calculate at what height above the earths surface a satellite must be placed if it is to remain over the same geographical point on the equator of the earth. what is the velocity of the satellite?
look up geostationary satellite
here
http://en.wikipedia.org/wiki/Geostationary_orbit
35,786 kilometers above earth
To determine the height above the Earth's surface at which a satellite must be placed to remain over the same geographical point on the equator, we can use the concept of a geostationary orbit.
In a geostationary orbit, a satellite orbits the Earth at the same rotational speed as the Earth, which allows it to remain fixed above a particular point on the equator. This means that the satellite will complete one orbit in the same amount of time as the Earth completes one rotation (approximately 24 hours).
To calculate the height, we can use the following formula:
h = (R * cos φ) - R
Where:
h = Height above the Earth's surface
R = Radius of the Earth
φ = Latitude of the desired point (in this case, it is 0° for the equator)
The radius of the Earth, R, is approximately 6,371 km.
Using the formula, we can calculate the height above the Earth's surface:
h = (6,371 km * cos 0°) - 6,371 km
h = (6,371 km * 1) - 6,371 km
h = 0 km
This means that the satellite must be placed at a height of 0 km above the Earth's surface, which is essentially at sea level.
Now, to calculate the velocity of the satellite, we can use the formula for the orbital velocity of a satellite in a circular orbit:
v = √(G * Me / r)
Where:
v = Velocity of the satellite
G = Gravitational constant (approximately 6.67430 x 10^-11 N(m/kg)^2)
Me = Mass of the Earth (approximately 5.972 x 10^24 kg)
r = Radius of the orbit (which is equal to the sum of the radius of the Earth and the height above the Earth's surface)
Using the given values, we can calculate the velocity:
r = 6,371 km + 0 km = 6,371 km
v = √((6.67430 x 10^-11 N(m/kg)^2) * (5.972 x 10^24 kg) / (6,371 km))
After performing the calculation, we would obtain the velocity of the satellite in m/s, which can be converted to other units if desired.
Please note that these calculations assume a simplified model of the Earth and neglect other factors such as atmospheric drag and the eccentricity of the satellite's orbit.