Answer:
[tex]a=4066435km[/tex]
Explanation:
Kepler's 3rd Law tells us that:
[tex]a^3=(\frac{GM}{4\pi^2})T^2[/tex]
where a is the semi-major axis, M the mass of the much more massive central object (in this case Jupiter, so [tex]M=1.898\times10^{27}Kg[/tex]), T the orbital period and [tex]G=6.67\times10^{-11}Nm^2/Kg^2[/tex] the gravitational constant.
If we count the days between July 21, 2019 and September 12, 2019 we get 53 days , from there we can calculate the seconds easily by doing 53(24)(60)(60), obtaining 4579200 seconds. We put this then in our equation:
[tex]a=\sqrt[3]{(\frac{GM}{4\pi^2})T^2}=\sqrt[3]{(\frac{(6.67\times10^{-11}Nm^2/Kg^2)(1.898\times10^{27}Kg)}{4\pi^2})(4579200s)^2}=4066435070m=4066435km[/tex]
