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4)How much below the surface of the earth the acceleration due to gravity i)reduce to 36 % ii)reduce by 36 % of its value on the surface of the earth Radius of earth=6400km |
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Answer» tion:Let the DEPTH at which acceleration due to gravity ( g) becomes 70% or 7g/10 = d.We know thatg_{d} = g(1 - \frac{d}{r} )g d =g(1− rd )Giveng_{d} = \frac{7g}{10}g d = 107g Also, R = 6400 km [ Radius of earth ]7g/10 = g [ 1 - d / 6400]7/10= 1-d/6400d/6400 = 1 - 7/10d/6400 = 3/10d =3 \times 6400 \div 103×6400÷10d= 1920 km.Therefore,1920 \: \: \: km1920km below the SURFACE of the earth does the acceleration due to gravity BECOME 70% of its value at the surface of earth |
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