Determine the velocity of a satellite orbiting Earth at a height of 3.2 x 10^3 meters.

To determine the velocity of a satellite orbiting Earth at a specific height, we can use the formula for the orbital velocity of a satellite:

V = sqrt(G * M / R)

Where:
V is the velocity of the satellite in meters per second,
G is the gravitational constant (approximately 6.67430 × 10^-11 N * (m/kg)^2),
M is the mass of the Earth (approximately 5.972 × 10^24 kg),
R is the distance between the center of the Earth and the satellite (which is the sum of the radius of the Earth and the height of the satellite).

Let's calculate it step by step:

Step 1: Convert the height to the distance from the center of the Earth.
Given height = 3.2 x 10^3 meters

The radius of the Earth is approximately 6.371 x 10^6 meters (mean sea level).

Hence,
R = radius of the Earth + height
= 6.371 x 10^6 meters + 3.2 x 10^3 meters
= 6.3710032 x 10^6 meters

Step 2: Substitute the values into the formula.

V = sqrt(G * M / R)
= sqrt((6.67430 × 10^-11 N * (m/kg)^2) * (5.972 × 10^24 kg) / (6.3710032 x 10^6 meters))

Step 3: Calculate the result.

V = sqrt(4.904948 x 10^14 N * m^2 / kg / meters)
= sqrt(4.904948 x 10^14 N * m / kg)
= 7.550109652 x 10^3 meters per second

Therefore, the velocity of the satellite orbiting Earth at a height of 3.2 x 10^3 meters is approximately 7.55 x 10^3 meters per second.