Solving the Rocket Equation: Confusion with Signs

In summary, the conversation discusses the derivation of the rocket equation and the confusion surrounding the use of signs. It explains how the conservation of momentum is used to derive the equation and the importance of understanding the direction of the velocity and mass changes in the equation. The conversation also mentions the use of unit vectors to avoid issues with dividing by a vector.
  • #1
calculus_jy
56
0
recenty i read the rocket equation, derivation of, however i think i have a slight confusion with signs
suppost initially a rocket has
mass= [tex]M[/tex]
velocity= [tex]\overrightarrow{v}[/tex]
then at a time dt later,
mass of rocket= [tex]M-dM[/tex]
velocity of rocket= [tex]\overrightarrow {v} +d\overrightarrow {v} [/tex]
mass of ejacted gas= [tex]dM[/tex]
velocity of gas= [tex]\overrightarrow{u}[/tex]

using conservation of momentum
[tex]\overrightarrow{v}M=(M-dM)(\overrightarrow{v}+d\overrightarrow{v})+\overrightarrow{u}dM[/tex]

[tex](\overrightarrow{u}-\overrightarrow{v})dM+Md\overrightarrow{v}=0[/tex]

but [tex](\overrightarrow{u}-\overrightarrow{v})[/tex]=velocity of gas relative to rocket
let [tex](\overrightarrow{u}-\overrightarrow{v})=\overrightarrow{U}[/tex]which is a constant

[tex]\overrightarrow{U}dM+Md\overrightarrow{v}=0[/tex]
[tex]-\int_{M_0}^{M}\frac{dM}{M}=\frac{1}{\overrightarrow{U}}\int_{\overrightarrow{v}_0}^{\overrightarrow{v}}d\overrightarrow{v}[/tex]

now [tex]-ln\frac{M}{M_0}=\frac{\overrightarrow{v}-\overrightarrow{v_0}}{\overrightarrow{U}}[/tex]

the problem is , when taking the velocity in the direciton rocket is travelling
[tex]\overrightarrow{U}<0[/tex]
[tex]-ln\frac{M}{M_0}>0[/tex]since [tex]\frac{M}{M_0}<1[/tex]
then
[tex]\overrightarrow{v}-\overrightarrow{v_0}<0[/tex] which is impossibe as the rocket is accelerating?
 
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  • #2
latex problem seems to be fixed now...
 
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  • #3
i don't see how you get [tex]
M- dM
[tex] is is because of [tex]E=mc^2[tex]
 
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  • #4
why won't the latex work
 
  • #5
Poincare1 said:
why won't the latex work

I hope this isn't too off topic but

i think you need [/tex] instead of [tex] at the end
 
  • #6
calculus_jy said:
[tex]\overrightarrow{U}dM+Md\overrightarrow{v}=0[/tex]
Good up to this point.
[tex]-\int_{M_0}^{M}\frac{dM}{M}
=\frac{1}{\overrightarrow{U}}
\int_{\overrightarrow{v}_0}^{\overrightarrow{v}}d\overrightarrow{v}[/tex]
This step is not valid. There is no such thing as the multiplicative inverse of a vector. Better is to define a unit vector v-hat directed along the rocket's delta-v vector. From the correct equation, this delta-v vector is directly opposed to the relative exhaust velocity vector. Thus

[tex]
\begin{aligned}
d\overrightarrow{v} &= dv \hat{v} \\
\overrightarrow{U} &= U \hat{v} & (U &\equiv \overrightarrow{U}\cdot \hat v)\\
&= -v_e \hat{v} & (v_e&\equiv -U)
\end{aligned}[/tex]

Note that ve is simply the magnitude of the relative velocity vector. With this, the vector differential equation becomes the scalar equation

[tex]-v_edM+Mdv=0[/tex]

from which

[tex]\int_{M_0}^{M}\frac {dM}{M} = \frac 1{v_e}\int_{v_0}^v dv[/tex]

or

[tex]\ln\frac{M}{M_0} = \frac{\Delta v}{v_e}[/tex]

You can use a vector formulation, but you can't divide by a vector like you did.
 
  • #7
however is the equation
[tex]v=v_0+\overrightarrow{U}ln\frac{M}{M_0}[/tex] not right?using the notion in the first post
in the step with the integrals, to prevent multiplicative of inverse of vector, simply put [tex]\overrightarrow{U}[/tex]
on the same side of the equation as [tex]\frac{dM}{M}[/tex]
i have been told that its the problem associated with dM such that the mass of rocket after dt is M+dM not M-dM
i don't get why minus can be used in scalar, but cannot be used in vercot derivation?
 
  • #8
Your error occurs much earlier than I stated earlier.

calculus_jy said:
using conservation of momentum
[tex]\overrightarrow{v}M=(M-dM)(\overrightarrow{v}+d\overrightarrow{v})+\overrightarrow{u}dM[/tex]
Here your dM is the quantity of mass ejected by the spacecraft . With this definition, a positive dM means the spacecraft loses mass. Things would have worked properly if you had used dM as positive meaning the spacecraft gains mass. Then the conservation of momentum equation becomes

[tex]\overrightarrow{v}M=(M+dM)(\overrightarrow{v}+d\overrightarrow{v})-\overrightarrow{u}dM[/tex]
 

1. What is the rocket equation and why is it important?

The rocket equation, also known as the Tsiolkovsky rocket equation, is a mathematical formula that describes the motion of a rocket in terms of its mass and the force it generates. It is important because it allows scientists and engineers to calculate the necessary parameters for a rocket to reach a desired velocity and altitude.

2. How is the rocket equation used in rocket science?

The rocket equation is used in rocket science to determine the amount of propellant needed for a rocket to reach a specific velocity and altitude. It is also used to calculate the efficiency of a rocket engine and to design the trajectory of a spacecraft.

3. What are the main components of the rocket equation?

The main components of the rocket equation are the velocity of the rocket, the mass of the rocket (including fuel), and the specific impulse of the rocket engine. These parameters are used to calculate the change in velocity (delta-v) that a rocket can achieve.

4. How does confusion with signs occur in the rocket equation?

Confusion with signs can occur in the rocket equation when the direction of velocity and acceleration are not clearly defined. This can lead to errors in calculations and can affect the accuracy of the results.

5. How can confusion with signs be avoided in the rocket equation?

To avoid confusion with signs in the rocket equation, it is important to clearly define the direction of velocity and acceleration. This can be done by using a consistent coordinate system and making sure that all parameters are represented with the correct signs (+ or -) in the equation.

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