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Orbital velocity V = sqrt((GM)/(R+h))...

Orbital velocity
`V = sqrt((GM)/(R+h))`
Suppose the orbit of a satellite is exactly 35780 km above the earth's surfce. Determine the tangential velocity of the satellite .

Text Solution

Verified by Experts

Given : h = 35780 km = 3578000m
`R = 6.4 xx 10^(6) m = 6400000m`
` M = 6 xx 10^(24) kg`
`G = 6.67 xx 10^(-11) Nm^(2)//kg^(2)`
To find v= ?
Formula : `V = sqrt((GM)/(R+h))`
Solution `V = sqrt((6.67xx 10^(-11) xx 6xx 10^(24))/((35780000 +6400000)))`
` = sqrt((40.02xx 10^(13))/(42180000))`
` = sqrt((40.02 xx 10^(13))/(42.18 xx 10^(6)))`
` = sqrt((40)/(42) xx 10^(7))`
` = sqrt(0.95 xx 10^(7))`
` = sqrt(9.5 xx 10^(6))`
` = 3.08 xx 10^(3)m//s or 3.08 km//s`
`therefore ` Tangential velocity will be `3.08 km//s`
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