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