Verified answer

You mean to say that for the first question, the square of the period is proportional to the cube of the semi-major axis...

Earth has a semi-major axis of 1 AU, and a period of 1 year. That makes the ratios easy to handle.

Mercury has a semi-major axis of 0.38 AU -- cubed = 0.059319 which is the square of the ratio of the periods, so take the square root and you'll have the ratio of the periods, or 0.243555. Since Earth's period is 1 year, Mercury's period is 0.243555 years or 88.956 days.

As to the second part, try the below link:

p^2/a^3 = constant (for "equal mass systems) ... and since the sun's mass DWARFS any difference in mass between Earth and Mercury

p^2/a^3 merc = p^2/a^3 earth

p= 1 earth period in years

"a" earth = 1 au ["au" = astronomic unit = 1 Earth orbit radius]

"a" merc = .39 au

p^2merc = (p^2/a^3 earth) * a^3 merc = 1/1 * .39^3 = .0593 earth years

p = sqrt (.0593) = 0.24355 years

0.39 to the power of 3 = x to the power of 2

so x will be 0.24 , this is around 88 days

All of these values of interest to you are well documented:

http://en.wikipedia.org/wiki/Mercury_(planet)

http://en.wikipedia.org/wiki/Mars

ihd

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