Severian596
Mar24-04, 01:22 PM
I'm independently studying SR right now. A practice problem in Hogg's text here (http://physics.nyu.edu/hogg/sr/sr.pdf), Chapter 4, problem 4-7 says (slightly modified):
In an interplanetary rocket race there are three rockets: X, Y, and Z. Slow team X is travelling in their old rocket at speed 0.9c relative to the finish line. They are passed by faster team Y, and X observes Y to pass at 0.9c relative to themselves. But team Y observes fastest team Z to pass Y's own rocket at 0.9c. What are the speeds of teams X, Y, and Z relative to the rest frame of the finish line?
Hint: The answer is not 0.9c, 1.8c, and 2.7c!
Just prior, Hogg gives the equation to find the velocity of a cantalope thrown by B who's moving with respect to rest frame A. The final measured velocity w of the cantalope (thrown by B at velocity v who is himself moving at velocity u) as measured by A is:
w = \frac{u+v}{1+\frac{uv}{c^{2}}}
I will state below how I solved the problem and would appreciate it if someone offers consent or denial, thanks!
Severian's Solution
Let X_R, Y_R, and Z_R be the velocities of rockets X, Y, and Z with respect to the Rest frame finish line.
Rocket X's velocity with respect to the finish line
We are told at the start that X_R = 0.9c.
Rocket Y's velocity with respect to the finish line
We must calculate Y's velocity with respect to the finish line.
Y_R=\frac{X_{R} + 0.9c}{1+\frac{(X_{R})(0.9c)}{c^{2}}}=0.99447514c
Rocket Z's velocity with respect to the finish line
We must first calculate Z's velocity with respect to Y, then finally use Y's percieved velocity of Z to find Z's velocity with respect to X.
Z_Y=\frac{Y_{R} + 0.9c}{1+\frac{(Y_{R})(0.9c)}{c^{2}}}=0.99970846c
therefore
Z_R=\frac{Z_{Y} + 0.9c}{1+\frac{(Z_{Y})(0.9c)}{c^{2}}}=0.99998466c
Final Results
X_R = 0.9c (this required no calculation)
Y_R = 0.99447514c (this required one calculation)
Z_R = 0.99998466c (this required two calculations)
In an interplanetary rocket race there are three rockets: X, Y, and Z. Slow team X is travelling in their old rocket at speed 0.9c relative to the finish line. They are passed by faster team Y, and X observes Y to pass at 0.9c relative to themselves. But team Y observes fastest team Z to pass Y's own rocket at 0.9c. What are the speeds of teams X, Y, and Z relative to the rest frame of the finish line?
Hint: The answer is not 0.9c, 1.8c, and 2.7c!
Just prior, Hogg gives the equation to find the velocity of a cantalope thrown by B who's moving with respect to rest frame A. The final measured velocity w of the cantalope (thrown by B at velocity v who is himself moving at velocity u) as measured by A is:
w = \frac{u+v}{1+\frac{uv}{c^{2}}}
I will state below how I solved the problem and would appreciate it if someone offers consent or denial, thanks!
Severian's Solution
Let X_R, Y_R, and Z_R be the velocities of rockets X, Y, and Z with respect to the Rest frame finish line.
Rocket X's velocity with respect to the finish line
We are told at the start that X_R = 0.9c.
Rocket Y's velocity with respect to the finish line
We must calculate Y's velocity with respect to the finish line.
Y_R=\frac{X_{R} + 0.9c}{1+\frac{(X_{R})(0.9c)}{c^{2}}}=0.99447514c
Rocket Z's velocity with respect to the finish line
We must first calculate Z's velocity with respect to Y, then finally use Y's percieved velocity of Z to find Z's velocity with respect to X.
Z_Y=\frac{Y_{R} + 0.9c}{1+\frac{(Y_{R})(0.9c)}{c^{2}}}=0.99970846c
therefore
Z_R=\frac{Z_{Y} + 0.9c}{1+\frac{(Z_{Y})(0.9c)}{c^{2}}}=0.99998466c
Final Results
X_R = 0.9c (this required no calculation)
Y_R = 0.99447514c (this required one calculation)
Z_R = 0.99998466c (this required two calculations)