An earth satellite revolves in a circular orbit at a height of 300 km above the earth's surface with a period of 90 min. What is its speed ? Calculate the radial acceleration of the satellite ? Radius of the earth is `6.38 xx 10^6`m.

A satellite revolves in a circular orbit at a height of 1.6 x 10^6 m above the surface of earth earth's radius is 6.4 x 10^6 m. then the orbital speed of the satelite will be nearly (g = 9.8 m/s^2)

A satellite is revolving round the earth in circular orbit

A satellite is revolving round the earth in circular orbit

Find the period of revolution of a satellite revolving the earth at a height of 200km above earth's surface ? Radius of earth = 6400 km

The time period of the earth's satellite revolving at a height of 35800 km is

A remote-sensing satellite of earth revolves in a circular orbit at a height of 0.25xx10^(6)m above the surface of earth. If earth's radius is 6.38xx10^(6)m and g=9.8ms^(-2) , then the orbital speed of the satellite is

An earth satellite of mass m revolves in a circular orbit at a height h from the surface of the earth. R is the radius of the earth and g is acceleration due to gravity at the surface of the earth. The velocity of the satellite in the orbit is given by

A satellite moving in a circular orbit at an altitude of 1000 km completes one revolution round the earth in 105 minutes. What is (z) its angular velocity and speed? Radius of the earth =6.4 xx 10^(6) m.

A satellite of mass 1000 kg is supposed to orbit the earth at a height of 2000 km above the earth's surface. Find a. its speed in the orbit b . its kinetic energy. c . The potential energy of the earth-satellite system and d . its time period. Mass of the earth =6xx10^24kg .

An artificial satellite is revolving in a circular orbit at a height of 1200 km above the surface of the earth. If the radius of the earth is 6400 km, mass is 6 xx 10^(24) kg find the orbital velocity (G = 6.67 xx 10^(-11)Nm^(2)//kg^(2))