A body weight $\mathrm{W}$, is projected vertically upwards from earth's surface to reach a height above the earth which is equal to nine times the radius of earth. The weight of the body at that height will be :

<p>The weight of an object varies with altitude due to the change in gravitational force. The force of gravity decreases with the square of the distance from the center of the Earth.</p> <p>The gravitational force (weight) at a height h from the surface of the Earth is given by:</p> <p>$W&#39; = W \left(\frac{R}{{R + h}}\right)^2$</p> <p>where:</p> <ul> <li>W&#39; is the weight at height h,</li> <li>W is the weight at the Earth&#39;s surface,</li> <li>R is the radius of the Earth,</li> <li>h is the height above the Earth&#39;s surface.</li> </ul> <p>In this problem, the height h is given as nine times the radius of the Earth (h = 9R). Substituting h = 9R into the equation gives:</p> <p>$W&#39; = W \left(\frac{R}{{R + 9R}}\right)^2 = W \left(\frac{1}{10}\right)^2$</p> <p>Simplifying this gives:</p> <p>$W&#39; = \frac{W}{100}$</p> <p>Therefore, the weight of the body at that height will be 1/100th of its weight on the surface of the Earth. <br/><br/>The correct answer is Option C: $\frac{W}{100}$.</p>