Rocket Problem
- Thread starter markclark
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Oh, so actually you have:
. . . . .v = V<sub>e</sub>ln(M<sub>0</sub>/m)
(It should have been mentioned in class that "ln" is not a stand-alone variable, as you have formatted it, to be multiplied by the other variables, but is the natural log function, containing, in this case, the argument "M<sub>0</sub>/m".)
To solve for "m", divide off the V<sub>e</sub>. Then raise both sides of the equation as powers on "e", the natural exponential. (If you don't know what logs are, though, you may not understand what I just said, in which case you might want to review a lesson on solving log equations.)
This will give you:
. . . . .e<sup>(v/Vo)</sup> = M<sub>0</sub>/m
Multiply through by m, and divide off the exponential factor to get the final answer.
Eliz.
. . . . .v = V<sub>e</sub>ln(M<sub>0</sub>/m)
(It should have been mentioned in class that "ln" is not a stand-alone variable, as you have formatted it, to be multiplied by the other variables, but is the natural log function, containing, in this case, the argument "M<sub>0</sub>/m".)
To solve for "m", divide off the V<sub>e</sub>. Then raise both sides of the equation as powers on "e", the natural exponential. (If you don't know what logs are, though, you may not understand what I just said, in which case you might want to review a lesson on solving log equations.)
This will give you:
. . . . .e<sup>(v/Vo)</sup> = M<sub>0</sub>/m
Multiply through by m, and divide off the exponential factor to get the final answer.
Eliz.
Let Mo = the ignition mass
Let Mb = te burnout mass
Let c = the exhaust velocity or the rocket gases
(c = gIsp where g = the acceleration due to gravity and Isp = the specific impulse of the rocket motor)
Let Vo = the initial velocity of the rocket
Let Vb = the burnout velocity of the rocket
(ln stands for natural logarithm)
Let r = the mass ratio of the rocket
The basic rocket equation is then Vb - Vo = deltaV = dV = (c)ln(Mo/Mb).
Still another way of expressing this basic law is Mo/Mb = r = e^(deltaV/c) or e = 2.71828 raised to the dV/(c) power.
You can solve for either Mo or Mb given the values of the other parameters.
Let Mb = te burnout mass
Let c = the exhaust velocity or the rocket gases
(c = gIsp where g = the acceleration due to gravity and Isp = the specific impulse of the rocket motor)
Let Vo = the initial velocity of the rocket
Let Vb = the burnout velocity of the rocket
(ln stands for natural logarithm)
Let r = the mass ratio of the rocket
The basic rocket equation is then Vb - Vo = deltaV = dV = (c)ln(Mo/Mb).
Still another way of expressing this basic law is Mo/Mb = r = e^(deltaV/c) or e = 2.71828 raised to the dV/(c) power.
You can solve for either Mo or Mb given the values of the other parameters.