2. Relativistic rockets and practice with separation of variables. Suppose a rocket of mass M ejects...

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2. Relativistic rockets and practice with separation of variables. Suppose a rocket of mass M ejects propellant at a constant velocity w with respect to the rocket, thus achieving an instantaneous velocity U with respect to the center of mass of the rocket-propellant system. (a) (35 points) Consider an infinitesimal amount of propellant with mass dm expelled from the rocket. Use conservation of energy, conservation of momentum, and velocity addition to derive the relativistic rocket equation: dM M dU w(1-U2/) (1) Feel free to refer to Morin Sec. 3.6 for inspiration, but note the difference here: our pro- pellant is massive and has velocity w Example (Relativistic rocket): Assume that a rocket propels itself by continually con- verting mass into photons and firing them out the back. Let m be the rocket's instantaneous mass, and let v be its instantaneous speed with respect to the ground. Show that (dropping the c's) dm dv + m 1-12 = 0. If the initial mass is M and the initial speed is zero, integrate Eq. (3.72) to obtain m = M =>> v = 1-(m/M) 1+ (m/M) (3.72) (3.73) First solution: The strategy of this solution will be to use conservation of momentum in the ground frame. Consider the effect of a small mass being converted into photons. The mass of the rocket goes from m to m + dm (where dm is negative). So in the frame of the rocket, photons with total energy E = -dm (which is positive) are fired out the back (we're dropping the c's). In the frame of the rocket, these photons have momentum Pr = dm (which is negative), since a photon has |p|= Elc E. With v being the instantaneous speed of the rocket with respect to the ground, the momen- tum of the photons in the ground frame, pg, can be found via the Lorentz transformation in Eq. (3.19), 2. Relativistic rockets and practice with separation of variables. Suppose a rocket of mass M ejects propellant at a constant velocity w with respect to the rocket, thus achieving an instantaneous velocity U with respect to the center of mass of the rocket-propellant system. (a) (35 points) Consider an infinitesimal amount of propellant with mass dm expelled from the rocket. Use conservation of energy, conservation of momentum, and velocity addition to derive the relativistic rocket equation: dM M dU w(1-U2/) (1) Feel free to refer to Morin Sec. 3.6 for inspiration, but note the difference here: our pro- pellant is massive and has velocity w Example (Relativistic rocket): Assume that a rocket propels itself by continually con- verting mass into photons and firing them out the back. Let m be the rocket's instantaneous mass, and let v be its instantaneous speed with respect to the ground. Show that (dropping the c's) dm dv + m 1-12 = 0. If the initial mass is M and the initial speed is zero, integrate Eq. (3.72) to obtain m = M =>> v = 1-(m/M) 1+ (m/M) (3.72) (3.73) First solution: The strategy of this solution will be to use conservation of momentum in the ground frame. Consider the effect of a small mass being converted into photons. The mass of the rocket goes from m to m + dm (where dm is negative). So in the frame of the rocket, photons with total energy E = -dm (which is positive) are fired out the back (we're dropping the c's). In the frame of the rocket, these photons have momentum Pr = dm (which is negative), since a photon has |p|= Elc E. With v being the instantaneous speed of the rocket with respect to the ground, the momen- tum of the photons in the ground frame, pg, can be found via the Lorentz transformation in Eq. (3.19),

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a To derive the relativistic rocket equation we start by applying conservation of energy and conserv... View the full answer