Why is the work done in moving a unit mass from infinity to a point (where gravitational field exists) negative? Gravitational force acts towards the center. Here, while performing work, force is towards the center and displacement is also towards the center, then why is the work performed on the body considered negative? "As in work, $W= -GM/r$"
where $G$ is the gravitational constant, $M$ is the mass of the Earth, $r$ is the distance between the center of the Earth and the object.
I understand the mathematical derivation(https://en.wikipedia.org/wiki/Gravitational_potential) but don't know the physical significance of negative sign.
Also,
"Work= -(Change in gravitational potential energy) = -(final gravitational PE - initial gravitational PE)"
Convention taken here is Gravitational PE at infinity to be zero, and nearer to the earth, gravitational PE is -x, we get above equation as,
Work = $-(-x -0) = x$ which is positive. I am confused. Please help me clearing my concepts.
Textbook Statement for Gravitational Potential:(Provided in the photo attached)

 A: Remember that work is a transition function - not a state function. You only do work over a distance, not at a point. So when using $W=F\Delta x$ in integral-form we must remember the edges of the integral:
$$W=\int_{r_i}^{r_f} F\;dx$$
Let's choose the positive axis outwards from Earth (force is negative $-F_g$ and displacement is $r_f<r_i$) and calculate work done on a falling object:
$$\begin{align}
W&=\int_{r_i}^{r_f} (-F_g)\; dx\\
&=\left[-\left(-\frac{GM}x\right)\right]_{r_i}^{r_f}=\left[\frac{GM}x\right]_{r_i}^{r_f}\\
&=\frac{GM}{r_f}-\frac{GM}{r_i}
\end{align}$$
$r_f<r_i$, and because they are in the denominators, $\frac{GM}{r_f}>\frac{GM}{r_i}$ and thus work $W$ is positive.
Or we can try with another choice of axis, e.g. from the starting point of the falling object and inwards towards Earth (force is positive $+F_g$ and displacement is $r_f>r_i$):
$$\begin{align}
W&=\int_{r_i}^{r_f} F_g\; dx\\
&\left[-\frac{GM}x\right]_{r_i}^{r_f}\\
&=\left(-\frac{GM}{r_f}\right)-\left(-\frac{GM}{r_i}\right)\\
&=\frac{GM}{r_i}-\frac{GM}{r_f}
\end{align}$$
In this case $r_i<r_f$, so $\frac{GM}{r_i}>\frac{GM}{r_f}$ and work is still positive. Choice of axis doesn't matter - a choice just has to be made to get the signs clear.

This can also be looked at from pure energy.
Gravitational potential energy is:
$$U=\int F \;dr=-\frac{GM}r$$
Work done by Earth on something falling is:
$$\begin{align}
W&=\Delta K=U_i-U_f\\
&=-\frac{GM}{r_i}-\left(-\frac{GM}{r_f}\right)\\
&=\frac{GM}{r_f}-\frac{GM}{r_i}
\end{align}$$
Again work $W$ will be positive.
A sign rule-of-thumb is always the formula: $W=Fr$: Work is positive if force and displacement are in the same direction, and negative if opposite. 
Gravity therefor always does positive work on a falling object. But was rising like a weatherballoon, the work done by gravity would be negative. Negative work just means that your efforts don't really work; the object still moves in another direction than in which you are pushing/pulling.
A: You have to think about what force is providing the work over a given distance. 
The gravitational force is an attractive force so that a heavy object will pull a lighter object towards itself (the heavier object). The work done on a point mass is done by a heavier gravitational force. 
It is common physics notation to denote work as negative when a system does work on itself and to denote work as positive when a an external force does work on the system.
Also, point masses move objects from an higher potential to lower potential energy.
A: Assume that $W$=Work is the energy that a source must put in the system to obtain a final configuration from a starting configuration. Let $K$ be the kinetic energy: you have
$$
W=K_f-K_i.
$$
This means that, to speed up a particle, some force must act on it by doing positive work: this work is added up to the energy. In conservative systems, you have potential energy $V$ such as, for any initial and final configuration, you have
$$
W=K_f-K_i=V_i-V_f.
$$
Let us take as system a gravitational system: we take the kinetic and gravitational energies to be
$$
K=\frac12 mv^2,\quad V=-G\frac{Mm}r,
$$
with obvious definitions. We start from an initial configuration that is often taken as the zero of energy: we start with a motionless particle $(v_i=0)$ at $(r_i\to\infty)$, very far from the attractor: this means $K_i=0$ and $V_i=0$. As a final position, we take the particle to be at some radius $r_f$ with velocity $v_f$. The work done by the gravitational field is
$$
W=-V_f=G\frac{Mm}{r_f}.
$$
This is a positive quantity, so this means that $K_f=W$ will be a positive kinetical energy: the particle speeds up when coming near to the center.
As you can see, negative difference of potential energies means positive work done on the system by the forces that you describe through your potential. This means that the gravitational force will attract the particle towards the center.
If we started with "repulsive gravity", $V=G\frac{Mm}r$, we would have had the opposite situation: we would have gotten $W=-V_f=-G\frac{Mm}{r_f}$, that is negative. Since the kinetic energy cannot be negative, this means that the particle will never get nearer to the origin in this situation, and an energy cost must be paid to make the particle become closer to the origin.
To close up everything, if work is positive you will have that the final configuration may be reachable from the starting position without having to give energy to the system.
A: It's the work done by us such that it starts from infinity (at zero speed) and ends at that "certain point" (also at zero speed).
If you do no work at all, then the mass will "fall" from infinity to that point, and have (potentially very significant) kinetic energy when it reaches the "end" point. But, that's not what we want; we want it to end at rest. So, we have to slow it down, which involves doing negative work on the mass.  Credit: Erick Anson from Quora
Logic for work-energy theorem:
Since both initial and final velocity of the mass is 0, there is no change in kinetic energy which prevents us from applying the theorem. The work done by us gets stored in the body in the form
 of Gravitational Potential Energy.
A: Here, we do not want to change the mass' kinetic energy. So in order to do that we have to apply a force opposite to the attractive gravitational force.
Therefore, in this case, the work done by us will be $-ve$ because the angle between the force that we are applying and the displacement is $180$ ($\cos 180 = -1$).
But the work done by the body with mass $M$ (which is pulling the body on which we are applying the force) will be $+ve$ because the angle between the force applied by this body and the displacement is $0$ ($cos0 = 1$).
A: "Work= -(Change in gravitational potential energy) = -(final gravitational PE - initial gravitational PE)"
Work in the above equavation is the work done by gravitational force.
When the object move from infinity to point "P". The displacement is in the same direction of gravitational force, gravitational force does positive work. Positive work really means kinetic energy is increased (at the cost of decrease in potential energy 0 to -x, where "-x" is the gravitational potential energy at point P). Potential energy decreases and converted to kinatic energy, when gravitation force does positive work.
Work = −(−x−0)=x which is positive.
Gravitation potential energy of an object at a point in the gravitational field is the work done to bring the object from infinity to that point without acceleration
To bring the object from infinity to point "P" without acceleration, we need to apply external force that is exactly equal and opposite to gravitational force alone the whole path. The work done by the external force is negative as the object is moving opposite to the direction of the force.
