At what height the acceleration due to gravity decreases by 51% of its value on the surface of the earth?

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Text Solution

Solution : As, `g_(h)=g((R)/(R+h))^(2) and g_(h)=g-51%` of `g=g-(51)/(100)g` <br> `:. g_(h)=(49)/(100)g` <br> `:. (49)/(100)g=g((R)/(R+h))^(2)` <br> `:.(7)/(10)=(R)/(R+h)` <br> `:. 7R+7h=10R` <br> `:. 7h=3RrArrh=(3R)/(7)`.