A remote - sensing satellite of earth revolves in a circular orbit at a height of 0.25 106 m above the surface of earth. If earth's radius is 6.38 106 m and g=9.8 ms2, then the orbital speed of the satellite is

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Q. A remote - sensing satellite of earth revolves in a circular orbit at a height of

0.25 \times 10^{6}

m above the surface of earth. If earth's radius is

6.38 \times 10^{6} m

and

g = 9.8 m s^{- 2}

, then the orbital speed of the satellite is:

Q. A remote-sensing satellite of earth revolves in a circular orbit at a height of

0.25 \times 10^{6}

m

above the surface of earth. If earth's radius is

6.4 \times 10^{6} m

and

g = 9.8 m s^{- 2}

, then the orbital speed of the satellite is

Q. A remote-sensing satellite of earth revolves in a circular orbit at a height of

0.25 \times 10^{6}

m

above the surface of earth. If earth's radius is

6.4 \times 10^{6} m

and

g = 9.8 m s^{- 2}

, then the orbital speed of the satellite is

A satellite is revolving round the earth in circular orbit at some height above surface of earth. It takes $5.26 \times 10^{3}$ seconds to complete a revolution while its centripetal acceleration is $9.92 m / s^{2}$ . Height of satellite above surface of earth is (Radius of earth $6.37 \times 10^{6}$ m)

Q. A satellite moves in a circular orbit around the earth at height

R / 2

from earth's surface where

R_{e}

is the radius of the earth. Calculate its period of revolution. Given

R_{e} = 6.38 \times 10^{6}

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