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{\bf ADVANCED DYNAMICS --- PHY 4241/5227}\\\vspace {2mm} 
{\bf SOLUTIONS -- SET 13}\\%
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\noindent
Problem 49: Momentum conservation in the rest frame ($m$ rest mass):
$$ c\,dm\ =\ m\,g\,d\tau\,.$$
Separation of variables gives (with initial mass $m_0$):
$$ \int_{m_0}^{m(\tau)} = - \frac{g}{c} \int_0^{\tau} d\tau'~~
   \Rightarrow~~m(\tau) = m_0\,e^{-g\tau/c}\,.$$
After $\tau =$ 20 years:
$$ m(20\,{\rm y}) = 1.09\times 10^{-9}\,m_0\,.$$
End of spacetrip journey.

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