newtonian gravity - How does Earth carry Moon with it, if it can not force Moon to touch it by gravitational force?

Earth's gravitational force is acting on its Moon in such a way that it forces the Moon to rotate round its orbit by centripetal force and carries it while rotating round the Sun by gravitational force. I don't understand why in this condition Moon doesn't come to the Earth while Earth is carrying it through gravitational force to rotate round the Sun?

Answer

I think I might understand another facet of your question besides what is addressed in the comments. Let me demonstrate a result in classical mechanics which I think might alleviate your concern.

The result is that

Given a system of particles, the center of mass of the system moves is though it were a point mass acted on by the net external force on the system.

So if you think of the Earth-Moon system as being acted on by a net external force which is simply the gravitational attraction to the Sun (to good approximation), then what's happening is that this entire system is orbiting (essentially freely falling) around the sun. The details of what's happening in the Earth-Moon system itself are described by the first link in the original comments, but for purposes of what's happening to the entire system consisting of the Earth+Moon when it orbits the Sun, the details of the internal interactions don't really matter.

Here is a proof of the statement above:

Consider a system of particles with masses $m_i$ and positions $\mathbf x_i$ as viewed in an inertial frame. Newton's second law tells us that the net force $\mathbf F_i$ on each particle is equal to its mass times its acceleration; $$ \mathbf F_i = m_i \mathbf a_i, \qquad \mathbf a_i = \ddot{\mathbf x}_i $$ Let $\mathbf f_{ij}$ denote the force of particle $j$ on particle $i$, and let us break up the force $\mathbf F_i$ on each particle into the sum of the force $\mathbf F^e_i$ due to interactions external to the system and the net force $\sum_j \mathbf f_{ij}$ due to interactions with all other particles in the system; $$ \mathbf F_i = \mathbf F_i^e + \sum_j \mathbf f_{ij} $$ Combining these two facts, we find that $$ \sum_i m_i\mathbf a_i = \sum_i \mathbf F_i^e + \sum_{ij} \mathbf f_{ij} $$ The last term vanishes by Newton's third law $\mathbf f_{ij} = -\mathbf f_{ji}$. The term on the left of the equality is just $M\ddot {\mathbf R}$ where $M$ is the total mass and $\mathbf R$ is the position of the center of mass of the system. Combining these facts gives $$ M\ddot{\mathbf R} = \sum_i \mathbf F_i^e $$