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What formula can be used in order to find the final velocity of a rocket if the time travelled by the rocket, exhaust velocity, initial and final masses of the rocket and the initial acceleration of the rocket is known (gravity is assumed to be constant)?

I came across the Tsiolkovsky rocket equation --> $V_f= V_e \ln \frac {m_0}{m_f}$. But why doesn't it include the time travelled by the rocket? is there another formula which also includes the time? Since the time travelled is also known to me in the particular case and I felt like I should also include it in a formula to reach the final velocity value.

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  • $\begingroup$ Hi Taylan, and welcome to the site! It looks like the equation you found satisfies your condition - that is, if you know the exhaust velocity, initial and final masses, and initial acceleration, it looks like you can use the equation you found to determine the final velocity. Could you elaborate more on why you think it shouldn't work? Or, if you really mean to ask why that equation doesn't include the time (which would be a good question that stands on its own), could you edit to make it more clear that that's the focus of your question? $\endgroup$
    – David Z
    Commented Nov 25, 2018 at 12:41
  • $\begingroup$ Hi David, thank you! I just edited the question to try to make it more clear. $\endgroup$ Commented Nov 25, 2018 at 12:48

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The mass of the rocket at a specific time is $m_0 - \alpha t$, where $\alpha$ is the rate of mass loss, so you can substitute this into the rocket equation to get $v(t) = v_e \ln \frac{m_0}{m_0 - \alpha t}$. The final velocity can then be obtained by checking the rocket at the time when it's ran out of fuel; this happens at $\alpha t = m_0 - m_f$.

The rate of mass loss is also known because you know the initial acceleration. From Newton's law, $m_0 a = \frac{dp}{dt}$, and in this case the momentum change comes from ejecting fuel, so $a = \frac{\frac{dm}{dt} v_e}{m_0}$, so $\alpha = \frac{m_0 a}{v_e}$.

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    $\begingroup$ Note this is only true if mass loss rate is constant. $\endgroup$ Commented Nov 25, 2018 at 14:30
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Actually this equation comes from force equation of a variable mass system. It assumes that mass leaks at constant rate with a certain and a constant velocity relative to rocket which provides the thrust. Now you must certainly know the rate at which mass leaks. And of course you know the initial mass as $(M=M_0-rt)$ where $r$ is the rate and $t$ is time

Suppose the rate is not constant, then neglect gravity and apply conservation of momentum in vertical direction at a later time $t$ and at initial time. Find the equation of motion by solving the equation.

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Here: a nice clean derivation will help make things a lot clearer; to show that under the condition of constant thrust (and throttling) in a constant direction, the vector $๐ฏ - ๐ฎ\ln\left({M/M_u}\right)$ is a constant of motion - i.e. independent of time, where $M$ is the rocket's (time-varying) mass, $M_u$ is some fixed reference mass (e.g. $M_u = 1\text{ kg}$), $๐ฏ$ the rocket's speed and $๐ฎ$ the the thrust velocity relative to the rocket. Apparently, your $๐ฏ_e = -๐ฎ$.

How is the Rocket Equation derived?
Like this. Let $๐$ and $๐ฉ$ be the vectors respectively denoting the momentum of the rocket and the total momentum of all the fuel thatโ€™s been expelled up to a given moment, let $M$ and $m$ denote their respective masses and let the vector $๐ฏ$ denote the rocketโ€™s velocity. For simplicity, ignore the effect of gravity or other external forces, even though weโ€™ll keep them in initially.

Propulsion occurs with part of the rocketโ€™s mass (the unused fuel) becoming the mass of the expelled fuel; similarly for the momentum. Let the respective rates of transfer be $ฮผ$ and $๐ฅ$. The latter quantity - the rate at which momentum goes from being part of the rocket (the unspent fuel) to becoming part of the spent fuel is one and the same as the force of propulsion of the rocket - oppositely directed (law of action and reaction).

Therefore, you can write the following system $$๐ = M๐ฏ, \quad \frac{dM}{dt} = โˆ’ฮผ, \quad \frac{dm}{dt} = ฮผ, \quad \frac{d๐}{dt} = โˆ’๐ฅ + ๐…, \quad \frac{d๐ฉ}{dt} = ๐ฅ + ๐Ÿ,$$ where $๐…$ and $๐Ÿ$ are, respectively, the external forces acting on the rocket and the spent fuel. For instance, in the presence of a (near-)constant gravitational acceleration $๐ $, we would have $๐… = M๐ $ and $๐Ÿ = m๐ $. But, weโ€™ll ignore the external force and just set $๐  = ๐Ÿฌ$ and $๐… = ๐Ÿฌ = ๐Ÿ$.

Notice that for the total system - rocket plus expelled fuel - we have $$\frac{d(M + m)}{dt} = 0, \quad \frac{d(๐ + ๐ฉ)}{dt} = ๐… + ๐Ÿ = ๐Ÿฌ.$$ So, past this point, we donโ€™t really need to consider $m$ or $๐ฉ$ anymore - though the people who paid for all that spent fuel do. So, they can be removed from the picture.

We will also assume that the fuel expelled from the rocket is expelled at a given โ€œthrustโ€ velocity $๐ฎ$ relative to the rocket, itself. That means its velocity is $๐ฏ + ๐ฎ$. The direction of the vector $๐ฎ$ is the direction of the thrust, and it is normally assumed that the corresponding speed $u = |๐ฎ|$ is fixed - though there may be some ability to control the level of throttling to adjust $u$. Then, we have the following additional relation $$๐ฅ = ฮผ(๐ฏ + ๐ฎ),$$ which expresses the rate of momentum expelled in the thrust in terms of the rate of expelled mass multiplied by the velocity of the thrust.

Substitute for $๐$ and $๐ฅ$ to obtain the following: $$\frac{dM}{dt} = โˆ’ฮผ, \quad \frac{d(M๐ฏ)}{dt} = โˆ’ฮผ(๐ฏ + ๐ฎ),$$ or upon combining the two: $$\frac{d(M๐ฏ)}{dt} = \frac{dM}{dt}(๐ฏ + ๐ฎ).$$

Use the product rule, $$\frac{d(M๐ฏ)}{dt} = \frac{dM}{dt}๐ฏ + M\frac{d๐ฏ}{dt},$$ substitute this in to get: $$\frac{dM}{dt}๐ฏ + M\frac{d๐ฏ}{dt} = \frac{dM}{dt}(๐ฏ + ๐ฎ),$$ and cancel out the $(dM/dt)๐ฏ$ from both sides to obtain: $$M\frac{d๐ฏ}{dt} = \frac{dM}{dt}๐ฎ.$$

Since $d(lnM) = dM/M$, we can re-write this as: $$\frac{d๐ฏ}{dt} = \frac{d(lnM)}{dt}๐ฎ,$$ So, if we assume the thrust has a constant throttling in a constant direction over a period of time - so that the vector $๐ฎ$ remains constant over that period - then we obtain (upon integration), the following: $$๐ฏ_1 โˆ’ ๐ฏ_0 = \left(\ln{M_1} โˆ’ \ln{M_0}\right)๐ฎ = \ln{\left(\frac{M_1}{M_0}\right)}๐ฎ,$$ where $\left(M_0, ๐ฏ_0\right)$ are the rocket's (mass, velocity) at the beginning of the interval and $\left(M_1, ๐ฏ_1\right)$ are the (mass, velocity) at the end of the interval. Note, in particular, that $M_0 โˆ’ M_1$ is the total mass of the fuel expelled during the thrust, so that $M_0 > M_1$ and $\ln\left(M_1/M_0\right) < 0$. Therefore, the net gain in the rocketโ€™s velocity $๐ฏ_1 โˆ’ ๐ฏ_0$ is in the opposite direction than the thrust velocity $๐ฎ$.

Thus, under the condition of constant thrust and throttling in a constant direction $$๐ฏ_1 - ๐ฎ\ln{M_1} = ๐ฏ_0 - ๐ฎ\ln{M_0},$$ i.e. $$๐ฏ - ๐ฎ\ln{M}$$ is a constant of motion, where - for dimensional consistency - we take the "$M$" in "$\ln{M}$" with reference to a fixed unit $M_u$, e.g. $M_u = 1\text{ kg}$ and write it as $\ln{M/M_u}$.

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    $\begingroup$ What does it look like when the thrust or throttling is not constant? $\endgroup$ Commented Oct 15, 2023 at 3:15
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    $\begingroup$ Good question! Since it was sorta the original question, I'll think about adding it in as an edit. $\endgroup$
    – NinjaDarth
    Commented Oct 16, 2023 at 20:56

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