A geostationary orbit, also referred to as a geosynchronous equatorial orbit (GEO), is a circular geosynchronous orbit 35,786 kilometres (22,236 miles) in altitude above Earth's equator (42,164 kilometers in radius from Earth's center) and following the direction of Earth's rotation.
An object in such an orbit has an orbital period equal to the Earth's rotational period, one sidereal day, and so to ground observers it appears motionless, in a fixed position in the sky. The concept of a geostationary orbit was popularised by the science fiction writer Arthur C. Clarke in the 1940s as a way to revolutionise telecommunications, and the first satellite to be placed in this kind of orbit was launched in 1963.
Communications satellites are often placed in a geostationary orbit so that Earth-based satellite antennas (located on Earth) do not have to rotate to track them but can be pointed permanently at the position in the sky where the satellites are located. Weather satellites are also placed in this orbit for real-time monitoring and data collection, and navigation satellites to provide a known calibration point and enhance GPS accuracy.
Geostationary satellites are launched via a temporary orbit, and placed in a slot above a particular point on the Earth's surface. The orbit requires some stationkeeping to keep its position, and modern retired satellites are placed in a higher graveyard orbit to avoid collisions.
Summary:: What is the maximum safe mass of an asteroid in geostationary orbit before it causes problems?
If there was an asteroid in geostationary orbit around the Earth, over the Pacific Ocean, what would be the highest mass it could have before it would start having...
A Satellite is brought up into a geostationary orbit (altitude 35800km measured from the surface of the earth). Satellite weights 1000.0kg. How much work is required to bring satellite from a surface of the Earth to
Newton's law of...
When considering a satellite in geosynchronous orbit, its speed is zero across (relative to) Earth's surface.
From Kepler's third Law: T2=(4π2r3)/(GM), we can derive that v2=GM/r
This would tell us that as the radius of a satellite to Earth's centre increases, its velocity...
I had this discussion while driving home to California from a trip to Washington state with a friend.
We were discussing the stability of a completely rigid, untethered, ring-structure around Earth, and I did not know how to explain to him that such a thing must be tethered by rigid towers lest...
rigid body dynamics
Calculate the height of a geo-stationary satellite of earth.
Gravitational force of earth=6.667 x 10^-11 nm^2/kgm^2
Mass of earth=6x10^24 kgm
Radius of earth=6400 km
The Attempt at a Solution
I plugged everything into the equation...