A few questions on Gravity What role does the unit vector $ \mathbf{e}_r $ play in $$ \mathbf{F} = - G \dfrac{Mm}{r^2} \mathbf{e}_r$$ and why there's a negative sign?
If the gravitational force always acting in the same direction then why do we use vector describing Law of Universal Gravitation? 
And how does one solve this differential equation? $$ m \dfrac{\mathrm d^2 \mathbf{r} }{\mathrm dt^2} = - G \dfrac{Mm}{r^2} \mathbf{e}_r $$
There's a lot to learn you see. I'm seeking a through explanation. 
 A: The Newtonian gravitational force is an attractive force between two massess. It is one experimental fact that this force is always atractive and in the direction connecting the masses.
This means that if we have two particles at positions $\mathbf{x}$ and $\mathbf{y}$ the force must be
$$\mathbf{F}=-F \frac{(\mathbf{y}-\mathbf{x})}{|\mathbf{y}-\mathbf{x}|}$$
Notice that we take the vector $\mathbf{y}-\mathbf{x}$ and divide by its length. The resulting vector is a unit vector and the idea is to capture just the direction connecting $\mathbf{x}$ and $\mathbf{y}$. The eventual dependence on the distance lies in $F$ which is just the magnitude of the force $F = |\mathbf{F}|$.
That said, it is also an experimental fact that $F$ has the two properties:


*

*It is proportional to both massess;

*It is inversely proportional to the square of the distance between them.
This means that there must be a constant, $G$ such that
$$F=G\frac{m_1m_2}{|\mathbf{y}-\mathbf{x}|^2}.$$
Putting it all together yields
$$\mathbf{F}=-G\frac{m_1m_2}{|\mathbf{y}-\mathbf{x}|^2}\frac{\mathbf{y}-\mathbf{x}}{|\mathbf{y}-\mathbf{x}|}$$
One usually simplifies it by denoting $\mathbf{r}=\mathbf{y}-\mathbf{x}$. In that case, $\mathbf{e}_r=\mathbf{r}/|\mathbf{r}|$ and this is just notation to write down:
$$\mathbf{F}=-G\frac{m_1m_2}{r^2}\mathbf{e}_r.$$
So in conclusion: the physics here is that the force is attractive (hence the minus sign) and in the direction connecting the masses (hence the vector $\mathbf{e}_r$ whose meaning is explained above).
A: 
In the diagram $\hat  {\mathbf{e}}_{\rm r} $ is a unit vector pointing from $M$ to $m$ along the line joining the centre of mass of $M$ to the centre of mass of $m$ with the origin is at the centre of mass of $M$ and $\mathbf r = r\, \hat  {\mathbf{e}}_{\rm r} $.  
The gravitational attractive force on mass $m$ due to mass $m$ is to the left ie in the $(-\hat  {\mathbf{e}}_{\rm r}) $ direction and has a magnitude of $G \dfrac{Mm}{r^2}$ so $\mathbf F_{\text{on m due to M}} = \dfrac{Mm}{r^2}\,(-\hat  {\mathbf{e}}_{\rm r}) = - \dfrac{Mm}{r^2}\,\hat  {\mathbf{e}}_{\rm r} $.   
The minus sign indicates that the force on mass $m$ due to mass $M$ is in the opposite direction to the unit vector $\hat  {\mathbf{e}}_{\rm r}$ ie it is an attractive force.  
Your second equation is the use of Newton's second law $\mathbf F = m \,\mathbf a$ applied to mass $m$ as the system and $ m \dfrac{\mathrm d^2 \mathbf{r} }{\mathrm dt^2} = - G \dfrac{Mm}{r^2} \hat{\mathbf{e}}_r $.  
