The acceleration of a body due to the attraction of the earth (radius R ) at a distance 2R from the surface of the earth is (.g= acceleration due to gravity at the surface of the earth)

Question

The acceleration of a body due to the attraction of the earth (radius R ) at a distance 2R from the surface of the earth is (.g= acceleration due to gravity at the surface of the earth) (A) g/9 (B) g/3 (C) g/4 (D) g

Tardigrade Admin · Accepted Answer

g'=g((R/R + h))2=g((R/R + 2 R))2=(g/9)

Q. The acceleration of a body due to the attraction of the earth (radius $R$ ) at a distance $2 R$ from the surface of the earth is $(g =$ acceleration due to gravity at the surface of the earth)

$g /9$

$g /3$

$g /4$

$g$

$g^{^{'}} = g (\frac{R}{R + h})^{2} = g (\frac{R}{R + 2 R})^{2} = \frac{g}{9}$

Q. The acceleration of a body due to the attraction of the earth (radius R ) at a distance 2R from the surface of the earth is (g= acceleration due to gravity at the surface of the earth)

g/9

g/3

g/4

g

g′=g(R+hR​)2=g(R+2RR​)2=9g​

Q. The acceleration of a body due to the attraction of the earth (radius $R$ ) at a distance $2 R$ from the surface of the earth is $(g =$ acceleration due to gravity at the surface of the earth)

$g /9$

$g /3$

$g /4$

$g$

$g^{^{'}} = g (\frac{R}{R + h})^{2} = g (\frac{R}{R + 2 R})^{2} = \frac{g}{9}$