1.

What is a period of revolution of earth satellite ? Ignore the height of satellite above the surface of earth. Given : The value of gravitational acceleration g= 10 m s^(-2) Radius of earta R_(E)= 6400 km. Take pi=3.14.

Answer»

85 minutes
156 minutes
83.73 minutes
90 minutes

Solution :`T = 2pisqrt((R)/(g)) = 2pisqrt((6400 XX 10^(3))/(10))`
`T = 5024s = 83.73` MIN


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