A remote-sensing satellite of earth revolves in a circular orbit at a hight 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

A remote - sensing satellite of earth revolves in a circular orbit at a height of 0.25 x 106 m above the surface of earth. It earth's radius is 6.38 xx 10^(6)m and g = 9.8 ms^(- 2) , then the orbital speed of the satellite is ..........

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

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

A remote-sensing satellite of earth revolves in a circular orbit at a height of 0.25xx10^6 m above the surface of earth. Find the orbital speed and the period of revolution of satellite . Given : Earth's radius R_e=6.38xx10^6 m and g=9.8 m//s^2 .

An artificial satellite is revolving in a circular orbit at height of 1200 km above the surface of the earth. If the radius of the earth is 6400 km and mass is 6xx10^(24) kg , the orbital velocity is

A remote sensing satellite of the Earth revolves in a circular orbit at a height of 250 km above the Earth's surface. What is the (i) Orbital speed and (ii) period of revolution of the satellite. Radius of the Earth, R=6.38xx10^(6)m , and acceleration due to gravity on the surface of the Earth, g=9.8ms^(-2)

A remote sensing satellite of the earth revolves in a circular orbit at a height of 250 km above the earth's surface. What is the (i) orbital speed and (ii) period of revolution of the satellite ? Radius of the earth, R=6.38xx10^(6) m, and acceleration due to gravity on the surface of the earth, g=9.8 ms^(-2) .