Simple explanation for the 2nd "cosmic speed"

Question

Many sources can explain about the first, second and third cosmic velocities, but the explanations contain difficult formulas and are not easy to understand.

I can check with a powerful magnet and an iron piece nearby will

• fall towards the magnet, (lower than first speed)
• fly nearby in circles, (first speed)
• or fly away - this being the second speed.

Is this right? If it is right, how about another way?

If I just grab the iron piece in one hand and the magnet in the other - and slowly, certainly lower than the first speed, separate them and move them away from each other. I don't use speed then - I just use my muscle power, which is larger than that of the magnet. Can we use only power, without a large speed, for overcoming gravity? And if yes, how I can calculate this power?

Answer 1

So the second cosmic speed is what is usually referred to in English as "escape velocity". This is defined as the velocity at which (if you are going straight up and above the atmosphere) you will not fall back to Earth, even if you have no further thrust.

You are absolutely correct that it is possible to leave Earth without reaching escape velocity by continuously thrusting. In practice though, this would be incredibly fuel inefficient. You want to thrust as much as possible as quickly as possible, so that you get out of the more intense parts of Earth's gravitational field more quickly, and waste less fuel fighting it.

As requested, in a comment, let's consider leaving Earth at a steady $1 km/s$. For simplicity I'll assume that the initial acceleration to $1 km/s$ is provided by something else and brings us to the top of the atmosphere, so we can ignore drag. So the high boost option would need a further $10.2 km/s$ of delta-V at this point. Escape velocity is given by $$v_{e}={\sqrt {\frac {2GM}{r}}}$$, so at a distance from Earth's centre of $$R = \frac{2GM}{1000^2}$$ escape velocity will be 1000 m/s, i.e. 1 km/s. This is $8\times 10^8 m$, i.e. $800\,000 km$, so we need to keep thrusting until we reach that distance. The amount of thrust reduces with distance, though, as gravity weakens. To keep a rocket of mass $m$ at constant velocity at distance $r$ we need a thrust of $\frac{GMm}{r^2}$. To put it another way, to get from $r$ to $r+dr$ we need an impulse (thrust times time) $$\frac{GMm}{r^2}\frac{dr}{1000} = \frac{4\times 10^{11} mdr}{r^2}$$.

We can integrate this from $r = 6\times 10^6$ to $R$ and get a total impulse of $$(4\times 10^{11} m)\left(\frac{1}{6\times 10^6} - \frac{1}{8\times 10^8}\right)$$ which is roughly $6.6\times 10^4 m$, corresponding to a delta-V of about 66 km/s -- six times the direct escape option.

Of course a delta-V six times as great doesn't mean six times as much fuel, because you also need to lift the fuel to varying extents. You need to use the rocket equation and know what your exhaust velocity is to calculate that.

An exception is certain types of propulsion which, by their nature, supply a very small thrust for a very long period, like an ion drive. Once you are in orbit, you can escape using an ion drive without ever getting close to the escape velocity at your original distance.

Answer 2

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.