Find the base of 121cm square
1 answer:
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To solve this problem we will use Kepler's third law of planetary motion. Formula for that law is:
Where:
T = orbital period
G = universal gravitational constant
M = mass of star
a = distance from planet to star
By rearranging this formula we get:
For Earth we have:
For Wasp-32b we have:
or
We can see that right sides are same. Left sides must be same. We can use this to solve for distance.
![\frac{365^{2}}{1.1* 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](https://tex.z-dn.net/?f=%5Cfrac%7B365%5E%7B2%7D%7D%7B1.1%2A%202.7%5E%7B2%7D%7D%20%3D%20%5Cfrac%7B1%7D%7Ba%5E%7B3%7D%20%7D%20%5C%5C%20%5C%5C%202.7%5E%7B2%7D%20%3D365%5E%7B2%7D%2Aa%5E%7B3%7D%20%5C%5C%20%5C%5C%20a%5E%7B3%7D%20%3D%5Cfrac%7B2.7%5E%7B2%7D%20%7D%7B365%5E%7B2%7D%7D%20%5C%5C%20%5C%5C%20a%3D%20%5Csqrt%5B3%5D%7B%5Cfrac%7B2.7%5E%7B2%7D%20%7D%7B365%5E%7B2%7D%7D%20%7D%20%5C%5C%20%5C%5C%20%5C%5C%20a%3D0.038AU)
Wasp-32b is at distance of 0.038AU.
Where:
T = orbital period
G = universal gravitational constant
M = mass of star
a = distance from planet to star
By rearranging this formula we get:
For Earth we have:
For Wasp-32b we have:
We can see that right sides are same. Left sides must be same. We can use this to solve for distance.
Wasp-32b is at distance of 0.038AU.