Why gravitational potential is negative, as displacement and force are in the same direction? Gravitational potential at a point in a gravitational field is defined as work done per unit mass in bringing a small test mass from infinity to the point.
I know that:


*

*By integration, we find that gravitation potential is negative.

*The maximum value of gravitational potential which is at infinity distance is 0, therefore minimum value should be negative.
But, what I do not understand is force is in the same direction as displacement so why work done is not positive? 
 A: This is something which puzzles a lot of students when they first study potential.
The confusion is due to the evaluation of the dot product to find the work done $\vec F \cdot d\vec r$, the sign of $d\vec r$ and the integration $\int \frac{1}{r^2}dr$ producing a negative sign.
Suppose a point mass $M$ at the origin of the coordinate system.
At position $\vec r$ the force on another point unit mass due to the gravitational field produced by mass $M$ is $-\dfrac{GM}{r^2}\hat r$.  
Consider the system as the two masses and apply an external force $\dfrac{GM}{r^2}\hat r$ on the unit mass and move the unit mass from position $r\; \hat r$ to $(r+dr) \hat r $.  
The work done by the external force is $\left(\dfrac{GM}{r^2}\hat r\right )\cdot (dr\;\hat r)=\dfrac{GM}{r^2} \; dr $ which is a positive quantity with the external force and the displacement both in the $\hat r$ direction.
The dot product has been evaluated and produced a scalar quantity.  
To find the work done by the external force over a finite change of position one has to do an integration.
Let the external force move the unit mass from $r$ to $\infty$.
As one might expect the integral $\displaystyle \int_r^\infty  \dfrac{GM}{r^2} \; dr = \dfrac{GM}{r}$ turns out to be positive.  
One can then say that for the external force to move the unit mass from infinity to $r$ it must do an amount of work $-\dfrac{GM}{r}$.
The reason for this is that starting at r going to infinity and the coming back to r should require no net work to be done by the external force.
So the potential at position $r \;\hat r$ is $-\dfrac{GM}{r}$.
Doing it this way one avoids possible conceptual problems about signs etc.
However it does illustrate a very important thing about evaluation of the dot product to produce a scalar and the subsequent integration.  
If the external force moved unit mass from infinity to r the integral $\displaystyle \int^r_\infty  \dfrac{GM}{r^2} \; dr = -\dfrac{GM}{r}$ leads directly to the expected result for the potential at position $r \;\hat r$  $=-\dfrac{GM}{r}$.  
It is the limits of the integration which determine the direction in which the external force moves or rather how $r$ changes and one should not prejudge that by assigning a negative sign to $dr \;\hat r$, evaluating the dot product as being negative and then doing the integration unless one realises that if $dr \;\hat r$ was assumed to be negative the limits of integration must be chosen to reflect that choice.
