# How can I caculate the time to traverse a distance with the Tsiolkovsky equation?

Using the Tsiolkovsky equation I know what the change in velocity is after burning x amount of fuel.

$$\Delta V = V_e \dot\ log(\frac{m_1}{m_1-m_2})$$

$m_1$ is my starting mass and $m_2$ is the amount of mass that I burn away.

And then knowing how quickly I burn fuel I can work out the average acceleration over this distance.

$$a = \frac{\Delta V}{(\frac{m_2}{b})}$$

where $b$ is the rate at which I burn fuel in my rocket, like mass per second.

First question is "is anything wrong in the above?"

So knowing this, I want to cross a distance $d$ and I want to initiate a burn of amount $m_2$ to get me moving, how do I work out the amount of time that will take?

My knowledge of math and physics isn't great (I'm learning as a hobby). So far I know that the time under constant acceleration can be found with:

$$t = \sqrt{\frac{2d}{a}}$$

But the problem is my rocket isn't under constant acceleration because the mass is constantly changing. So I'm assuming that this wouldn't work for me.

What I've done so far is attempted to put the $t$ into the Tsiolkovsky equation:

$$\Delta V = V_e \dot\ log(\frac{m_1}{m_1-(m_2 \dot\ t)})$$

so that I can reverse the equation (is reverse the correct word?) like this (thanks wolfram):

$$t = \frac{m_1e^{-\frac{d}{V_e}}(e^{\frac{d}{V_e}}-1)}{m_2}$$

and this gives me the $t$ it will take to cross a distance $d$ after initiating a burn of mass $m_2$? Is this correct?

No you can't use mean acceleration in the way you propose, because the equation $t = \sqrt{\frac{2\,s}{a}}$ assumes constant acceleration.

You need to describe the system with a differential equation that takes account of the system's dynamics: since you're learning as a hobby, you may not have seen much of this. Your last paragraph is correct reasoning and is nearer to what you need. The correct wording is "solve" or "invert" or "re-arrange" the equation, but "reverse" is pretty evocative and nearest to "invert".

You need further information to solve your problem: you need a model of how your rocket's mass decreases with time. The simplest (and probably quite accurate model) is that the rate of decrease of mass is some constant mass flow rate: let's call this $q$.

Let's go back to the differential equation whence the Tsiolkovsky equation is derived. We calculate the rocket's change in velocity $\mathrm{d}\,v$ after it has thrown a mass $\mathrm{d}\,m$ out the back at speed $v_e$ relative to it: relative the frame at some instant, before the mass is thrown, the total system's linear momentum is nought: so this must the the momentum relative to this frame after the mass is thrown. The rocket's increase in momentum is $m\,\mathrm{d} v$, which must be balanced by the thrown mass's momentum in the opposite direction so that:

$$m\,\frac{\mathrm{d}v}{\mathrm{d} m} = v_e$$

This is the differential equation which is solved to get the Tsiolkovsky equation. With some juggling, we re-arrange it to:

$$\frac{\mathrm{d}v}{\mathrm{d} t} = \frac{1}{2} \frac{\mathrm{d}v^2}{\mathrm{d} s} = \frac{v_e}{m}\,\frac{\mathrm{d}\,m}{\mathrm{d}t} = \frac{v_e\,q}{m}\tag{1}$$

The first step is a standard identity that converts the acceleration - i.e. the rate of change $\frac{\mathrm{d}v}{\mathrm{d} t}$ of the velocity with respect to time $t$, into a rate of change with respect to the distance travelled $s$. Now, from the Tsiolkovsky equation we have $m(v) = m_0\,\exp\left(-\frac{v-v_0}{v_e}\right)$, where $v_0$ is the beginning velocity and $m_0$ the beginning mass: when we put this into equation (1) we get:

$$\frac{1}{2} \frac{\mathrm{d}v^2}{\mathrm{d} s} = v\,\frac{\mathrm{d}v}{\mathrm{d} s}=\frac{v_e\,q}{m_0}\,\exp\left(\frac{v-v_0}{v_e}\right)\tag{2}$$

This is the differential equation you must integrate to get the distance travelled as a function of $v$. Let me know how you go with this one. Also from (1), we get in the above way from the inverted Tsiolkovsky equation:

$$\frac{\mathrm{d}v}{\mathrm{d} t} = \frac{v_e}{m}\,\frac{\mathrm{d}\,m}{\mathrm{d}t} = \frac{v_e\,q}{m_0}\,\exp\left(\frac{v-v_0}{v_e}\right)\tag{3}$$

which is the differential equation you must solve to get $v$ as a function of time.

Time as a function of distance comes from this last equation. On integrating this last equation, you get

$$v(t) = v_o+v_e\log\left(\frac{m_0}{m_0 - q\, t}\right)$$

and then you need to integrate this, because you now have the differential equation $\frac{\mathrm{d}\,s}{\mathrm{d}\,t}=v_0+ v_e\log\left(\frac{m_0}{m_0 - q\, t}\right)$. This last integration leaves you with:

$$s(t) = v_e\, \left(t-\frac{m_0}{q}\right) \log \left(\frac{m_0}{m_0-q\, t}\right)+t\, (v_0+v_e)$$

To find time to travel a certain distance will need to be done numerically, as, given $s$, you have a transcendental equation in $t$.

• Thank you very much for your help. I'm unsure how getting the distance as a function of velocity or velocity as a function of time will get me the time as a function of distance. Or would I simply invert these? Jan 27 '15 at 3:00
• @DavidMurphy See my last edits Jan 27 '15 at 3:59
• thanks for the additional info. I may have bitten off more than I can chew here. I'm firstly unsure where you've retrieved the initial differential equation that's solved for the Tsiolkovsky equation? Next, how has any amount of juggling re-arranged it into your first equation? I don't understand how we've gone from having mass and momentum into distance? In your first step we're converting acceleration into rate of change with respect to distance but the initial equation doesn't have a distance or time... Do you have a book or website you'd recommend? Jan 28 '15 at 2:10
• @DavidMurphy The first DE is a statement of conservation of momentum: one is simply looking at very small "blobs" of exhaust gas at a time. Can you reason through the following problem? "I stand on a stationary skateboard with no friction. My total mass is $M$. I throw a medicine ball with mass $m$ at speed $v_e$. As a result, I begin to move in the opposite direction to the throw with some unknown speed $v$. Write down an equation that expresses conservation of momentun, i.e. states that the system's momentum before (i.e. nought) equals the system's momentum after the throw .... Jan 28 '15 at 3:17
• @DavidMurphy ....(assuming the skateboard's wheels have zero rolling friction). You should see that $m\,v_e = (M-m)\,v$. Can you see this? Jan 28 '15 at 3:18