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)`.
Text Solution
AI Generated Solution
Topper's Solved these Questions
GRAVITATION
SL ARORA|Exercise Problem From Competitive Examinations|14 Videos
GRAVITATION
SL ARORA|Exercise Exercise|480 Videos
FLUIDS IN MOTION
SL ARORA|Exercise All Questions|117 Videos
HEAT
SL ARORA|Exercise Problem For Self Practice|72 Videos
Similar Questions
Explore conceptually related problems
A remote sensing satellite of the Earth in a circular orbit at a height of 400 km above the surface of Earth. What is the (a) orbital speed, and (b) period of revolution of satellite ? Radius of Earth = 6 xx 10^(6) m and acceleration due to gravity the surface of Earth is 10 m//s^(2) .
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 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)
Calculate the escape speed for an atmospheric particle 1600 km above the Earth's surface, given that the radius of the Earth is 6400 km amd acceleration due to gravity on the surface of Earth is 9.8 ms^(-2) .
An artificial satellite is revolving in a circular orbit at a height of 600 km from the earth's surface. If its time of revolution be 100 minutes, calculate the acceleration due to gravity acting on it towards the earth. Radius of earth = 6.4 xx 10^(6) m
An artificial satellite is orbiting at a height of 1800 km from the earth's surface. The earth's radius is 6300 km and g=10m/s^(2) on its surface. What is the radial acceleration of the satellite?
A satellite moves in a circular orbit around the earth at height (R_(e))//2 from earth's surface where R_(e) is the radius of the earth. Calculate its period of revolution. Given R = 6.38 xx 10^(6) m .
Calculate the period of revolution of a satellite orbiting above the surface of the Earth at a distance equal to 9 times the radius of the Earth.
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
An artificial satellite is describing an equatorial orbit at 3600 km above the earth's surface. Calculate its period of revolution? Take earth radius 6400km .
