A remote-sensing satellite of earth revo...

A remote-sensing satellite of earth revolves in a circular orbit at a hight of `0.25xx10^(6)m` above the surface of earth. If earth's radius is `6.38xx10^(6)m` and `g=9.8ms^(-2)`, then the orbital speed of the satellite is

`6.67 km//s`

`7.76 km//s`

`8.56 km//s`

`9.13 km//s`

Text Solution

Verified by Experts

The correct Answer is:

Here `r = 0.25 xx 10^(6) + 6.38 xx 10^(6) = 6.63 xx 10^(6)m`
`R = 6.38 xx 10^(6), g = 9.8 m//s^(2)`
as `upsilon = sqrt((GM)/(r )) = sqrt((GM xx R^(2))/(R^(2) xxr)) = sqrt((gR^(2))/(r ))`
`upsilon = sqrt((9.8 xx 6.38 xx 6.38 xx 10^(12))/(6.63 xx 10^(6))) = sqrt(60 xx 10^(6))m//s`
`= 7.756 km//s = 7.76 km//s`

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