How to Calculate Time and Uncertainty for Rotating Pulsars

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The discussion focuses on calculating the rotation time and associated uncertainty for a pulsar that rotates every 1.400 806 448 872 75 ms, with a ±5 ms uncertainty. The correct calculation for the number of rotations in 30 days is 2,592,000. For the time taken to rotate 2.0 x 10^6 times, the accurate result is approximately 2,801.6129 seconds. The uncertainty in this time can be determined by calculating the time using the maximum and minimum rotation periods derived from the given uncertainty.

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Hello, perhaps you could help with a problem.

Suppose a pulsar rotates once every 1.400 806 448 872 75 + or - 5 ms, where the trailing + or - 5 indicates the uncertainty in the last decimal place (it does not mean + or - 5ms).

a. How many times does the pulsar rotate in 30.0 days?
I solved this one, 30days(60s/min)(60min/hr)(24hr/day) = 1.85(10^9)

b. How much time does the pulsar take to rotate 2.0(10^6) times?
(Give your answer to at least 4 decimal places.)

c. What is the associated uncertainty of this time?

It's these last two that are giving me issues, for b. I figured you just multiply the amount of rotations by the time it takes to rotate, I came up with 2801.61 s but according to webassign is wrong. I even stuffed it with 2 zeros to go out 4 decimal places. I'm just not sure about c, though I need the answer to b. to start.
 
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scmejla said:
Hello, perhaps you could help with a problem.

Suppose a pulsar rotates once every 1.400 806 448 872 75 + or - 5 ms, where the trailing + or - 5 indicates the uncertainty in the last decimal place (it does not mean + or - 5ms).

a. How many times does the pulsar rotate in 30.0 days?
I solved this one, 30days(60s/min)(60min/hr)(24hr/day) = 1.85(10^9)
Your notation is misleading! (30)(60)(60)(24)= 2592000. Then you multiplied by 1000 ms/s and divided by 1.400 806 448 872 75 ms.

b. How much time does the pulsar take to rotate 2.0(10^6) times?
(Give your answer to at least 4 decimal places.)

c. What is the associated uncertainty of this time?

It's these last two that are giving me issues, for b. I figured you just multiply the amount of rotations by the time it takes to rotate, I came up with 2801.61 s but according to webassign is wrong. I even stuffed it with 2 zeros to go out 4 decimal places.
What do you mean "stuffed it with 2 zeros? 2(10^6)= 2000000 and since you are multiplying by approximately 1.4, you should get about 2800000.
I get 2801612.8977455.

I'm just not sure about c, though I need the answer to b. to start.
1.400 806 448 872 75 + or - 5 ms, could be as large as 1.400 806 448 872 75 +5 = 1.400 806 448 872 80 ms or a small as 1.400 806 448 872 75 -5 = 1.400 806 448 872 70 ms. Do your calculation using each of these, then find the average of the two extremes. The uncertainty will be that average + or - something.
 
Thank you for your help, you've led me to the correct answer. It is 2801.6128977455, you must have left out the (ms) or 10^-3 when multiplying. My calculator was only set up to display 6 digits. Thanx again.
 

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