Gravitational potential energy in Escape Velocity

Question

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?

Answer 1

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:

$$E\approx\sum_n\,\mathbf{F}(\mathbf{r}_n)\cdot\boldsymbol{\Delta}\mathbf{r}$$

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:

$$E=\lim_{\boldsymbol{\Delta}\mathbf{r}\rightarrow0}\left[\sum_n\,\mathbf{F}(\mathbf{r}_n)\cdot\boldsymbol{\Delta}\mathbf{r}\right]$$

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:

$$U(\mathbf{R}_f)-U(\mathbf{R}_i)=-\frac{GMm}{R_f}-\left(-\frac{GMm}{R_i}\right)$$

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: $$E_i=E_f$$ $$U(\mathbf{R}_i)+K_i=U(\mathbf{R}_f)+K_f$$ $$K_i-K_f=U(\mathbf{R}_f)-U(\mathbf{R}_i)$$ $$-\Delta K=\Delta U$$

Answer 2

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: $$\frac{GM}{r} - \frac{GM}{\infty} = \frac{1}{2}v^2 - 0 \implies \frac{GM}{r} = \frac{1}{2}v^2$$