## anonymous 3 years ago 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?

1. anonymous

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

2. anonymous

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

3. anonymous

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

4. anonymous

will it be about 1.05 ?

5. anonymous

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

6. anonymous

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

7. anonymous

because they want to find T right?

8. anonymous

yea my answer is wrong -_- lol

9. anonymous

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