91Ó°ÊÓ

We can assume the data listed below as,

• The mass of the book is,$m_1=m_2=5 kg$.
• The distance between the books is, $r=10 cm$ .
• The books are assumed as a material point.

The force exerted by a body on another body is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.

The magnitude of the force that the books exert on each other is given by

$F_{tt}=G\frac{m_1â‹\dots m_2}{r^2}$

Here,= Gravitational Constant

$m_1$= mass of first book

$m_2$= mass of second book

r = distance between the two books

Substitute all the values in the above equation.

$$\begin{gathered} \mathrm{F}=6.7×{10}^{-11}\frac{{\mathrm{Nm}}^2}{{\mathrm{kg}}^2}\frac{\left(5\mathrm{kg}\right)·\left(5\mathrm{kg}\right)}{{\left(0.1\mathrm{m}\right)}^2} \\ =16.75×{10}^{-8}\mathrm{N} \end{gathered}$$

From an approximate equation, the force on the books on the surface of Earth can be expressed as,

$F=mâ‹\dots g$

Here, m is the mass of the book, g is the acceleration due to gravity..

Substitute all the values in the above equation.

$$\begin{gathered} F_{ET}=9.8\frac{N}{kg}·5kg \\ =49N \end{gathered}$$

Now, compare the two values of forces

The ratio of these forces is:

$$\begin{gathered} \frac{F_{tt}}{F_{ET}}=\frac{16.75×{10}^{-8}N}{49N} \\ ≈0.341×{10}^{-8} \end{gathered}$$

Thus, from the above ratio, we can see that the force exerted by the books on each other is negligible compared to the gravitational force on a book by the earth.