QJEEM23Ph6d46: JEE Main 2023 Physics - Class 11 JEEM23Ph6d46 Solution: The weight of a body on the surface of the earth is . The…

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:

$F = G \frac{m_1 m_2}{r^2}$

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.

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:

$F_\text{surface} = G \frac{m_\text{body} m_\text{earth}}{R_\text{earth}^2}$

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 $r_\text{new} = R_\text{earth} + \frac{1}{4} R_\text{earth} = \frac{5}{4} R_\text{earth}$ .

Now the gravitational force acting on the body at this new height is:

$F_\text{new} = G \frac{m_\text{body} m_\text{earth}}{r_\text{new}^2}$

$F_\text{new} = G \frac{m_\text{body} m_\text{earth}}{\left(\frac{5}{4} R_\text{earth}\right)^2}$

To find the ratio between the new gravitational force and the original force on the surface, we can write:

$\frac{F_\text{new}}{F_\text{surface}} = \frac{G \frac{m_\text{body} m_\text{earth}}{\left(\frac{5}{4} R_\text{earth}\right)^2}}{G \frac{m_\text{body} m_\text{earth}}{R_\text{earth}^2}}$

Canceling out the common terms, we get:

$\frac{F_\text{new}}{F_\text{surface}} = \frac{R_\text{earth}^2}{\left(\frac{5}{4} R_\text{earth}\right)^2} = \frac{1}{\left(\frac{5}{4}\right)^2} = \frac{1}{\left(\frac{25}{16}\right)} = \frac{16}{25}$

Now, since the weight of the body on the surface is $100\,\text{N}$ , we can find the new gravitational force as:

$F_\text{new} = \frac{16}{25} \times 100\,\text{N} = 64\,\text{N}$

So, the gravitational force on the body when taken at a height equal to one-fourth the radius of the Earth is $64\,\text{N}$ .