Question

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?

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

will it be about 1.05 ?

Answer 5

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

Answer 6

T = K a^3

sqrt/3.35E18 * (3.2E8)^3

Answer 7

because they want to find T right?

Answer 8

yea my answer is wrong -_- lol

Answer 9

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