problem is not correct. For solutions need to know either the ##u##or acceleration of the rocket ##\ddot z## .
I believe there is enough information if you assume a constant value for the energy density of the fuel.
 
  • #12
I solved this promblem and get:

##m=m_0\left[1+\frac{m_0}{2P}\int_{0}^t{\left(\ddot z + g\right) ^2 dt}\right]^{-1}##

haruspex said:
I believe there is enough information if you assume a constant value for the energy density of the fuel.

I think that if the rocket engine power is constant ##-\frac{\dot m u^2}{2}##, the astronaut who has to manage a rocket, need to change the velocity of the propelant ##u## . Thus, the astronaut can control the rocket, ie change the velocity by your own (##\ddot z##). So to solve the problem, we must assume that ##u## (or ##\ddot z##) is independent variable. I assumed that ##\ddot z## is such.
 
  • #13
sergiokapone said:
I solved this problem and get:

##m=m_0\left[1+\frac{m_0}{2P}\int_{0}^t{\left(\ddot z + g\right) ^2 dt}\right]^{-1}##
I think that if the rocket engine power is constant ##-\frac{\dot m u^2}{2}##, the astronaut who has to manage a rocket, need to change the velocity of the propelant ##u## .
I don't see how the astronaut can change the velocity of the exhaust if both the power and the energy density are fixed. Isn't the energy density ##\frac 12 u^2##? In other words, it seems to me ##\dot m## is constant.
 
  • #14
haruspex said:
I don't see how the astronaut can change the velocity of the exhaust if both the power and the energy density are fixed. Isn't the energy density ##\frac 12 u^2##? In other words, it seems to me ##\dot m## is constant.

As stated in the problem statement - ##\dot m \frac 12 u^2## is the power of the rocket engine, and it is constant. Thus, astronaut can manage by changing ##u##.
 
  • #15
sergiokapone said:
As stated in the problem statement - ##\dot m \frac 12 u^2## is the power of the rocket engine, and it is constant.
Yes, I understand that.
Thus, astronaut can manage by changing ##u##.
That was not stated in post #5. You added that at post #12 as an opinion. I'm saying that if the energy density is constant then there is no apparent ability to vary u. Let that density be ##\rho##:
##\rho (-\dot m) = power = -\frac 12 \dot m u^2##
 
  • #16
haruspex said:
I'm saying that if the energy density is constant then there is no apparent ability to vary u.
##\rho## - is not an energy density, because it has dimension ##J\cdot kg^{-1} = (m/s)^2## - specific kinetic energy of the fuel or velocity squared.
 
  • #17
sergiokapone said:
##\rho## - is not an energy density, because it has dimension ##J\cdot kg^{-1} = (m/s)^2## - specific kinetic energy of the fuel or velocity squared.
I don't understand your point. My ##\rho## is an energy density because I defined it that way. In my equation it has dimension of velocity squared, as you say. Where's the inconsistency?
 

1. How is the energy of a rocket determined in a uniform gravity field?

The energy of a rocket in a uniform gravity field is determined by the sum of its kinetic energy and potential energy. Kinetic energy is the energy of motion, while potential energy is the energy that an object has due to its position or configuration.

2. What factors affect the energy of a rocket in a uniform gravity field?

The energy of a rocket in a uniform gravity field is affected by its mass, velocity, and position in the field. The greater the mass and velocity of the rocket, the more energy it will have. Additionally, the higher the rocket is in the field, the greater its potential energy will be.

3. How does the energy of a rocket change during flight in a uniform gravity field?

The energy of a rocket changes during flight in a uniform gravity field due to the conversion between kinetic and potential energy. As the rocket accelerates upwards, its kinetic energy decreases and its potential energy increases. When the rocket falls back down, the opposite occurs.

4. How does the energy of a rocket affect its flight trajectory in a uniform gravity field?

The energy of a rocket affects its flight trajectory in a uniform gravity field because it determines the rocket's speed and height at any given point. A rocket with more energy will have a higher speed and be able to reach greater heights, while a rocket with less energy will have a lower speed and reach lower heights.

5. Can the energy of a rocket in a uniform gravity field be conserved?

The energy of a rocket in a uniform gravity field can be conserved in an ideal scenario where there is no external force acting on the rocket. In this case, the total energy (kinetic + potential) will remain constant throughout the rocket's flight. However, in reality, there are always external forces such as air resistance that will cause the energy to decrease over time.

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