Verified answer

Explanation:

The time period of a satellite in orbit depends on its distance from the center of the Earth. It can be calculated using Kepler's third law of planetary motion, which states:

T^2 ∝ r^3

Where:

T = Time period (in seconds)

r = Distance from the center of the Earth (in meters)

For a geostationary satellite at a height of 36,000 km, the total distance from the Earth's center is:

r = Earth's radius (6,400 km) + Satellite's altitude (36,000 km) = 42,400 km = 42,400,000 meters

Now, let's find the time period for the geostationary satellite:

T_geostationary^2 ∝ (42,400,000)^3

T_geostationary^2 ∝ 77,894,144,000,000,000,000,000

Now, let's calculate the time period for a spy satellite orbiting very close to Earth's surface (R = 6,400 km):

r_spy = Earth's radius (6,400 km) = 6,400,000 meters

Using Kepler's third law, we can find the time period for the spy satellite:

T_spy^2 ∝ (6,400,000)^3

T_spy^2 ∝ 262,144,000,000,000,000

Now, we can compare the time periods:

T_geostationary^2 / T_spy^2 = (77,894,144,000,000,000,000,000) / (262,144,000,000,000,000) ≈ 297,053,760

Taking the square root of this ratio to find the time period of the spy satellite:

T_spy ≈ √297,053,760 ≈ 17,240 seconds

So, the time period of the spy satellite orbiting very close to Earth's surface is approximately 17,240 seconds, or about 4.79 hours.