A geosynchronous satellite is a satellite whose orbital period matches the rotation of the Earth. Calculate the height above the surface of the Earth which a satellite must have in order to be in a geosynchronous orbit. (Please enter your answer in units of kilometers)
http://www.schoolphysics.co.uk/age16-19/Mechanics/Gravitation/text/Geostationary_satellite/index.html
by the way p is pi, 3.14159 etc
Thank you
To calculate the height above the surface of the Earth for a geosynchronous satellite, we first need to determine the orbital period that matches the rotation of the Earth.
The rotation period of the Earth is approximately 24 hours. Therefore, the orbital period of a geosynchronous satellite should also be 24 hours.
The formula to calculate the orbital period (T) of a satellite is given by:
T = 24 hours
Next, we can use the formula for the orbital period to determine the satellite's height (h) above the surface of the Earth. The formula is:
h = (GM(T^2)/(4π^2))^(1/3) - R
Where:
- G is the gravitational constant (approximately 6.67430 x 10^-11 m^3/kg/s^2)
- M is the mass of the Earth (approximately 5.972 × 10^24 kg)
- T is the orbital period in seconds
- π is a mathematical constant (approximately 3.14159)
- R is the radius of the Earth (approximately 6,371 km)
Converting the orbital period from hours to seconds, we have:
T = 24 hours x 60 minutes x 60 seconds = 86,400 seconds
Substituting the values into the formula, we have:
h = (6.67430 x 10^-11 m^3/kg/s^2)*(5.972 × 10^24 kg)*((86,400 seconds)^2)/(4π^2)^(1/3) - 6,371 km
Simplifying the formula gives us:
h = (3.986004418 x 10^14 m^3/s^2)^(1/3) - 6,371 km
Now we can solve for h:
h ≈ 42,164 km
Therefore, a geosynchronous satellite must have a height of approximately 42,164 kilometers above the surface of the Earth.