The percentage change in acceleration due to gravity at an altitude equal to radius of earth compared to that surface of earth is given by ??
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Answered by
19
Heya.....!!
On surface of Earth " g = GM/R^2
At height R new g' = GM/(R+R)^2
g'= GM/4R^2
∆g = GM/4R^2 - GM/R^2
∆g = - g[3/4]
∆g = -3/4g
Percentage change ( %) =. ∆g/g×100
= -3/4×100 = -75 % decrease
>> 75% decrease is the answer .
HOPE IT HELPS U
On surface of Earth " g = GM/R^2
At height R new g' = GM/(R+R)^2
g'= GM/4R^2
∆g = GM/4R^2 - GM/R^2
∆g = - g[3/4]
∆g = -3/4g
Percentage change ( %) =. ∆g/g×100
= -3/4×100 = -75 % decrease
>> 75% decrease is the answer .
HOPE IT HELPS U
Answered by
4
it will Decrease by 75 %
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