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A satellite revolves around the earth at...

A satellite revolves around the earth at a height of 1000km. The radius of earth is `6.38 xx 10^(3)km`. Mass of earth is `6 xx 10^(24)kg and G= 6.67 xx 10^(-11)N m^(2) kg^(-2)`. Find the orbital velocity and period of revolution.

Text Solution

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We are given: `h= 1000km = 10^(6)m`
`R= 6.38 xx 10^(3)km = 6.38 xx 10^(6)m`
`R+ h = (6.38 xx 10^(6) + 10^(6))m= 7.38 xx 10^(6)m`
`M= 6 xx 10^(24)kg`
We are to calculate: orbital velocity `v_(o)` and period of revolution T. Using the relation
`v_(o)= sqrt((GM)/(R+h)) = sqrt((6.67 xx 10^(-11) xx 6 xx 10^(24))/(7.38 xx 10^(6)))= 7364m//s`
Also, period of revolution
`T= (2pi (R+h))/(v_(o)) = (2pi xx 7.38 xx 10^(6))/(7364)= 6294s`
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