Time Period Of A Satellite Revolving Earth Is Given As T Height At Which It Orbits The Earth

The semi synchronous orbit is a near circular orbit low eccentricity 26 560 kilometers from the center of the earth about 20 200 kilometers above the surface.

Time period of a satellite revolving earth is given as t height at which it orbits the earth.

What is the time period of a satellite orbiting at a height of 2 5 r above the earth s surface. The speed of the satellite is 7 519 km s and the period of revolution of the satellite is 5930 s. Radius of earth 6400 km. The period of a satellite is the time it takes it to make one full orbit around an object.

Where r is the radius of the earth. Distance of moon from the earth 3 84 x 10 5 km. T 5071 3600 1 408 h. Radius of earth r 6400 km 6 4 x 10 8 m radius of orbit of moon r 3 84 x 10 5.

Now if the height of the satellite h from the surface of the earth is negligible with respect to the radius of the earth then we can write r r h r as h is negligible. This gives the height of a satellite above the earth s surface whose time period is same as that of earth s. If you know the satellite s speed and the radius at which it orbits you can figure out its period. The semi synchronous orbit and the molniya orbit.

From equation 3 the fundamental form of orbital velocity equation we get an equation of nearby orbit s orb. Then r gm e t 2 4π 2 1 3 42400 km and h 36000 km. Two medium earth orbits are notable. If time period of a satellite is 24 hrs.

Such a satellite appears to be stationary when observed from the earth s surface and is hence known as geostationary satellite. In this case you add the distance from the center of the earth to the surface of the earth 6 38 10 6 meters to the satellite s height above the earth. The mean angular velocity of the earth around the sun is 1 per day. The time period of an earth stellite in circuarl orbit is independet fo.

The period of the earth as it travels around the sun is one year. The equation assumes that the satellite is high enough off the ground that it orbits out of the atmosphere. From the following data calculate the period of revolution of the moon around the earth. A geostationary satellite is orbiting the earth at a height of 6r above its surface.

The time period of a satellite orbiting around the earth is given by. T 2πr v c 2 x 3 142 x 6400 7 931 5071 s. You can calculate the speed of a satellite around an object using the equation. Ncert dc pandey sunil batra hc verma pradeep errorless.

R 1 r 6r 7r r 2 r 2 5 r 3 5 r time period of geostationary satellite t 1 24 hours. G 9 8 m s 2. A satellite at this height takes 12 hours to complete an orbit.