This is an addendum to Steve Linton's answer which adds some information about fuel use.
The rocket equation says that $\Delta v = v_e \ln(m_0/m_f)$, where: $\Delta v$ is the change in velocity, $v_e$ is the exhaust velocity, $m_0$ is the initial mass and $m_f$ the final mass.
For our purposes we can think about
- $\Delta v_{S}$ – the 'slow' $\Delta v$;
- $\Delta v_{F}$ - the 'fast' $\Delta v$;
- $m_{0,S}$, the 'slow' $m_0$;
- $m_{0,F}$, the 'fast' $m_0$.
Where by 'slow' I mean the 'just climb at $1\,\mathrm{km/s}$' and 'fast' means the direct escape option. I'll assume that $v_e$ and $m_f$ are the same. What we want to do is get $m_{0,S}$ in terms of $m_{0,F}$.
We know (from Steve Linton's answer)
$$
\begin{align}
\frac{\Delta v_S}{\Delta v_F} &= 6\\
&= \frac{\ln\left(\frac{m_{0,S}}{m_f}\right)}{\ln\left(\frac{m_{0,F}}{m_f}\right)}
\end{align}
$$
or
$$\ln\left(\frac{m_{0,S}}{m_f}\right) = 6 \ln\left(\frac{m_{0,F}}{m_f}\right)$$
or
$$\frac{m_{0,S}}{m_f} = \left(\frac{m_{0,F}}{m_f}\right)^6$$
So this tells you how bad the fuel requirements are. Note that this is independent of $v_e$: it sucks just as badly for a very high $v_e$ engine like an ion drive (or, well, it sucks less, but only because $\langle\text{small number}\rangle^6$ is less horrible than $\langle\text{large number}\rangle^6$
As an example, if the the exhaust velocity is $4500\,\mathrm{ms^{-1}}$, then for a direct-escape $\Delta v$ of $11\,\mathrm{kms^{-1}}$ we get $m_{0,F}/m_f \approx 11.5$: so ours would be $m_{0,S}/m_f \approx 2.3\times 10^6$: about one two-millionth of the rocket would be payload. This is ... substantially worse.