Michelson Morley experiment? [closed]

• Its not that I question the conclusions reached concerning the Michelson–Morley experiment, however I would like to know how the following issue was addressed please?

If I could pass bob through a beam splitter, and have each copy of him pace out each leg of the interferometer at say $2~\rm{km/hr}$. However if along one of the legs, there was an escalator aiding his initial progress at $1~\rm{km/hr}$.

Okay, so they each leave the beam splitter at the same time, however aided by the escalator one of them moves at $3~\rm{km/hr}$, and reaches the mirror at the end of his leg earlier than the other. But on the return journey he is inhibited by progression of the escalator, and moves at $1~\rm{km/hr}$. So the other bob who travelled at $2~\rm{km/hr}$ the whole time, makes up ground on the other bob on the return journey, and they arrive home at the same time as each other.

• With so many clever people working on this experiment over the years, I know there must be a contingency for this issue. If somebody can inform me please?

closed as off-topic by CuriousOne, ACuriousMind♦, user10851, Bill N, MartinMar 14 '16 at 17:53

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – CuriousOne, ACuriousMind, Martin
If this question can be reworded to fit the rules in the help center, please edit the question.

• Trouble with homework? Tell us what you have tried on your own and we can try to help you. – CuriousOne Mar 14 '16 at 5:49
• What trouble are you having with calculating this scenario? $v=ds/dt$, from which you get $dt=ds/v$. Substitute the velocities and you can calculate the total time that your Bob would take along the arms of your interferometer. – CuriousOne Mar 14 '16 at 6:26
• Please take what @CuriousOne says seriously and calculate the travel time of the scenarios you suggest. The two walkers don't take the same time in the course of the trip. Really. Do the math in detail. – dmckee Mar 14 '16 at 6:58
• @BillAlsept: The velocities are in the denominator, not in the numerator. 1/3+1/1=4/3, which is not the same as 1/2+1/2=1. – CuriousOne Mar 14 '16 at 7:24
• In your example set the escalator speed to 2km/hr. Bob now takes an infinite time to return on the leg with the escalator because his net return speed is zero. – John Rennie Mar 14 '16 at 7:34