# This can't be wrong...

• Aug 23rd 2006, 11:55 AM
Jameson
This can't be wrong...
I'm doing my physics homework online, and this unit conversion problem is getting to me. Here it is along with my work.

Time standards are now based on atomic clocks. A promising second standard is based on pulsars, which are rotating neutron stars (highly compact stars consisting only of neutrons). Some rotate at a rate that is highly stable, sending out a radio beacon that sweeps briefly across Earth once with each rotation, like a lighthouse beacon. Suppose a pulsar rotates once every 1.342 806 448 872 75 6 ms, where the trailing $\displaystyle \pm$ 6 indicates the uncertainty in the last decimal place (it does not mean $\displaystyle \pm$6 ms).

Now my thoughts:

It rotates one time every 1.342806448872756ms, so I need to convert this to one # of times per 30 days.

$\displaystyle {1.342806448872756ms} \times \frac{10^3ms}{1s} \times \frac{60s}{1min}$$\displaystyle \times \frac{60min}{1hour} \times \frac{24hours}{1day} \times 30 days which equals 3480554315 times, which is incorrect. I know I'm having a brain fart. Someone tell me what it is. • Aug 23rd 2006, 12:00 PM topsquark Quote: Originally Posted by Jameson I'm doing my physics homework online, and this unit conversion problem is getting to me. Here it is along with my work. Time standards are now based on atomic clocks. A promising second standard is based on pulsars, which are rotating neutron stars (highly compact stars consisting only of neutrons). Some rotate at a rate that is highly stable, sending out a radio beacon that sweeps briefly across Earth once with each rotation, like a lighthouse beacon. Suppose a pulsar rotates once every 1.342 806 448 872 75 6 ms, where the trailing \displaystyle \pm 6 indicates the uncertainty in the last decimal place (it does not mean \displaystyle \pm6 ms). Now my thoughts: It rotates one time every 1.342806448872756ms, so I need to convert this to one # of times per 30 days. \displaystyle {1.342806448872756ms} \times \frac{10^3ms}{1s} \times \frac{60s}{1min}$$\displaystyle \times \frac{60min}{1hour} \times \frac{24hours}{1day} \times 30 days$

which equals 3480554315 times, which is incorrect. I know I'm having a brain fart. Someone tell me what it is.

You want #rot/30 days. You have 1 rot/1.342806448872756 ms. Thus
$\displaystyle \frac{1 \, rot}{1.342806448872756 ms} \times \frac{1000 \, ms}{1 \, s}$, etc.

-Dan
• Aug 23rd 2006, 12:02 PM
Jameson
Ah yes. With all of that converting I missed that detail. Thanks :)
• Aug 23rd 2006, 12:06 PM
topsquark
Quote:

Originally Posted by Jameson
Ah yes. With all of that converting I missed that detail. Thanks :)

It's an easy mistake to make.

I'm glad to see you are using the "factor-label" method. It's alot easier to catch mistakes that way. I had to bribe my students to get them to use it.

-Dan
• Aug 23rd 2006, 12:53 PM
Jameson
And as a followup to my last question, if you could double check me on this part of the same question.

Part b) said (b) How much time does the pulsar take to rotate $\displaystyle 3.0 \times 10^6$ times? (Give your answer to at least 4 decimal places.)

I got the correct answer of 4028.4193 s.

Now here's part c) What is the associated uncertainty of this time?

Here's my reasoning: the uncertainty for one rotation is $\displaystyle 6 \times 10^{-14}ms$ plus or minus.

In seconds the uncertainty is $\displaystyle 6 \times 10^{-14}ms \times \frac{10^{-3}}{1ms}$ which is $\displaystyle 6 \times 10^{-17}s$

Now multiply this by $\displaystyle 3.0 \times 10^6$ times and I get: $\displaystyle 1.8 \times 10^{-10}s$

Check out?
• Aug 23rd 2006, 02:44 PM
topsquark
Quote:

Originally Posted by Jameson
And as a followup to my last question, if you could double check me on this part of the same question.

Part b) said (b) How much time does the pulsar take to rotate $\displaystyle 3.0 \times 10^6$ times? (Give your answer to at least 4 decimal places.)

I got the correct answer of 4028.4193 s.

Now here's part c) What is the associated uncertainty of this time?

Here's my reasoning: the uncertainty for one rotation is $\displaystyle 6 \times 10^{-14}ms$ plus or minus.

In seconds the uncertainty is $\displaystyle 6 \times 10^{-14}ms \times \frac{10^{-3}}{1ms}$ which is $\displaystyle 6 \times 10^{-17}s$

Now multiply this by $\displaystyle 3.0 \times 10^6$ times and I get: $\displaystyle 1.8 \times 10^{-10}s$

Check out?

Both answers look good to me. :)

-Dan