This leads to a paradox known as the Gibbs paradox, after Josiah Willard Gibbs. The paradox allows for the entropy of closed systems to decrease, violating the second law of thermodynamics. A related paradox is the "mixing paradox".
Barium cation has +2 charge and oxide anion has −2 charge
We will define the Total mass to calculate the force, so our values are:
Total Mass 
The Weight is,
Through the hook's Law we calculate X.
, where x is the lenght of compression and K the Spring constant.
We don't have a K-Spring, but we can assume a random value (or simply let the equation in function of K)
I assume a value of 
When using the right-hand rule to determine the direction of the magnetic force on a charge, which part of the hand points in the direction that the charge is moving? The answer is <span>thumb.
</span>One way to remember this is that there is one velocity, represented accordingly by the thumb. There are many field lines, represented accordingly by the fingers. The force is in the direction you would push with your palm. The force on a negative charge is in exactly the opposite direction to that on a positive charge. Because the force is always perpendicular to the velocity vector, a pure magnetic field will not accelerate a charged particle in a single direction, however will produce circular or helical motion (a concept explored in more detail in future sections). It is important to note that magnetic field will not exert a force on a static electric charge. These two observations are in keeping with the rule that <span>magnetic fields do no </span>work<span>.</span>
To solve this problem we will use the related concepts in Newtonian laws that describe the force of gravitational attraction. We will use the given value and then we will obtain the proportion of the new force depending on the Radius. From there we will observe how much the force of attraction increases in the new distance.
Planet gravitational force
Distance between planet and star
Gravitational force is
Applying the new distance,
Replacing with the previous force,
Replacing our values
Therefore the magnitude of the force on the star due to the planet is 
