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Obviously, we can make rockets to go very fast, but what is a reasonable top speed? Assume that a rocket is fired from rest at a space station in deep space, where gravity is negligible. (a) If the rocket ejects gas at a relative speed of 2000 $\mathrm{m} / \mathrm{s}$ and you want the rocket's speed eventually to be $1.00 \times 10^{-3} c$ , where $c$ is the speed of light, what fraction of the initial mass of the rocket and fuel is not fuel? (b) What is this fraction if the final speed is to be 3000 $\mathrm{m} / \mathrm{s} ?$

(a) $f=7.17 \times 10^{-66}$

(b) $f=0.223$

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{'transcript': "So in this exercise we are dealing with a classical rocket propulsion problems on which, at some time, t we have a rocket traveling in space with some velocity. V be viewed by someone in the herbs reference frame and later on at a time T plus DT on which D T is an infinite Tess Amo. Amount of time the rocket ejects a particle off full off mass minus D m in order to gain a amount of velocity. Devi Okay, so this exercise asked the ship. Calculate what ISS? The fraction between the final mess off the rocket and the initial mass off the rocket on which the initial mass off the rocket is the mess on which the time is equal to zero and the rocket hasn't expelled any fool. So it wants us to calculate this fraction for a expelled of full velocity off 200 m per second. And this velocity is on the Rockets reference frame and a final a final rocket velocity off three times sent to the 5 m per second. Okay, so the book has equation that relates the final velocity off the rocket with this with this fraction So it's equation 8 40. And this equation says that the final velocity off the rocket minus the initial velocity of the rocket is equal to the velocity off the full, viewed by the rockets reference frame times. Ah, natural logarithms off the initial mass over the finalists. Okay, so we know that and the initial time the velocity of the rocket is zero because it hasn't expelled any any fool. And we can also right, Uh uh, This this fraction we can put this fraction in here by multiplying both sides of the equation by minus one. And we have that minus VF minus V x l and and zero m. Because the natural algorithm has a property on which we have on which when we have the minus off the natural logarithms, we can invert the term that is inside the parenthesis such we have. So we have that taking the expelled velocity in putting this on the other side of the equation we have minus VF is equal to the X. Uh, sorry. Madness we have over Vieques is equal to the natural logarithms off the final mass over the initial mass. Okay, this is a property off the That's her logarithms. Um, okay, so we can just, uh, take the exponential off both sides. Such we have that, uh, the exponential off minus the final velocity off the rocket over the expelled velocity off the full is equal to. So the exponential off the off the natural logarithms is equal to the term inside of the natural logarithms. So is equal to em over m zero. Okay, so let's just be stupid. The values that we know. So we had a am over, um, zero has to be able to e minus three times 10 to the five over two times. Stand to the three, and this is equal to 7.2 times 10 to the minus 66. Okay, so this is this has to be the fraction between the final mass of the rocket and the initial mass off the rocket such that the rocket arrives to a final velocity off three times. 10 to the five. Okay, uh, now, question be asked us to once again calculated this fraction, but now, considering a different final velocity off the rocket. So a final velocity off 3000 m per second. Yeah, So let's just, uh use this expression again, and we have that I am over. Um zero in our case is equal to e uh, minus three times 10 to the 3/2 times Stand 2 to 3, and this is equal to 0.223 Okay, so in this case, this has to be diffraction between the final minutes of the rocket and the initial mess. If we want the rocket to arrive a final velocity off 3000 m per second."}