At what height from the surface …
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Naveen Sharma 8 years, 9 months ago
Ans. Let Acceleration due to gravity at Height h is g' and Acceleration due to gravity at surface is g.
\(g' = g - {35g\over 100} = g- {7g \over 20} = {13g\over 20}\)
We know that Acceleration due to gravity at Height h from the Earth Surface is Given By :
\(g' = g ({1 - {2h\over R}}) \space \space \space \space \space \space \space ... (1) \)
Where (R = 6400 Km ) is Radius of Earth
=> \({13g\over 20 }= g({1- {2h\over R})}\)
=> \({13\over 20 }= ({1- {2h\over R})}\)
=> \(1- {13\over 20 }= {2h\over R}\)
=> \({7\over 20 }= {2h\over R} => h = {7R\over40}\)
=> \( h = {7\times 6400\over40} = 1120\space Km\)
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