If the value for k for planets moving around our sun is 3.35 *10^18 m^3s^2, what would the period of a planet be if its average radius to the sun is 3.2*10^8?

Question

anonymous · Answer

I believe you're familiar with Kepler's third law, $\frac{T^2 }{r^3}=k$?

anonymous · Answer

T^2 = 4pi^2/GM a^3?

anonymous · Answer

Yeah, but you don't need that complicated equation here since k is given as 3.35 *10^18 m^3s^2.

anonymous · Answer

will it be about 1.05 ?

anonymous · Answer

You used this? $r^3 = \frac{10.24 \times 10^{16}}{3.35 \times 10^{18}}$

anonymous · Answer

T = K a^3 

sqrt/3.35E18 * (3.2E8)^3

anonymous · Answer

because they want to find T right?

anonymous · Answer

yea my answer is wrong -_- lol

anonymous · Answer

lol yh sorry about that. Eyes get blurry in a hurry. $1.05 \times 10^{21} s$ yup you're right.