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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 the height, then
`therefore a = (V^(2))/(r) = g_(h) rArr (V^(2))/(r) = 8 = (64 xx 10^(6))/(R+h) = 8`
`rArr R + h = 8 xx 10^(6)`
`h = 8 xx 10^(6) - 6 xx 10^(6) = 2 xx 10^(6)m`
= 2000 km
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