An object is taken to a height above the surface of earth at a distance ${5 \over 4}$ R from the centre of the earth. Where radius of earth, R = 6400 km. The percentage decrease in the weight of the object will be :

<p>The weight of an object at a distance d from the center of the Earth is given by:</p> <p>$W&#39; = W \left(\frac{R}{d}\right)^2$</p> <p>where:</p> <ul> <li>W&#39; is the weight at distance d,</li> <li>W is the weight at the Earth&#39;s surface,</li> <li>R is the radius of the Earth,</li> <li>d is the distance from the center of the Earth.</li> </ul> <p>In this problem, we are asked to find the percentage decrease in weight, which can be calculated by:</p> <p>$\Delta W = \frac{W - W&#39;}{W} \times 100\%$</p> <p>where ΔW is the percentage change in weight. Substituting the weight formula into the percentage change formula gives:</p> <p>$\Delta W = \left(1 - \left(\frac{R}{d}\right)^2\right) \times 100\%$</p> <p>The radius of the Earth R is 6400 km and the distance d from the center of the Earth is given as $\frac{5}{4}R$. Substituting these values into the equation gives:</p> <p>$$\Delta W = \left(1 - \left(\frac{6400~km}{\frac{5}{4} \cdot 6400~km}\right)^2\right) \times 100\% = \left(1 - \left(\frac{4}{5}\right)^2\right) \times 100\% = 36\% $$</p> <p>Therefore, the percentage decrease in the weight of the object when taken to a height of $\frac{1}{4}R$ above the surface of the Earth is 36%.</p>