A planet has a single moon that is solely influenced by the gravitational interaction between the two bodies. We will assume that the moon is moving in a circular orbit around the planet and that the moon travels with a constant speed in that orbit. The mass of the planet is mp = 3.03 × 1025 kg. The mass of the moon is mm = 9.65 × 1022 kg. The radius of the orbit is R = 2.75 × 108 m.

What is the period of the moon's orbit around the planet in earth days (1 earth day = 24 hours).

To find the period of the moon's orbit around the planet, we need to use Kepler's third law of planetary motion. This law states that the square of the period of a planet's orbit is proportional to the cube of its average distance from the sun (or in this case, the planet).

The formula for the period of an orbit is:

T^2 = (4π^2 * R^3) / (G * (mp + mm))

where T is the period of the orbit, R is the radius of the orbit, G is the gravitational constant, and mp and mm are the masses of the planet and moon, respectively.

Given that the radius of the orbit (R) is 2.75 × 10^8 m, the mass of the planet (mp) is 3.03 × 10^25 kg, the mass of the moon (mm) is 9.65 × 10^22 kg, and the gravitational constant (G) is approximately 6.67 × 10^-11 m^3/(kg*s^2), we can substitute these values into the equation and solve for T.

T^2 = (4π^2 * (2.75 × 10^8)^3) / (6.67 × 10^-11 * (3.03 × 10^25 + 9.65 × 10^22))

Now we can calculate T using this equation.

To determine the period of the moon's orbit around the planet, we can use the formula for the period of a circular orbit:

T = 2π * √(R^3 / G * (mp + mm))

where T is the period in seconds, R is the radius of the orbit, G is the gravitational constant, mp is the mass of the planet, and mm is the mass of the moon.

The gravitational constant is approximately 6.67430 × 10^(-11) N m^2/kg^2.

First, let's substitute the given values into the formula:

T = 2π * √((2.75 × 10^8)^3 / (6.67430 × 10^(-11)) * (3.03 × 10^25 + 9.65 × 10^22))

Now we can simplify this equation:

T = 2π * √((2.75 × 10^8)^3 / (6.67430 × 10^(-11)) * (3.03 × 10^25 + 9.65 × 10^22))
≈ 2π * √(2.525 × 10^24)

Finally, we can calculate the period T:

T ≈ 2π * √(2.525 × 10^24)
≈ 2π * 5.031618986
≈ 31.60366 seconds

To convert this into earth days, we need to divide by the number of seconds in one earth day (24 hours * 60 minutes * 60 seconds):

T = 31.60366 seconds / (24 hours * 60 minutes * 60 seconds)
≈ 0.3662 earth days

Therefore, the period of the moon's orbit around the planet is approximately 0.3662 earth days.