For any object orbiting the Sun, Kepler's Law may be written T2 = kr3. If T is measured in years and r in unit

For any object orbiting the Sun, Kepler's Law may be written T2 = kr3. If T is measured in years and r in units of the Earth's distance from the Sun, then k = 1. What, therefore, is the time (in years) for Mars to orbit the Sun if its mean radius from the Sun is 1.5 times the Earth's distance from the Sun?

k = T²/ r^3
Since k = 1

T² = r^3 where T is in years and r is in units of earth distance to the sun

Given r = 1.5 units
T² = 1.5 ^3

T = 1.84 years
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1.8 years. Since the number of significant digits in the least accurate measurement is two, any number with more significant figures is incorrect. (ie. 1.837117307 years)

T^2 = k * r^3
For the earth, T = 1 year, k = 1, and r = radius of earth
For mars, radius = 1.5 * r

T^2 = 1 * 1.5^3 = 3.375
T = √3.375 ≈ 1.837 years

T^2 = (1.5)^3

1.5*1.5*1.5 = 2.25*1.5 = 3.375

T = sqrt(3.375)

plug that into a calculator for the answer

T² = Kr³

Now for mars it should be,
T = √(Kr³)
= √(1*r²*r)
= r√r
Now put the radius of earth to get the orbit time for mars.

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