Final answer:
The height of a geostationary satellite above the Earth's surface is 35,880 km, and its orbital velocity can be calculated using the formula v = √(gr), with 'g' being the acceleration due to gravity and 'r' being the orbital radius.
Step-by-step explanation:
To find the height and orbital velocity of a geostationary satellite, we need to use the provided radius of Earth and the acceleration due to gravity. A geostationary satellite orbits in such a way that it remains stationary with respect to a point on the Earth's equator. This type of orbit has a specific radius and period that we can calculate based on the given information.
The height of a geostationary satellite above the Earth's surface can be calculated by subtracting the Earth's radius from the satellite's orbital radius. As stated in another problem reference provided, a geosynchronous satellite orbits Earth at a distance of 42,250 km (the orbital radius from the center of the Earth). Since the Earth's radius is 6370 km, the height above the surface is:
Height = Orbital Radius - Earth's Radius = 42,250 km - 6370 km = 35,880 km
To calculate the orbital velocity of the geostationary satellite, we apply the formula for circular orbital velocity, which is derived from balancing gravitational force and the necessary centripetal force for circular motion. The formula is:
v = √(G*M_e/r)
Where 'v' is the orbital velocity, 'G' is the gravitational constant, 'M_e' is the mass of the Earth, and 'r' is the orbital radius. However, in our scenario where we are making an approximation using acceleration due to gravity and ignoring G and M_e, we can rearrange the formula for acceleration due to gravity (g = GM_e/r^2), to solve for the velocity (v) as follows:
v = √(gr) = √(10 m/s^2 * 42,250,000 m) = √(422,500,000) m/s
Performing the calculation gives us the orbital velocity.