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The centripetal acceleration of a satellite that circles the earth at an altitude 400 km above see level is (g on the surface of earth is `10m//s^(2)`, Radius of the earth is `6.4xx10^(6)m`)A. `8.75m//s^(2)`B. `9.2m//s^(2)`C. `10m//s^(2)`D. `7.5m//s^(2)` |
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Answer» Correct Answer - a `g_(2)=g[1-(2h)/(R)]=10[1-(2xx400)/(6400)]` `=10[1-(1)/(8)]` `=10[(8-1)/(8)]=(70)/(8)=8.75m//s^(2).` |
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