A body weight W Newton at the surface of the earth. Its weight at a height equal to half the radius of the earth will be

The correct option is 4W/9.

Let's derive the solution step-by-step using the law of gravitation.

The weight of a body of mass $m$ at the surface of the Earth (radius $R$) is given by:
$W=mg=\frac{GMm}{R^2}$
where $G$ is the universal gravitational constant, and $M$ is the mass of the Earth.

At a height $h$ above the surface of the Earth, the distance of the body from the center of the Earth becomes $r=R+h$. Therefore, the weight $W^{\prime }$ of the body at height $h$ is given by:
$W^{\prime }=\frac{GMm}{{(R+h)}^2}$

According to the problem, the height is equal to half the radius of the earth, so:
$h=\frac{R}{2}$

Substituting this value of $h$ into the expression for $W^{\prime }$:
$W^{\prime }=\frac{GMm}{{(R+\frac{R}{2})}^2}$
$W^{\prime }=\frac{GMm}{{\left(\frac{3R}{2}\right)}^2}$
$W^{\prime }=\frac{GMm}{\frac{9}{4}{R}^2}$
$W^{\prime }=\frac{4}{9}(\frac{GMm}{R^2})$

Since the weight at the surface is $W=\frac{GMm}{R^2}$, we can substitute $W$ in the equation:
$W^{\prime }=\frac{4}{9}W$

Thus, the weight of the body at a height equal to half the radius of the Earth will be 4W/9.