For a satellite to be in a circular orbit 780 km above the surface of the earth, what is the period of the orbit (in hours)?A 1.98 hr B 1.65 hr C 1.78 hr D 1.88 hr

To determine the period of a satellite in a circular orbit 780 km above the surface of the earth, we can use Kepler's third law of planetary motion. This law states that the square of the period of an orbit is proportional to the cube of the semi-major axis of the orbit.

First, we need to find the radius of the orbit. The radius of the orbit is equal to the radius of the Earth plus the altitude of the satellite. The radius of the Earth is approximately 6,371 km, so the radius of the orbit is 6,371 km + 780 km = 7,151 km.

Next, we can calculate the period of the orbit using the formula:

T^2 = (4π^2 / GM) * r^3

Where T is the period of the orbit, G is the gravitational constant (approximately 6.67430 × 10^-11 m^3 kg^-1 s^-2), and M is the mass of the Earth (approximately 5.972 × 10^24 kg).

Converting the radius of the orbit to meters, we have r = 7,151 km * 1,000 m/km = 7,151,000 m.

Plugging in the values into the formula, we get:

T^2 = (4π^2 / (6.67430 × 10^-11 * 5.972 × 10^24)) * (7,151,000)^3

Simplifying the equation, we find:

T^2 ≈ 1.88 hours

Taking the square root of both sides, we get:

T ≈ √1.88 ≈ 1.37 hours

Therefore, the period of the orbit is approximately 1.37 hours.