From the derivation of formula of Escape Velocity we know that

Minimum kinetic energy = Gravitational Potential Energy

But in this circumstance, isn't it the value of gravitational potential energy should be infinity since Energy = Force x Displacement and the object travels an infinity distance from the planet.

I don't understand why my reference book shows that Gravitational potential energy $= -GMm/r .$

Why does the $r$ is taken as the distance between the center of the planet and the object since the force is applied for an infinity distance?

  • $\begingroup$ The object is not yet at infinity when computing that number. Also, energy is not force times displacement, but rather the integral, because now the force is going to vary as you move around. $\endgroup$ Nov 28, 2023 at 13:14
  • 1
    $\begingroup$ You've discovered that if the gravitational force didn't diminish as r increases, then it would take infinite energy to escape. $\endgroup$
    – PM 2Ring
    Nov 28, 2023 at 13:17
  • $\begingroup$ The change in kinetic energy is $\int_{t_1}^{t_2} \vec{F}\cdot\vec{v} \,dt$. It's only force times distance if $\vec{F}\cdot\vec{v}$ is constant the entire time. $\endgroup$
    – Chad K
    Nov 28, 2023 at 13:37
  • $\begingroup$ Some integrals over an infinite domain converge to a finite number. $\endgroup$
    – Andrew
    Nov 28, 2023 at 14:05
  • $\begingroup$ Have you done the integration? It’s instructive. $\endgroup$
    – my2cts
    Nov 28, 2023 at 18:48

2 Answers 2


The long answer is that the equation you provided (force times displacement, $E=\mathbf{F}\cdot\boldsymbol{\Delta}\mathbf{r}$) is for a specific type of energy, but this energy is work (not potential energy) and the force is not constant. So to find the potential energy we need to add up all the instantaneous forces. An approximation can be:


Where at each step (literally a step in space of size $\boldsymbol{\Delta}\mathbf{r}$) we add the force at that point. The summation is along the path we imagine our rocket or particle traveling, say, from an initial position, $\mathbf{R}_i$, to a final position, $\mathbf{R}_f$. However, for this to be correct (i.e., exact) we need the steps to be infinitely small (called infinitesimal steps). We achieve this when $\boldsymbol{\Delta}\mathbf{r}\rightarrow0$. Thus:


If you have taken calculus this is the integral:

$$E=\int_{\mathbf{R}_i}^{\mathbf{R}_f}\mathbf{F}(\mathbf{r})\cdot d\mathbf{r}$$

The gravitational force is: $$\mathbf{F}(\mathbf{r})=-\frac{GMm}{r^2}\mathbf{\hat{r}}$$

Thus the energy is: $$E=\int_{\mathbf{R}_i}^{\mathbf{R}_f}\left(-\frac{GMm}{r^2}\right)\mathbf{\hat{r}}\cdot d\mathbf{r}=-GMm\int_{R_i}^{R_f}\frac{dr}{r^2}=GMm\left(\left.\frac{1}{r}\right|_{R_i}^{R_f}\right)=GMm\left(\frac{1}{R_f}-\frac{1}{R_i}\right)$$ $$E=\frac{GMm}{R_f}-\frac{GMm}{R_i}$$

This energy is called work and its relation to potential energy is given by $\Delta U(\mathbf{r})=-W$ or $U(\mathbf{R}_f)-U(\mathbf{R}_i)=-W$, which gives:


Where we can see that: $$U(\mathbf{r})=-\frac{GMm}{r}$$

This function is the potential energy and it tells us that it's difference is the negative of the work. Another way of looking at this is through energy conservation: $$E_i=E_f$$ $$U(\mathbf{R}_i)+K_i=U(\mathbf{R}_f)+K_f$$ $$K_i-K_f=U(\mathbf{R}_f)-U(\mathbf{R}_i)$$ $$-\Delta K=\Delta U$$

  • $\begingroup$ what does the Δr stand for in the equation E≈∑nF(rn)⋅Δr. If it's the distance of the object travel, then why in U(r)=−GMm/r it is the radius of the planet ? $\endgroup$
    – ZhangJin
    Dec 2, 2023 at 12:02
  • $\begingroup$ why are we taking the sum of all instantaneous force instead of taking the average value, for exp. A person pulls a car 5m by a force 100N In this situation, the Work will be 500J because 5m x 100N If I didn't misunderstood, the force of 100N keep apply on the car along the 5m journey. If the force apply is not constant, shouldn't we take the average value instead of summing it up? $\endgroup$
    – ZhangJin
    Dec 2, 2023 at 12:03
  • $\begingroup$ @ZhangJin if we start at $R_i$ and we take a bunch of steps to $R_f$, we define the size of each step as $\Delta r$, so after the first step we are at $R_i+\Delta r$ and after the second we are at $R_i+2\Delta r$. As for why we take the instantaneous force, what is the average force of $F\propto -1/r$? The multiplication only works in that example because the force of 100N is constant, so $E=\int_{0m}^{5m} 100N\,dx=100N\times\left(5m-0m\right)=500J$. We are just doing this for an infinitely small step so that the difference in force during this step is negligible, then we sum these small E $\endgroup$
    – QPhysl
    Dec 2, 2023 at 14:21

The gravitational force acting on the body is $$\mathbf F_g = -\frac{GMm\mathbf{\hat r}}{r^2}$$ If it is the net force, we can apply the Newtonian second law: $$-\frac{GMm\mathbf{\hat r}}{r^2} = m\mathbf a$$ Making a dot product at both sides by an infinitesimal vector displacement $\mathbf {dr}$: $$-\frac{GM\mathbf{\hat r\cdot dr}}{r^2} = \frac{\mathbf {dv}}{dt}\cdot \mathbf {dr} = \mathbf {dv}\cdot \frac{\mathbf {dr}}{dt} = \mathbf {dv\cdot v} = \frac{1}{2}d(v^2)$$ On the other hand, for the left side: $$-\frac{GM\mathbf{\hat r\cdot dr}}{r^2} = -\frac{GM\mathbf {r\cdot dr}}{r^3} = -\frac{GM(xdx+ydy+zdz)}{(x^2+y^2+z^2)^\frac{3}{2}} = d\left(\frac{GM}{\sqrt{x^2+y^2+z^2}}\right)=d\left(\frac{GM}{r}\right)$$ The minimum requirement for velocity is to go to zero when $r$ go to infinity. So: $$\frac{GM}{r} - \frac{GM}{\infty} = \frac{1}{2}v^2 - 0 \implies \frac{GM}{r} = \frac{1}{2}v^2$$


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.