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)?

To find the period of the orbit, we can use Kepler's third law, which 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 calculate the semi-major axis of the orbit. The altitude of the satellite above the surface of the Earth is given as 780 km. Since the radius of the Earth is approximately 6371 km, we can calculate the semi-major axis as the sum of the radius of the Earth and the altitude of the satellite:

Semi-major axis = Radius of the Earth + Altitude = 6371 km + 780 km = 7151 km

Next, we can use the formula for the period of a circular orbit:

Period = 2π * √(semi-major axis^3 / GM)

where G is the gravitational constant and M is the mass of the Earth.

The gravitational constant G is approximately 6.67430 × 10^-11 m^3 kg^-1 s^-2, and the mass of the Earth M is approximately 5.972 × 10^24 kg.

Converting the semi-major axis to meters:

Semi-major axis = 7151 km * 1000 m/km = 7,151,000 m

Now we can substitute the values into the formula:

Period = 2π * √((7,151,000 m)^3 / (6.67430 × 10^-11 m^3 kg^-1 s^-2 * 5.972 × 10^24 kg))

Calculating this expression will give us the period of the orbit in seconds. To convert it to hours, we can divide the result by 3600 (since there are 3600 seconds in an hour).

Finally, we can calculate the period of the orbit in hours.