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A satellite is revolving round the earth...

A satellite is revolving round the earth in a circular orbit with a velocity of 8km/s. at a height where acceleration due to gravity is `8m//s^(2)`. How high is the satellite from the earth ? (Take R = 6000 km)

Text Solution

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As centripetal acceleration equals to acceleration due to gravity at that height , then
`:.a=V^2/r=g_(h)impliesV^2/r=8=(64xx10^(6))/(R+h)=8`
`implies R + h = 8 xx10^(6)`
`h = 8 xx10^(6) - 6 xx10^(6)= 2 xx10^(6) m = 2000 km `
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