The height at which the acceleration due to gravity becomes $$\frac{g}{9}$$ (where $$g=$$ the acceleration due to gravity on the surface of the earth) in terms of $$R,$$ the radius of the earth, is:
A.
$$\frac{R}{{\sqrt 2 }}$$
B.
$$\frac{R}{2}$$
C.
$$\sqrt 2 \,R$$
D.
$$2\,R$$
Answer :
$$2\,R$$
Solution :
we know that $$\frac{{g'}}{g} = \frac{{{R^2}}}{{{{\left( {R + h} \right)}^2}}}$$
$$\eqalign{
& \therefore \frac{{\frac{g}{9}}}{g} = {\left[ {\frac{R}{{R + h}}} \right]^2} \cr
& \therefore h = 2\,R \cr} $$
Releted MCQ Question on Basic Physics >> Gravitation
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