How much below the surface of the earth does the acceleration due to gravity becomes 70% of its value at the surface of earth?
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ve^2=2RM/R WHERE ve=escape speed , R = radius of earth , M = mass of earth , G = universal G constant
Vo^2=GM/R WHERE Vo=orbital speed , R = radius of earth , M = mass of earth , G = universal G constant
Ve=(2^1/2 )Vo
Vo^2=GM/R WHERE Vo=orbital speed , R = radius of earth , M = mass of earth , G = universal G constant
Ve=(2^1/2 )Vo
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