Calculate the time required in millions of years for one orbit.

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Homework Help Overview

The problem involves calculating the time required for the solar system to complete one orbit around the center of the Milky Way galaxy, given its distance from the center and its orbital speed.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to calculate the orbital time by converting light years to kilometers and using the circumference formula. Some participants suggest checking the conversion factor and the significance of significant figures in the calculations.

Discussion Status

Participants are exploring the calculations involved in determining the orbital time. There is a focus on ensuring the accuracy of unit conversions and the appropriateness of significant figures, with no explicit consensus reached on the correctness of the calculations.

Contextual Notes

The original poster notes uncertainty in how to approach the problem, and there is mention of homework constraints regarding significant figures and the required format for the answer.

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Homework Statement



The solar system is situated 26,000 light years from the centre of our galaxy, the Milky Way, and it orbits the centre at a speed of 220 km/sec.

Calculate the time required in millions of years for one orbit.
(NOTE: The circumference of a circle = 2 pi r where r is its radius.)

Homework Equations


The Attempt at a Solution


I am not really sure how to do it, please help thanks
 
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welcome to pf!

hi quah13579! welcome to pf! :smile:

(have a pi: π :wink:)

find how to convert light years into km, and then calculate the length of the circumference of the orbit …

what do you get? :smile:
 


tiny-tim said:
hi quah13579! welcome to pf! :smile:

(have a pi: π :wink:)

find how to convert light years into km, and then calculate the length of the circumference of the orbit …

what do you get? :smile:



9,460,730,472,580.8(light-year in km) * 26000(of light-years around) * 2 * 3.1415926535 = 1545531590213152572.8104183193436(km) / 220(km/sec) / 60(sec/min) / 60 (min/hrs) / 24 (hrs/days) / 365.25(days/years) = 222613367.04559 years
is it right?
 
quah13579 said:
9,460,730,472,580.8(light-year in km) * 26000(of light-years around) * 2 * 3.1415926535 = 1545531590213152572.8104183193436(km) / 220(km/sec) / 60(sec/min) / 60 (min/hrs) / 24 (hrs/days) / 365.25(days/years) = 222613367.04559 years
is it right?

hmmm … never never never use all those figures (you'll lose marks in the exam if you do) … 3 sig figs should usually be more than enough

and the question asked for the answer in millions of years :rolleyes:

but apart from that (and i haven't checked your lightyear/km figure), that looks fine :smile:
 

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