A body weight 48 N on the surface of the earth. The gravitational force experienced by the body due to the earth at a height equal to one-third the radius of the earth from its surface is:

The correct answer is 27 N.

To find the gravitational force (weight) experienced by the body at a height equal to one-third the radius of the Earth, we can use the formula for the variation of acceleration due to gravity with altitude.

The weight of a body on the surface of the Earth is given by:
$W_0=mg=48\text{N}$
where $m$ is the mass of the body and $g$ is the acceleration due to gravity at the surface of the Earth.

The acceleration due to gravity $g^{\prime }$ at a height $h$ above the surface of the Earth is related to $g$ by the relation:
$g^{\prime }=g{(\frac{R}{R+h}}^2$
where $R$ is the radius of the Earth.

Given that the height is one-third the radius of the Earth:
$h=\frac{R}{3}$

Substitute the value of $h$ into the denominator:
$R+h=R+\frac{R}{3}=\frac{4R}{3}$

Now, substitute this back into the formula for $g^{\prime }$:
$g^{\prime }=g{\left(\frac{R}{4R/3}\right)}^2=g{\left(\frac{3}{4}\right)}^2=\frac{9}{16}g$

The gravitational force (weight $W^{\prime }$) experienced by the body at height $h$ is:
$W^{\prime }=m{g}^{\prime }=m\left(\frac{9}{16}g\right)=\frac{9}{16}{W}_0$

Substitute the surface weight of the body ($W_0=48\text{N}$):
$W^{\prime }=\frac{9}{16}\times 48=9\times 3=27\text{N}$