Sum of gravitational force $W$ and contact force $N$ on a book at rest on table is not 0 There is a book on the table fixed rigidly to the earth. The book is at rest with respect to the earth. The acceleration of this book with respect to the earth is 0. The forces on this book are   
a) Gravitational force W exerted by earth.  
b) Contact force N exerted by the table.  
Is the vector sum of both the forces 0?  
A renowned physics book - Concepts of physics by H.C Verma states that "on a very accurate measurement, the above answer is no. The sum of the forces is not 0 although the book is at rest. The earth is not strictly an inertial frame. However, the sum is not too different from 0 and we can say that earth is an inertial frame of reference to a good approximation." 
Can someone please elaborate on this? Why the sum is strictly not 0? 
 A: Because, as they said, Earth is not an inertial frame, there are also some pseudo-forces acting on the book, like the centrifugal force related to the rotation of Earth. The book is at rest in this non-inertial frame, which means that the sum of all forces and pseudoforces in this frame is equal to zero. The sum of all real forces is only almost zero.
A: The brief answer is, the earth is spinning, and your book is rotating together with the earth. There need to be some force to maintain the rotation of the book, the centripetal force. Without that force on it, the book cannot just rotate by itself. That's where the net force comes from.
You can actually calculate the net force for the book's rotation with earth. Let's take approx. $v \sim 460m/s$ as the speed of the spinning earth on the equator, where you will need max centripetal force, and radius of earth $R \sim 6400km$. Use $F = mv^2/R$, we will get the centripetal acceleration $0.03 m/s^2$, only 0.3% of the gravitational acceleration, $10 m/s^2$. 
That is, you will have maximumly about 0.3% of W left, after summing W and N. 
I hope the calculation is correct.
A: 
If there exists a book on a table the book is exerted a force on the table through weight due to gravity

The force that the book exerts on the table is not a gravitational force, it's a normal force.
The table react with an equal and opposite force which is also a normal force. So the book exerts a normal force on the table, and the table exerts a normal force on the book.

But is the force acting on the table due to gravity?

No, it's not, and in fact this force (the normal force) is only indirectly due to gravity. 
The only relevant gravitational force is the force exerted by the Earth on the book. The book also exerts a gravitational force back on the Earth, but because the Earth is so heavy, that force has no noticeable effect. (The Earth also exerts a gravitational force on the table, and the table on the Earth, but those don't matter so much in this particular scenario.)
A: There are two forces acting on a mass $m$ at the Equator.
The force due to the gravitational attraction of the Earth $F_{\rm G}=\dfrac{GM_{\rm E}m}{R^2_{\rm E}}$ and the force due to the mass being in contact with the Earth, $Z'$.  
 
If the mass was not accelerating then its equation of motion would be $F_{\rm G} - Z' = m\,0$ and so those two forces would be equal in magnitude but opposite in direction.  
However the Earth is spinning about its axis and so is the mass with an angular speed $\omega$ so the equation of motion of the mass is $F_{\rm G} - Z' = m\,R_{\rm E} \,\omega^2$.
This shows that $F_{\rm G} \ne Z'$.
You can think of the gravitational force having two components $X'$ and $Y'$.
The $X'$ component takes care of the centripetal acceleration $X' = m\,R_{\rm E}\, \omega^2$ and the $Y'$ component take care of the contact force due to the Earth.  
The effect is a bit more complex at other positions on the Earth but non existent at the Poles where the mass would undergo no centripetal acceleration.  
A: A brief statement would be:-The earth is only an approximate inertial reference frame as it rotates about its axis with a very small angular velocity which makes it a non inertial reference frame but it would be a good approximation to treat it as an inertial frame.
