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Attn. Oh great helpers

  • Thread starter orbity
  • Start date
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Simple Physics question.

I thInk this is physics? anyhow.. :confused:

I found that the distance from the earth to the sun is 1.5*10^11m , and
that the mass of the sun is 1.9*10^30kg,

and i found Keplers's Law of Periods here
http://hyperphysics.phy-astr.gsu.edu/hbase/kepler.html#c6

but I do not understand how to do the rest... :shy: :yuck: :confused: :cry:

1.use Kepler’s Law of Periods to find the period of the earth’s orbit, recorded in seconds
2.show how to express your answer in years; and
3.if your answer does not agree with the accepted earth orbital period of 365.25 days, explain the discrepancy.
 
Last edited:

Curious3141

Homework Helper
2,830
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orbity said:
I thInk this is physics? anyhow.. :confused:

I found that the distance from the earth to the sun is 1.5*10^11m , and
that the mass of the sun is 1.9*10^30kg,

and i found Keplers's Law of Periods here
http://hyperphysics.phy-astr.gsu.edu/hbase/kepler.html#c6

but I do not understand how to do the rest... :shy: :yuck: :confused: :cry:

1.use Kepler’s Law of Periods to find the period of the earth’s orbit, recorded in seconds
2.show how to express your answer in years; and
3.if your answer does not agree with the accepted earth orbital period of 365.25 days, explain the discrepancy.
1. Just apply the formula. To help you out, write down the values of the variables and constants.

2. How many seconds in a year? (1 Year = 365 days*24 hours/day*60 minutes/hour*60 seconds/minute)

3. Think of the simplifications involved. Some hints : what's the reason for the leap year? Make a recalculation taking the leap year cycle into account. Can that alone explain the discrepancy? Why does the mass of the Earth not appear in the formula? Is that an acceptable simplification? Do you think these astronomical measurements of distance and mass are made or given to the requisite precision? Justify your concern about data precision by estimating the relative uncertainty in the final answer caused by relative uncertainties in each of the variables.
 
Last edited:
4
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Curious3141 said:
1. Just apply the formula. To help you out, write down the values of the variables and constants.

2. How many seconds in a year? (1 Year = 365 days*24 hours/day*60 minutes/hour*60 seconds/minute)

3. Think of the simplifications involved. Some hints : what's the reason for the leap year? Make a recalculation taking the leap year cycle into account. Can that alone explain the discrepancy? Why does the mass of the Earth not appear in the formula? Is that an acceptable simplification? Do you think these astronomical measurements of distance and mass are made or given to the requisite precision? Justify your concern about data precision by estimating the relative uncertainty in the final answer caused by relative uncertainties in each of the variables.
thanks for your help but i dont know how to do any of that, its ok
 

Curious3141

Homework Helper
2,830
86
orbity said:
thanks for your help but i dont know how to do any of that, its ok
Why is it OK? Isn't this stuff you're supposed to learn to do?:confused:

Start with the formula.[tex]T^2 = \frac{4\pi^2}{GM}a^3[/tex]. You're required to find the value of T. If you use standard (SI) units for all the other constants and variables, you'll get T in seconds as required for the first part.

G is the Universal Constant of Gravitation, given by 6.673*10^(-11) Nm^2/kg^2.

M is given as 1.9*10^30 kg. a (technically the semi-major axis of an elliptical orbit, but don't worry about that now) is given as 1.5*10^11 m. [tex]\pi[/tex] is approx. 3.1416, or you can use the calculator value.

So where's the difficulty, at least in the first part?
 

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