Period of moon=273 days

Radius of moon =3,9x10^5 km
Radius of satellite = 7200km
Period of satellite=?

To find the period of the satellite, we can use Kepler's third law of planetary motion, which relates the period of a satellite to its distance from the object it is orbiting.

Kepler's third law states that the square of the period of a satellite is proportional to the cube of its average distance from the object it is orbiting.

The formula for Kepler's third law is:

T^2 = (4π^2 / G * (M1 + M2)) * r^3

Where:
T = Period of the satellite
G = Gravitational constant (approximately 6.67430 x 10^-11 m^3/kg/s^2)
M1 = Mass of the object the satellite is orbiting
M2 = Mass of the satellite
r = Distance between the center of the object and the center of the satellite

Before we apply the formula, we need to convert the given parameters into the appropriate units.

Period of the moon = 273 days = 273 * 24 * 3600 seconds (since 1 day = 24 hours = 24 * 60 * 60 seconds)
Radius of the moon = 3.9 x 10^5 km = 3.9 x 10^5 * 1000 meters (since 1 km = 1000 meters)
Radius of the satellite = 7200 km = 7200 * 1000 meters

Next, we need to determine the mass of the object the satellite is orbiting. Assuming it is the Earth, the mass of Earth is approximately 5.972 × 10^24 kg.

Now, let's substitute the values into the formula and solve for T:

T^2 = (4π^2 / G * (M1 + M2)) * r^3

T^2 = (4π^2 / (6.67430 × 10^-11) * (5.972 × 10^24 + M2)) * (7200 * 1000)^3

Simplifying the equation further, we can calculate the period of the satellite.

Keep in mind that the mass of the satellite (M2) is not provided in the question, so you would need to know or assume its mass in order to calculate the period accurately.