The weight of a body on the surface of the earth is $100 \mathrm{~N}$. The gravitational force on it when taken at a height, from the surface of earth, equal to one-fourth the radius of the earth is:

<p>To find the gravitational force on the body when taken at a height equal to one-fourth the radius of the Earth, we can use the formula for gravitational force:</p> <p>$F = G \frac{m_1 m_2}{r^2}$</p> <p>where $F$ is the gravitational force, $G$ is the gravitational constant, $m_1$ and $m_2$ are the masses of the two objects, and $r$ is the distance between their centers.</p> <p>The weight of the body on the surface of the Earth is given as $100\,\text{N}$, which is also the gravitational force acting on it:</p> <p>$F_\text{surface} = G \frac{m_\text{body} m_\text{earth}}{R_\text{earth}^2}$</p> <p>When the body is taken to a height equal to one-fourth the radius of the Earth, the distance between the centers of the body and the Earth becomes $$r_\text{new} = R_\text{earth} + \frac{1}{4} R_\text{earth} = \frac{5}{4} R_\text{earth}$$.</p> <p>Now the gravitational force acting on the body at this new height is:</p> <p>$F_\text{new} = G \frac{m_\text{body} m_\text{earth}}{r_\text{new}^2}$</p> <p>$$F_\text{new} = G \frac{m_\text{body} m_\text{earth}}{\left(\frac{5}{4} R_\text{earth}\right)^2}$$</p> <p>To find the ratio between the new gravitational force and the original force on the surface, we can write:</p> <p>$$\frac{F_\text{new}}{F_\text{surface}} = \frac{G \frac{m_\text{body} m_\text{earth}}{\left(\frac{5}{4} R_\text{earth}\right)^2}}{G \frac{m_\text{body} m_\text{earth}}{R_\text{earth}^2}}$$</p> <p>Canceling out the common terms, we get:</p> <p>$$\frac{F_\text{new}}{F_\text{surface}} = \frac{R_\text{earth}^2}{\left(\frac{5}{4} R_\text{earth}\right)^2} = \frac{1}{\left(\frac{5}{4}\right)^2} = \frac{1}{\left(\frac{25}{16}\right)} = \frac{16}{25}$$</p> <p>Now, since the weight of the body on the surface is $100\,\text{N}$, we can find the new gravitational force as:</p> <p>$F_\text{new} = \frac{16}{25} \times 100\,\text{N} = 64\,\text{N}$</p> <p>So, the gravitational force on the body when taken at a height equal to one-fourth the radius of the Earth is $64\,\text{N}$.</p>