Change in potential energy is $$\Delta U = -\text{Work Done} \, .$$ While deriving the equation for potential energy we just equate the work done by gravitational force with the potential energy and do not consider $\Delta U$, i.e. $U(r) – U(\infty)$, where $r$ is the distance of the object from the gravitating source. If we do that then the gravitational potential energy would come out positive. Why is that we don't take $\Delta U$ in this case?
[Physics] Derivation of gravitational potential energy
conventionsforcesnewtonian-gravitypotential energywork
Related Solutions
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.
1) point 1: We do not need to put the $0$ point at infinity. Since it is potential energy, we can set it to $0$ anywhere we want. This is because adding a constant to potential energy does not change the force involved (since $F=-\frac{dU}{dr}$). The reason so many people choose to do this is because what we are usually interested in is a change in potential energy, rather than an absolute value of it. If we set $U = 0$ at infinity then things work out nicely. For example, in the case of gravity at large scales, $U=-\frac{Gm_1m_2}{r}$. This expression goes to $0$ as $r$ goes to infinity. Therefore we can look at $U(r)$ as a change in potential energy from infinity, and we don't need to keep track of some arbitrary constant. The change of potential energy between two points in space becomes $$\Delta U=U(r_2)-U(r_1)=-\frac{Gm_1m_2}{r_2}+\frac{Gm_1m_2}{r_1}$$
If we did want to set $U=0$ somewhere else, then we always have an arbitrary constant following us around, but then it goes away when we find changes in potential energy. Let's say $U=0$ at some point $r = R_0$. Then our function of the potential energy is $U(r)=-\frac{Gm_1m_2}{r}+\frac{Gm_1m_2}{R_0}$. This is perfectly valid physically, but we get the same result as before for the change in potential energy between two points in space:$$\Delta U=U(r_2)-U(r_1)=-\frac{Gm_1m_2}{r_2}+\frac{Gm_1m_2}{R_0}+\frac{Gm_1m_2}{r_1}-\frac{Gm_1m_2}{R_0}$$$$\Delta U=-\frac{Gm_1m_2}{r_2}+\frac{Gm_1m_2}{r_1}$$ Therefore, we just set $U=0$ at infinity since it is the simplest thing to do (similar to why we put potential energy to $0$ when a spring is at its resting state).
1) point 2: This is not how you find the work done to push something against gravity in general, but we can make some assumptions to get to where it seems like you need to be. In general the work done by any force is $\int \vec F\cdot d \vec r$. So if you push on the object with some force, then this is how you determine the work done by yourself if you do not know anything else. If you know that the only forces acting on the object is yours and gravity, then you can use energy conservation as another way to get the work you have done: $$W_{tot}=\Delta K=W_{me}+W_{grav}=W_{me}-\Delta U$$ where K is the kinetic energy. If the object starts and stops at the same speed, then we can go further:$$W_{me}=\Delta U$$ And this is probably where you getting confused. In the case where we are close to the Earth, then $W_{me}=mg \Delta h$. In the case where we are farther from the Earth, $W_{me}=-\frac{Gm_1m_2}{r_2}+\frac{Gm_1m_2}{r_1}$. So your issue might be from trying to apply the first (an approximation) when you really need to be applying the second.
2) I am kind of confused on the wording here, but if I understand what you are asking, you are just wondering if you can change where $U=0$ is defined. As already stated, this is perfectly fine! You just have to make sure you stay consistent. The absolute values of your potential energies can change, as long as the relative values between points remains constant.
(Side note: You can run into issues if you try to put $U=0$ at infinity if your mass distribution is nonzero at infinity as well, but since we are just looking at gravity around the Earth we should be fine)
Best Answer
is to be read as
Change in potential energy is $\Delta U = -$Work Done by gravitational forces
So imagine that you have two point masses as the system that you are considering and define the gravitational potential energy to be zero when their separation is infinite, $U\infty) = 0$
The separation of the masses is then decrease to $r$ and so the gravitational potential energy of the masses is now $U(r)$.
In going from infinity to separation $r$ the change in gravitational potential energy of the system of two masses is $\Delta U = U(r) - U(\infty) = U(r)$
So that is the left hand side of your equation dealt with.
The gravitational force is attractive and along the line joining the two point masses.
Imagine that something (you?) is holding the masses apart but allows them to start at rest when at an infinite separation and finish at rest when the separation is $r$.
Whilst the separation is decreasing the gravitational forces are moving and doing work.
This work is positive because the direction of the gravitational forces and their direction of movement are the same.
So minus the work done by the gravitational forces is negative which means that the change in gravitational potential energy of the system is negative ie the gravitational potential energy has decreased.
Another way of defining the change in gravitational potential energy is $\Delta U = $Work Done by external forces
Now the external forces are opposite in direction to the gravitational forces so the direction of the external forces is opposite to the direction of movement of the external forces.
Hence the work done by the external forces is negative as is the change in gravitational potential energy of the system.
If the system os two masses is isolated then the centre of mass of the system cannot move.
This means that if the masses are allowed to move closer together due to their mutual attraction their centre of mass does not move.
This in turn means that both masses move towards the centre of mass and the gravitational forces (equal in magnitude by Newton's third law) on both bodies do work.
However in the case of a system like the Earth and an object on the Earth with the mass of the object being very much less than that of the Earth the movement of the Earth towards the centre of mass of the system is negligible compared with the movement of the object.
The Earth is assumed not to move at all.