[Bit a long question]

Here in this question, I need to know clear my doubts on Gravity, more precisely Work done by a Conservative Force(here, gravitational force).

Ok, suppose there is an object lying. Neglect any damping force like friction, etc.
Now, suppose a force(it could be consevative or non-conservative force) acts on the object. It displaces from it's position.

Now, if the displacement is in the direction of that force, then the work done by the force is Positive.
And, if the displacement is in opposite direction of that force, then the work done by the force is Negative.

I think the above statements are true for any kind of force(conservative/non-conservative).

Now, we'll come to the main part of the question.

1st Part

Suppose a mass m is at point P, r distance away from another mass M.
The mass m will experience gravitational force, $\vec{F_g}$ towards mass M.

Now , an external force $\vec{F}$ acts on the object of mass m which moves it to $\infty$.
Assume a small displacement vector $d\vec{r}$ in directon of external force $\vec{F}$ for which there is no change in the gravitational force, $\vec{F_g}$. Also, consider a unit vector $\hat{r}$ in the direction of external force $\vec{F}$.

Let dWg be the work done by gravitation for this little displacement, $d\vec{r}$.

dWg = $\vec{F_g}\cdot d\vec{r}$ = Fg dr cos 180°

dWg = - Fgdr -----------Eq.(1)

So, I think the gravitational force, $\vec{F_g}$ is doing negative work and the external force, $\vec{F}$ is doing the same positive work due to the sole reason that the displacement of object is in the direction of external force and opposite to gravitational force.

Now, $\vec{F_g}$ = - $\vec{F}$ = $\frac{GMm}{r^2}$$\hat{r}$

So, in equation (1)

$$\int \, dW_g = \int_r^\infty - F_g \, dr$$
$$W_g= \int_r^\infty \frac{-GMm}{r^2} \, dr$$
$$W_g= -GMm \int_r^\infty \frac{1}{r^2} \, dr$$
$$W_g= -GMm \biggl[\frac{-1}{r}\biggr]_r^\infty$$
$$W_g= -GMm \biggl[\frac{-1}{\infty}-\frac{-1}{r}\biggr]$$
$$W_g= \frac{-GMm}{r}$$

So, $$W_g = {\color{red}{\int_r^\infty - F_g \, dr}} = {\color{green}{\frac{-GMm}{r}}}$$

The above 2 equations are in agreement. Both equations, one in red and another in green colour will give negative work done by the gravitational force.

2nd Part

Now, we'll consider that the object of mass m is at $\infty$. Also, the object is taken from $\infty$ to point P which is again, at distance r from mass, M.
This time, the little displacement vector $d\vec{r}$ will be in the direction of gravitational force $\vec{F_g}$. Also, the unit vector $\hat{r}$ is in the same direction as well.Thus,

$\vec{F_g} = \frac{GMm}{r^2}\hat{r}$

So, the work done by gravitational force will be

dWg = $\vec{F_g}\cdot d\vec{r}$ = Fg dr cos 0°

dWg = Fgdr -----------Eq.(2)

Here, the gravtational force, $\vec{F_g}$ will do positive work because the displacement of object is in the direction of gravitational force.

So , in equation (2),

$$\int \, dW_g = \int_\infty^r F_g \, dr$$
$$W_g = \int_\infty^r \frac{GMm}{r^2} \, dr$$
$$W_g = GMm \int_\infty^r \frac{1}{r^2} \, dr$$
$$W_g = GMm \biggl[\frac{-1}{r}\biggr]_\infty^r$$
$$W_g = GMm \biggl[\frac{-1}{r}-\frac{-1}{\infty}\biggr]$$
$$W_g = \frac{-GMm}{r}$$

So, $$W_g = {\color{blue}{\int_\infty^r F_g \, dr}} = {\color{green}{\frac{-GMm}{r}}}$$

So, finally my doubt:-

The blue equation tells that the work done by gravitational force is postive(beacuse the displacement vector is in same direction as that of gravitational force).
But, after integrating it, we have the same green equation which we derived it earlier which tells us that the work done by gravitational force is negative.

Now, if we look at the red and blue equations, both are same, just integrating the term in opposite order.
And we know that the gravitational force is a conservative force so whether it is the blue one or red one, the work by gravity will be same which is given by the green equation.

But at the end, my doubt is why the blue and green equations are inconsistent to each other (blue saying work by gravity is positive whereas green saying work by gravity is negative) ?


2 Answers 2


Lets place the stationary mass $M$ a the origin. In both cases the gravitational force is towards the origin, and is given by $$\vec{F}_{g}=-\frac{GMm}{r^{2}}\hat{r}$$

Thus in the first case, the work done by the gravitational force is $$W_{r\rightarrow\infty}=\int_{r}^{\infty}\vec{F}_{g}\cdot\vec{dr}=-\int_{r}^{\infty}\frac{GMm}{r^{2}}dr=-\frac{GMm}{r}$$

On the other hand, in the second case we just need to sweep the limits of integration $$W_{\infty\rightarrow r}=\int_{\infty}^{r}\vec{F}_{g}\cdot\vec{dr}=-\int_{\infty}^{r}\frac{GMm}{r^{2}}dr=\frac{GMm}{r}$$

which is exactly the opposite, and the signs are just what you've anticipated. An important thing to note is that also in both cases $$\vec{dr}=dr\hat{r}$$

and the sign is set by the integration limits. So in the first case $dr>0$ and in the second $dr<0$ (in a very informal manner). You ignored this fact when you wrote $\cos\left(180^{\circ}\right)$ in the dot product in the second case.

  • $\begingroup$ I have written cos 0 in the second case. In the second case, I have taken the direction of $d\vec{r}$ along the force $\vec{F_g}$ because the force is acting towards mass,M(at origin) and the displacement of mass, m is also towards mass, M.So, why didn't you change the direction of $d\vec{r}$ in the 2nd case? Also, if I get agree with you completely, then the conservative nature of gravity is not satisfied in your post?? $\endgroup$
    – lakhi
    Nov 12, 2017 at 6:50
  • $\begingroup$ My doubt is that the green equation is satisfying the conservative property of the gravitational force but at the same time it is not consistent with the blue equation $\endgroup$
    – lakhi
    Nov 12, 2017 at 7:04
  • $\begingroup$ @lakhi Regarding your first comment, I didn't change the direction of $dr$ because that's what the limit of integration do. $dr$ is a line element in the radial direction, and its sign (direction) is: plus for integration from $r$ to infinity, and minus for integration from infinity to $r$. If you, then, agree with what I've written, why isn't gravity conservative? You can clearly see that the work done on closed loops is zero, since $W_{r\rightarrow\infty}+W_{\infty\rightarrow r}=0$. Regarding your second comment, I don't completely understand what you mean. $\endgroup$
    – eranreches
    Nov 12, 2017 at 8:31
  • 1
    $\begingroup$ The blue and green equations are consistent (though your calculation in the second case is not correct), because you integrate 'backwards'. Observe that in general $\int _{a}^{b}=-\int_{b}^{a}$. As I said, the scalar product in both cases is the same, but the direction of integration is reversed. The direction of integration encapsules the information about whether both vectors point along the same direction or the opposite. $\endgroup$
    – eranreches
    Nov 12, 2017 at 16:44
  • 1
    $\begingroup$ You did two things that cancel each other and led you to have the same answer in both cases. You need to be more careful with coordinates. First, you say there is a minus sign because one time both force and path point in the same direction and in the other time in opposite directions. Second, there is a minus sign because you swept the integration order. You can't do both. That's why you get the same answer in both cases. $\endgroup$
    – eranreches
    Nov 12, 2017 at 16:57

In the second case at first you did scalar multiplication and got the sign. Now all becomes magnitude and no additional signs are allowed from integration. You might say "why?" Remember the definition of scalar multiplication. ${\vec A }{\cdot}{\vec B}={ABcos\theta}$ See the magic! After scalar multiplication you just get the magnitude and the cosine gives sign. So in your second case perform the integration inside a modulus . And just keep $cos\theta$ outside. Hence you get positive work

Alert: This method may not have mathematical rigor but it's powerful


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.