Since the period of the supposed planet is less than the period of the earth orbiting the sun, That means the mass of the central star is greater than the mass of the sun. The answer is option C
According to Kepler's third law of planetary motion for planets and satellites, he described circular orbit about the same center body stating that:
The squared of the period is directly proportional to the cube of the radius. That is,
[tex]T^{2} \alpha r^{3}[/tex]
The orbital period formula is expressed as :
[tex]T^{2}[/tex] = 4[tex]\pi ^{2}[/tex] [tex]r^{3}[/tex] ÷ GM
Suppose a planet orbits a star at the same distance as Earth (1 AU) but with a period of roughly 200 Earth days. The mass of the central star will be greater than the mass of our sun based on the formula given above.
In a geostationary orbit, whatever the mass of the satellite may be, they will travel at the same speed in a particular orbit. mass M in the above formula becomes the mass of the sun or the center mass.
Since the period of the supposed planet is less than the period of the earth orbiting the sun, That means the mass of the central star is greater than the mass of the sun.
Therefore, the correct answer is option C
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