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The Tsiolkovsky rocket equation is derived for a rocket under no external forces. No drag, no gravity.

$$\frac{\Delta v}{v_{exhaust}}=ln\frac{m_{initial}}{m_{final}}$$

While sometimes g appears, that only happens when someone is using specific impulse instead of exhaust velocity.

I remember seeing the derivation of a variant of this equation where a constant external force is included. That could be for example gravity near the surface of a body during a vertical take-off (ok ok... at the north pole).

I don't want to debate the usefulness of this variant, I would just like to see the derivation and possibly a link to its use. I know it was closed form, and required $t_{burn}$ explicitly.

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  • $\begingroup$ $...-g_0 \Delta t$ ? That would be for vertical polar launch. Note how the original equation is entirely time- and space-independent. If you want to include influence of gravity, it will be either over time or over a distance. $\endgroup$ – SF. Jun 1 '16 at 7:25
  • $\begingroup$ @SF. I also came to the conclusion that no more than that is required. As for the choice of gravity over distance or time, time is usually what you want as the burn time of the rocket is known in most cases. $\endgroup$ – Hohmannfan Jun 1 '16 at 7:34
  • $\begingroup$ @Hohmannfan: That is if you assume gravity constant. If you want to apply gravity dropping off with distance and a gravity turn trajectory, you'll want an integral over a curve. $\endgroup$ – SF. Jun 1 '16 at 7:51
  • $\begingroup$ @SF. Yup, but he has stated that he actually want to consider it constant. You can cover it yourself if you want to. $\endgroup$ – Hohmannfan Jun 1 '16 at 8:01
  • $\begingroup$ @Hohmannfan: I doubt I have the skill... $\endgroup$ – SF. Jun 1 '16 at 8:21
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where a constant external force is included

In the vertical take off case, the external force is not constant, as the gravitational force decreases somewhat with altitude. But as you only require it to be valid near the surface, I am going to assume you actually want the force to be constant1.

We have to sources of change in velocity here.

  1. The rocket engine, given by the Tsiolkovsky equation and
  2. The acceleration due to the external acceleration, that is simply acceleration times duration, in this case $g \cdot t_{burn}$

So, if the direction of the external force is exactly opposite of the direction we thrust, like gravity, it is not more complicated than subtracting:

$$\Delta v=\left|ln\left(\frac{m_{initial}}{m_{final}}\right) \cdot v_{exhaust} -g \cdot t_{burn}\right|$$

We need to take the absolute value, as the external force can be stronger than our own rocket's acceleration, but $\Delta v$ is a scalar.

In general, if you have an angle $u$ between the two forces, you can express it as

$$\Delta v=\sqrt{\left(ln\left(\frac{m_{initial}}{m_{final}}\right) \cdot v_{exhaust}+cos(u) \cdot g \cdot t_{burn}\right)^2+(sin(u)\cdot g \cdot t_{burn})^2}$$

The derivation should be pretty self explanatory, as this is only adding two vectors by decomposing the $g \cdot t_{brun}$ component, and get the length of the resulting vector.

(I am still representing the acceleration of the outside force as $g$)

I am not sure how this is going to be useful though, as launch analysis usually requires numerical simulation anyway.

1Err, acceleration really, as the system has a changing mass.

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  • $\begingroup$ Did Soviets use numerical simulation for Vostok? $\endgroup$ – SF. Jun 1 '16 at 7:27
  • $\begingroup$ @SF. In a sense, yes, as there are no analytical solutions. $\endgroup$ – Hohmannfan Jun 1 '16 at 7:29
  • $\begingroup$ Yep, you're right - "near surface" meant constant fo... Ah!!! Right! acceleration - not force. Good call! $\endgroup$ – uhoh Jun 1 '16 at 7:30

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