assuming earth to be a sphere of radius 6400km,calculate the height above the earth's surface at which the value of acceleration due to gravity reduces to half it's value on the earth surface
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value of g above the surface of the earth can be given by expression
g'= g(1-(h/R))
where g' is value of accelaration at that point and h is the height at which we are going and R is the radius of earth as g' = g/2
g/2 = g( 1 - (h/R) )
1/2 = 1 - (h/R)
- 1/2 = -(h/R)
so h = R/2
so at a height half of the radius of earth which becomes 3200 km the value of g reduces to half
g'= g(1-(h/R))
where g' is value of accelaration at that point and h is the height at which we are going and R is the radius of earth as g' = g/2
g/2 = g( 1 - (h/R) )
1/2 = 1 - (h/R)
- 1/2 = -(h/R)
so h = R/2
so at a height half of the radius of earth which becomes 3200 km the value of g reduces to half
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