Question

Wasp-32b orbits with a period of only 2.7 days a star with a mass that is 1.1 times that of the sun. how many au from the star is this planet? assume the orbital period of earth is 365 days.

Answer 1

To solve this problem we will use Kepler's third law of planetary motion. Formula for that law is:
$T^{2} = \frac{4 \pi ^{2} }{GM} a^{3}$
Where:
T = orbital period
G = universal gravitational constant
M = mass of star
a = distance from planet to star

By rearranging this formula we get:
$\frac{T^{2}}{a^{3} } = \frac{4 \pi ^{2} }{GM}$
For Earth we have:
$\frac{365^{2}}{1^{3} } = \frac{4 \pi ^{2} }{GM}$
For Wasp-32b we have:
$\frac{2.7^{2}}{a^{3} } = \frac{4 \pi ^{2} }{G*1.1M}$ or
$1.1* \frac{2.7^{2}}{a^{3} } = \frac{4 \pi ^{2} }{GM}$ 

We can see that right sides are same. Left sides must be same. We can use this to solve for distance.
$\frac{365^{2}}{1^{3} } = 1.1* \frac{2.7^{2}}{a^{3}$
$\frac{365^{2}}{1.1* 2.7^{2}} = \frac{1}{a^{3} } \\ \\ 2.7^{2} =365^{2}*a^{3} \\ \\ a^{3} =\frac{2.7^{2} }{365^{2}} \\ \\ a= \sqrt[3]{\frac{2.7^{2} }{365^{2}} } \\ \\ \\ a=0.038AU$

Wasp-32b is at distance of 0.038AU.