Respuesta :
Explanation:
It is given that,
A planet were discovered between the sun and Mercury, with a circular orbit of radius equal to 2/3 of the average orbit radius of Mercury.
Mass of the Sun, [tex]M=1.99\times 10^{30}\ kg[/tex]
Radius of Mercury's orbit, [tex]r=5.79\times 10^{10}\ m[/tex]
Radius of discovered planet, [tex]R=\dfrac{2}{3}r[/tex]
[tex]R=\dfrac{2}{3}\times 5.79\times 10^{10}\ m=3.86\times 10^{10}\ m[/tex]
Let T is the orbital period of such a planet. Using Kepler's third law of planetary motion as :
[tex]T^2\propto R^3[/tex]
[tex]T^2=\dfrac{4\pi^2R^3}{GM}[/tex]
[tex]T^2=\dfrac{4\pi^2\times (3.86\times 10^{10})^3}{6.67\times 10^{-11}\times 1.99\times 10^{30}}[/tex]
[tex]T=\sqrt{1.71\times 10^{13}}[/tex]
T = 4135214.625 s
or
T = 47.86 days
So, the orbital period of such a planet is 47.86 days. Hence, this is the required solution.
