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Calculate the height at which the value of acceleration due to gravity becomes 50% of that at the surface of the earth. (Radius of the earth = 6400 km) |
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Answer» Solution :G at height , `g_h=g/(1+h/R)^2` In this problem , `g_h=50/100`g=0.5 g 0.5g= `g/(1+h/R)^2 rArr (1+h/R)^2 =g /(0.5g)=2` `1+h/R= SQRT2, h/R =sqrt2-1 rArr` the height , h=0.414 R =0.414 x 6400 = 2650 KM . |
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