Doubt by Devika
Solution :
R'=R-2% of R
R'=R-(2/100)R
R'=R-0.02R
R'=R-(2/100)R
R'=R-0.02R
R'=R(1-0.02)
R'=R(0.98)
R'=0.98R
R'=R(0.98)
R'=0.98R
We know,
g=GM/R²
g'=GM/R'²
g'=GM/(0.98R)²
g'=GM(0.98)²R²?
g'=GM/(0.9604)R²
g'=g/0.09604
g'=1.0412g
Percentage change in acceleration due to gravity
=[(g'-g)/g]×100%
=[1.0412g-g]/g×100%
=[0.0412g/g]×100%
=[0.0412]×100%
=4.12%(apprxo)
g=GM/R²
g'=GM/R'²
g'=GM/(0.98R)²
g'=GM(0.98)²R²?
g'=GM/(0.9604)R²
g'=g/0.09604
g'=1.0412g
Percentage change in acceleration due to gravity
=[(g'-g)/g]×100%
=[1.0412g-g]/g×100%
=[0.0412g/g]×100%
=[0.0412]×100%
=4.12%(apprxo)
Hence, the value of acceleration due to gravity increases by 4.12% (approx).
Note : We can also solve this question by using logarithmic differentitation which will give an approximate answer of around 4%.