JEE Main 2023 — Gravitation Question with Solution

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 $$64\text{N}$$.