# Some exercise questions about Tsiolkovsky's Rocket Equation and Thrust Equation

I have found 6 of the question's answers, but I don't know they are correct. The last question about time of the burn I get negative result -0.42 seconds. Time could not be a negative value, so I checked my answer more than 10 times but can't get where I do mistake.

### Questions

$$1atm = P{_{o}} = 101300Pa$$ $$F{_{nozzle}} = (P{_{e}} - P{_{o}}) \cdot A{_{e}}$$ $$F{_{nozzle}} = (101325 - 101300) \cdot 1 = 25N$$

$$Po = 0Pa$$ $$F{_{nozzle}} = (P{_{e}} - P{_{o}}) \cdot A{_{e}}$$ $$F{_{nozzle}} = (101325 - 0) \cdot 1 = 101325N$$

$$F{_{nozzle}} = 25N, \dot{m} = 1kg/sec, v{_{e}} = 400m/sec$$ $$F{_{thrust}} = \dot{m}v{_{e}} + (P{_{e}} - P{_{o}}) \cdot A{_{e}}$$ $$F{_{thrust}} = \dot{m}v{_{e}} + F{_{nozzle}}$$ $$F{_{thrust}} = 1 \cdot 400 + 25 = 425N$$

$$F{_{nozzle}} = 101325N, \dot{m} = 1kg/sec, v{_{e}} = 400m/sec$$ $$F{_{thrust}} = \dot{m}v{_{e}} + (P{_{e}} - P{_{o}}) \cdot A{_{e}}$$ $$F{_{thrust}} = \dot{m}v{_{e}} + F{_{nozzle}}$$ $$F{_{thrust}} = 1 \cdot 400 + 101325 = 101725N$$

$$I{_{sp}} = 363sec, I = 2MN = 2000000N$$ $$C = g \cdot I{_{sp}}$$ $$C = 9,8 \cdot 363 = 3557,4m/s$$

$$I{_{sp}} = 363sec, I = 2MN = 2000000N, C = 3557,4m/sec$$ $$F{_{thrust}} = \dot{m} \cdot C$$ $$\dot{m} = \frac{F{_{thrust}}}{C}$$ $$\dot{m} = \frac{2000000}{3557} = 562,2kg/sec$$

6th question's answer is, $$I{_{sp}} = 363sec, I = 2MN = 2000000N, C = 3557,4m/sec, \dot{m} = 562,2kg/sec, \Delta v = 7700m/sec$$ $$\frac{M{_{full}}}{M{_{empty}}} = e^{\frac{\Delta v}{C}}$$ $$\frac{M{_{full}}}{M{_{empty}}} = e^{\frac{7700}{3557,4}} = 8,7$$ $$MR = 8,7$$

$$I{_{sp}} = 363sec, I = 2MN = 2000000N, C = 3557,4m/sec, \dot{m} = 562,2kg/sec, \Delta v = 7700m/sec, MR = 8,7$$ $$t{_{b}} = ln(\frac{M{_{full}}}{M{_{empty}}}) \cdot I{_{sp}} - \frac{\Delta v}{g}$$ $$t{_{b}} = ln(8,7) \cdot 363 - \frac{7700}{9,8} = -0,42sec??$$

Why the 7th question's answer is coming negative value as a time value? Also, are other answers are true?

Thanks!

• I looked at this a little bit, and I'm questioning whether equation 3.37 (the burn time equation) is really valid. The basic rocket equation is dv = Isp g0 ln(MR) so if you divide through by g0, you get dv / g0 = Isp ln(MR) and that seems to mean that equation 3.37 is always going to be zero. Can you provide the derivation of equation 3.37 from your book? The burn time equation I know requires you to know the value of the initial mass, not just a ratio, and this makes sense to me intuitively. Equation 1.21 here braeunig.us/space/propuls.htm gives a burn time, you need m0. – Organic Marble May 2 '19 at 21:38
• Equation 3.37 is the burn time equation, not the delta v equation. I want to see the derivation of the burn time equation. – Organic Marble May 2 '19 at 22:03
• What book is this from? Maybe I can find it online. – Organic Marble May 2 '19 at 22:15
• I was afraid you were going to say that. I am not a fan of that book. That makes me more inclined to think that equation is wrong. But I will look into it. – Organic Marble May 2 '19 at 23:23
• This very site has a great list of sources:space.meta.stackexchange.com/questions/249/… The best book I know is Sutton but only the 4th edition is truly great. The later edition online is not nearly as good. But still better than the Taylor one. – Organic Marble May 2 '19 at 23:34

1. Wikipedia gives sea level pressure as 101325 Pa, that would make the answer zero, otherwise OK, what you have is close enough.
2. Looks good to me
3. Looks good to me
4. Looks good to me
5. Looks good to me
6. Looks good to me
7. I don't like that equation. Let's check it with some real world data from here. The burn took 156.92 seconds. We can calculate mass flow from Isp and thrust, it's 29.63 kg/s. So total prop burned is 4650 kg and the mass-ratio is 1.3838. Isp is 314 so the first term in the burn time equation is 102.0. Delta-V is 999.4 m/s and g0 is 9.8, that makes the 2nd term in the burn time equation 101.9. Subtract the two terms and I get essentially zero, just like you did. Please check my numbers, it is literally an exercise for the student

I'm gonna suppress my true feeling about that equation and just say

I don't understand the derivation of that equation or the conditions under which it is supposed to apply. I recognize that the flaw may be in my analysis and not the equation.

• Oh, thanks! You tried so hard, i shouls sorry about this, you gave your hours by answering me. Thank you again and thank you. Have a good day and you are true teacher to me! – ICCQBE May 3 '19 at 0:41
• I wouldn't do it if I didn't enjoy it. The world needs more aerospace engineers. Please run the numbers on that Apollo case in #7 with the real world data and let me know if you get the same results. – Organic Marble May 3 '19 at 0:42
• I will do it, tomorrow and i will surely feedback you for this exercise. My eyes began to hurt, here is 3:45am, I won't forget this exercise and I will start working with book that you suggested tomorrow. Have a good day, sir! – ICCQBE May 3 '19 at 0:50
• I get the same results. $I{_{sp}} = \frac{I}{\Delta m{_{propellant}}g}$ we get $\Delta m{_{propellant}} = 29,6071kg/s$ and $v{_{exhaust}} = C = gI{_{sp}} = 3080,34m/s$ and using $\Delta t = \frac{M{_{L}}}{E{_{V}}}\cdot(1-e^{-(\frac{\Delta v}{E{_{V}}})}) = 156.92s$. But I have a question, With $I{_{sp}} = \frac{I{_{sp}}}{\Delta v{_{propellant}}}g$ we found $\Delta m{_{propellant}} = 29,6071kg/s$, why this gives us mass flow rate instead of $m{_{propellant final}} - m{_{propellant first}}$ so change in propellant mass, I couldn't get this, I haven't solve any problem using this method. – ICCQBE May 3 '19 at 10:18
• Thanks for checking my math! I calculated mass flow rate from $\dot m = T / (I_{sp} g_0)$ and multiplied this mass flow rate by the burn time (which we already knew for this real world problem) to get the delta mass. I will have to think about the equation you show. – Organic Marble May 3 '19 at 12:18