(i) If the radius of the orbit of Planet P is equal to that of the Earth's orbit, then the gravitational force acting on Planet P can be expressed as:
F = G * Mp * Sp / r^2
where G is the gravitational constant, Mp is the mass of Planet P, and Sp is the mass of the Sun in Planet P's orbit.
The gravitational force acting on the Earth can be expressed as:
F = G * Me * Sg / r^2
where Me is the mass of the Earth and Sg is the mass of the Sun in the Earth's orbit.
Since the radius is the same for both orbits, we can equate the two expressions for the gravitational force:
G * Me * Sg / r^2 = G * Mp * Sp / r^2
Solving for the ratio of the masses of the Suns, we get:
Sg / Sp = Me / Mp
Since the period of a planet in orbit is given by T = 2π √ (r^3 / (G * M)), and Planet P completes its orbit in half the time it takes the Earth, it follows that:
Tp / Te = 2
where Tp is the period of Planet P and Te is the period of the Earth.
Therefore, Tp = 2 * Te
Substituting the expression for T in terms of r and M, we get:
2 * 2π √ (r^3 / (G * Me)) = 2π √ (r^3 / (G * Mp))
Dividing both sides by 2π, we get:
√ (r^3 / (G * Me)) = √ (r^3 / (G * Mp))
Squaring both sides, we get:
r^3 / (G * Me) = r^3 / (G * Mp)
Solving for the ratio of the masses of the Suns, we get:
Sg / Sp = Me / Mp
Since the mass of the Earth is much smaller than the mass of the Sun, we can approximate Me / Mp as 0, and we get:
Sg / Sp = 0
Therefore,
Sg = 4 * Sp
So, the mass of the Sun in the Earth's orbit is four times the mass of the sun in Planet P's orbit.
(ii) If the period of Planet P is given by T = 2π √ (r^3 / (G * Mp)), then the period can be expressed as:
T = 2π √ (r^3 / (G * Mp)) = 2₁√ √/GM₂