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Is the radius of Planet X if its period is l.3118 seconds?
G= 667-16" and my= 19x10° kg

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Answer: We can use Kepler's third law to find the radius of Planet X. Kepler's third law states that the square of the period of a planet's orbit is proportional to the cube of its semi-major axis.

Mathematically, we can write this as:

T^2 = (4π^2 / GM) a^3

where T is the period of the planet's orbit, G is the gravitational constant, M is the mass of the central body (in this case, my = 19 x 10^12 kg), and a is the semi-major axis of the planet's orbit.

Solving for a, we get:

a = [T^2GM / (4π^2)]^(1/3)

Substituting the given values, we get:

a = [(1.3118 s)^2 x 6.67 x 10^-11 m^3/(kg s^2) x 19 x 10^12 kg / (4π^2)]^(1/3)

Simplifying, we get:

a = 3.166 x 10^11 meters

Since the radius of the planet is equal to half of the length of its semi-major axis, we can find the radius of Planet X by dividing a by 2:

radius = a / 2 = 1.583 x 10^11 meters

Therefore, the radius of Planet X is approximately 1.583 x 10^11 meters.

Explanation:

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