What is the orbital speed of a satellite of mass 490 kg in geostationary orbit?
the distance in one orbit: 2PI(re+h)
period=24 hrs (convert to seconds).
v=2PI(re+h)/period
gravity acceleration= centripetalacc
9.8 re^2/(re+h)^2 = v^2 (above)/(re+h)
solve for h re is radius earth.
then solve for V
To find the orbital speed of a satellite, we need to use the equation for orbital velocity:
v = √(GM/r)
where:
- v is the orbital velocity (speed),
- G is the gravitational constant,
- M is the mass of the Earth, and
- r is the radius of the orbit.
In the case of a geostationary orbit, the satellite remains stationary with respect to a fixed point on the Earth's surface. This means that the satellite's orbital radius is equal to the radius of the Earth.
The mass of the Earth, M, is approximately 5.98 x 10^24 kg.
The radius of the Earth, r, is approximately 6,371 km, or 6,371,000 meters.
Plugging in these values into the orbital velocity equation:
v = √((6.67 x 10^-11 N m^2/kg^2) * (5.98 x 10^24 kg) / (6,371,000 m))
Now, let's calculate the orbital velocity:
v = √((39.79306 x 10^13 N m^2 / kg) / 6,371,000 m)
v = √(6.24409268765 x 10^6 N m^2/kg / m)
v = √(9.80530240325 m^2/s^2)
v = 3.13 x 10^3 m/s
So, the orbital speed of a satellite of mass 490 kg in geostationary orbit is approximately 3.13 x 10^3 meters per second.