- #1
Thales Castro
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Problem statement:
A rocket propels itself rectilinearly by giving portions of its mass a constant (backward) velocity ## u ## relative to its instantaneous rest frame. It continues to do so until it attains a velocity ## v ## relative to its initial rest frame. Prove that the ratio of the initial to the final rest mass of the rocket is given by
$$ \frac{m_{i}}{m_{f}} = \left( \frac{c+v}{c-v} \right) ^{\frac{c}{2u}} $$Attempt at a solution:
Consider, first, the instantaneous rest frame of the rocket at a given time.
The rocket ejects a small mass ## dm ## with backward velocity ## u ## and has an increase ##dv'## in its velocity. It's rest mass has a new value ## m' ##, which can be calculated by mass conservation law (considering that ## dv' ## is small and ##\gamma(dv') \approx 1##):
$$ m = dm \gamma(u) + m' \implies m' = m - dm \gamma(u)$$
Then, by conservation of mommentum, we have:
$$ 0 = -dmu + \left( m-dm \gamma(u) \right) dv' $$
Ignoring second order differentials, we have:
$$ dm u = m dv' $$
Now, we need to transform ##dv'## to the velocity measured by the inertial frame in which the rocket had velocity 0 when ##t=0## (lab frame).
Assume the rocket had velocity ## v ## at ## t = t_{1} ## and had an increase ## dv ## after ejecting material. Then, by relativistic velocity transformation (ignoring second order differentials):
$$ v + dv = \frac{ v + dv' } { 1+ \frac{vdv'}{c^{2}} } \implies dv' = \frac{1}{1-\frac{v^{2}}{c^{2}} } dv $$
So we have:
$$ \frac{u}{m} dm = \frac{1}{1-\frac{v^{2}}{c^{2}} } dv $$
$$ \int_{m_{i}}^{m_{f}} \frac{u}{m} dm = \int_{0}^{v} \frac{1}{1-\frac{(v')^{2}}{c^{2}} } dv' $$
Finally:
$$ \frac{m_{f}}{m_{i}} = \left| \frac{v+c}{v-c} \right| ^{\frac{c}{2u}} $$
Problem is, the left side should be ## \frac{m_{i}}{m_{f}} ##. The mistake is most likely on the first two equations, since the relation between the differetials shoud have a minus on one side (since velocity gets bigger when mass is released), but I can't find it. Can anyone help me?
A rocket propels itself rectilinearly by giving portions of its mass a constant (backward) velocity ## u ## relative to its instantaneous rest frame. It continues to do so until it attains a velocity ## v ## relative to its initial rest frame. Prove that the ratio of the initial to the final rest mass of the rocket is given by
$$ \frac{m_{i}}{m_{f}} = \left( \frac{c+v}{c-v} \right) ^{\frac{c}{2u}} $$Attempt at a solution:
Consider, first, the instantaneous rest frame of the rocket at a given time.
The rocket ejects a small mass ## dm ## with backward velocity ## u ## and has an increase ##dv'## in its velocity. It's rest mass has a new value ## m' ##, which can be calculated by mass conservation law (considering that ## dv' ## is small and ##\gamma(dv') \approx 1##):
$$ m = dm \gamma(u) + m' \implies m' = m - dm \gamma(u)$$
Then, by conservation of mommentum, we have:
$$ 0 = -dmu + \left( m-dm \gamma(u) \right) dv' $$
Ignoring second order differentials, we have:
$$ dm u = m dv' $$
Now, we need to transform ##dv'## to the velocity measured by the inertial frame in which the rocket had velocity 0 when ##t=0## (lab frame).
Assume the rocket had velocity ## v ## at ## t = t_{1} ## and had an increase ## dv ## after ejecting material. Then, by relativistic velocity transformation (ignoring second order differentials):
$$ v + dv = \frac{ v + dv' } { 1+ \frac{vdv'}{c^{2}} } \implies dv' = \frac{1}{1-\frac{v^{2}}{c^{2}} } dv $$
So we have:
$$ \frac{u}{m} dm = \frac{1}{1-\frac{v^{2}}{c^{2}} } dv $$
$$ \int_{m_{i}}^{m_{f}} \frac{u}{m} dm = \int_{0}^{v} \frac{1}{1-\frac{(v')^{2}}{c^{2}} } dv' $$
Finally:
$$ \frac{m_{f}}{m_{i}} = \left| \frac{v+c}{v-c} \right| ^{\frac{c}{2u}} $$
Problem is, the left side should be ## \frac{m_{i}}{m_{f}} ##. The mistake is most likely on the first two equations, since the relation between the differetials shoud have a minus on one side (since velocity gets bigger when mass is released), but I can't find it. Can anyone help me?