keplers 3rd law

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KevinE

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Dec 3, 2014
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Hi guys i'm pretty sure this might be in the wrong section but it was the best fit. Excuse the pun

I was given this question

Question 1
Kepler’s law of harmonies states that the ratio T^2/R^3


holds for all planets where
T is the period, the time taken to orbit the sun once.
R is the orbital distance the planet is from the sun.


The average orbital distance of Mars is 1.52 times the average orbital distance of the Earth.
Knowing that the Earth orbits the sun in approximately 365 days, use Kepler's law of
harmonies to predict the time for Mars to orbit the sun.

Can someone give me an idea of where to begin please. :confused:
 
I've found this answer but although the guy seems fairly convinced he's got it
it just doesn't seem to work out when i plug in the numbers

Given: Rmars = 1.52 • Rearth and Tearth = 365 daysUse Kepler's third law to relate the ratio of the period squared to the ratio of radius cubed
(Tmars)2 / (Tearth)2 • (Rmars)3 / (Rearth)3(Tmars)2 = (Tearth)2 • (Rmars)3 / (Rearth)3
(Tmars)2 = (365 days)2 * (1.52)3
​
(Note the Rmars / Rearth ratio is 1.52)
Tmars = 684 days

I could probably give this answer but it's not much good if i don't even understand it
 
KevinE said:
I've found this answer but although the guy seems fairly convinced he's got it
it just doesn't seem to work out when i plug in the numbers

Given: Rmars = 1.52 • Rearth and Tearth = 365 daysUse Kepler's third law to relate the ratio of the period squared to the ratio of radius cubed
(Tmars)2 / (Tearth)2 • (Rmars)3 / (Rearth)3(Tmars)2 = (Tearth)2 • (Rmars)3 / (Rearth)3
(Tmars)2 = (365 days)2 * (1.52)3
​
(Note the Rmars / Rearth ratio is 1.52)
Tmars = 684 days

I could probably give this answer but it's not much good if i don't even understand it
Click to expand...

From Wikipedia:

"Mars's average distance from the Sun is roughly 230 million km (1.5 AU, or 143 million miles), and its orbital period is 687 (Earth) days. The solar day (or sol) on Mars is only slightly longer than an Earth day: 24 hours, 39 minutes, and 35.244 seconds. A Martian year is equal to 1.8809 Earth years, or 1 year, 320 days, and 18.2 hours.[7] "

So that answer is probably correct.

Since all the steps are spelled out in the answer, exactly where you are getting lost?
 
Ok well consider the 1st and 2nd parts of the equation

(Tmars)2 / (Tearth)2 • (Rmars)3 / (Rearth)

(Tmars)2 = (Tearth)2 • (Rmars)3 / (Rearth)3

Why all of a sudden in the second part does Tmars become equal to (Tearth)2 • (Rmars)3 / (Rearth)3

 
KevinE said:
Ok well consider the 1st and 2nd parts of the equation

(Tmars)2 / (Tearth)2 • (Rmars)3 / (Rearth)

(Tmars)2 = (Tearth)2 • (Rmars)3 / (Rearth)3

Why all of a sudden in the second part does Tmars become equal to (Tearth)2 • (Rmars)3 / (Rearth)3
Click to expand...

Because first statement is wrong and you are missing a "=" sign in the first statement. That should be:


(Tmars)2 / (Tearth)2 = (Rmars)3 / (Rearth)3
 
Hmmm ok so if I had the square root of Tearth^2 and the cube root of Rearth^3 would they be the equal?

Is Tearth = to Rearth?
 
KevinE said:
Hmmm ok so if I had the square root of Tearth^2 and the cube root of Rearth^3 would they be the equal?

Is Tearth = to Rearth?
Click to expand...

No.... Tearth has units of day and Rearth has unit of miles. Those could not equal to each other!!
 
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