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Would a rocket burn more fuel to get from Earth's surface to Low Earth Orbit, or to get from LEO to Geosynchronous Earth Orbit?

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The delta-v required to get into LEO is significantly higher. This diagram shows the delta-vs that are required to move between some "locations" in the inner Solar System. –  mmc Sep 11 '12 at 19:49
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To calculate fuel consumption, you can typically use the Tsiolkovsky Rocket Equation shown here (without taking Relativity into account):

$$ \Delta V = v_e * ln(\frac{m_0}{m_1}) $$

  • $m_0$ is the initial total mass, including propellant,
  • $m_1$ is the final total mass,
  • $v_e$ is the effective exhaust velocity, and
  • $\Delta V$ is the maximum change in the speed of the vehicle (with no external forces)

The $\Delta V$ from Earth's surface to LEO from Kennedy Space Center is $9.3 - 10\;km/s$, and LEO (KSS) to GEO is $4.24\;km/s$. Source: Delta-V Budgets, Earth-Moon Space, High Thrust

Assuming an exhaust velocity of 4.5 km/s, single stage rocket, we get

$$ \frac{m_0}{m_1} = e^{\Delta V/v_e} = e^{9.3/4.5} = 7.90 $$

$$ m_{propellant} = (1 - \frac{m_1}{m_0})m_0 = (1 - \frac{1}{7.90})m_0 = (1 - 12.66\%)m_0 = 87.3\%\;m_0 $$

Just getting the rocket to LEO takes 87.3% of your initial mass. Now let's try this again for LEO -> GEO:

$$ \frac{m_1}{m_2} = e^{\Delta V/v_e} = e^{4.24/4.5} = 2.57 $$

$$ m_{propellant} = (1 - \frac{m_2}{m_1})m_1 = (1 - \frac{1}{2.57})m_1 = (1 - 38.98\%)m_1 = 61.02\%\;m_1 $$

However, this is a percentage of the initial mass at LEO, which is 12.66% of the original. Let $m_1$ be the new initial mass for the Rocket Equation achieved after reaching LEO (from above) and $m_2$ be the final mass after reaching GEO. The fraction of launch mass $m_0$ would be

$$ \frac{m_2}{m_0} = \frac{m_2}{m_1}*\frac{m_1}{m_0} = \frac{1}{2.57}*\frac{1}{7.90} = .3898 * .1266 = 4.93\% $$

$$ m_{propellant} = (\frac{m_1}{m_0} - \frac{m_2}{m_0})m_0 = (12.66\% - 4.93\%)m_0 = 7.73\%\;m_0 $$

Thus, getting from Earth's Surface -> LEO takes about 87.3% of the mass of whatever you launch ejected at 4.5 km/s, where LEO -> GEO takes only 7.73% of of that mass at 4.5 km/s, leaving you in GEO with 4.93% of what you started with.

Note: because of the difficulty from Earth->LEO, we typically use multi-staged rockets which can ease the fuel required. Because you said 'a rocket', I took that literally and did the calculation with a single-stage-to-orbit configuration.

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+1, but note that staging doesn't get you out the exponential nature of the Tsiolkovsky equation. It only makes it easier to achieve a given "overall mass ratio". –  mmc Sep 13 '12 at 17:27
    
Completely agreed, which is why I used the word 'ease' - it backs the percentage of fuel requirement down by some non-trivial amount, but it's still a large percentage of your starting mass. –  Ehryk Sep 13 '12 at 17:30
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It should also be noted that $v_e$, by contrast, will exponentially relieve fuel requirements ($m_p = 1 - e^{-\frac{\Delta V}{e_v}}$. If we could accelerate the exhaust to 300 km/s instead of just 4.5 km/s (1% of c), then it would only take 3% of $m_0$ to get from the surface to LEO. –  Ehryk Sep 13 '12 at 17:36
    
What is m2? Is it the final mass at GEO? –  Everyone Sep 13 '12 at 19:02
    
Yes, sorry. I didn't want to repeat $m_0$ and $m_1$. –  Ehryk Sep 13 '12 at 22:09
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The Wikipedia article Delta-V budget discusses the energy needed to get between various locations. The delta-v from the surface of the Earth to LEO is given as 9.3 to 10 km/sec while the delta-v from LEO to GEO is around 4 km/sec depending on what you mean by LEO. So Earth to LEO takes around twice as much fuel as LEO to GEO.

Delta-v is a difficult concept for anyone who hasn't come across it before. It's a general guide as to how much fuel is needed. I answered a question about delta-v in Difference between deltaV and specific impulse, or there's a Wikipedia article here. You can think of the delta-v as proportional to the amount of fuel needed.

If you calculate the gravitational potential energy for a fixed mass you find it changes far more going from LEO to GEO than it does from the Earth's surface to LEO. Therefore you might think getting from LEO to GEO takes more fuel. The difference is that as the rocket rises it burns fuel so it's mass decreases. By the time the rocket reaches LEO its mass is a small fraction of the launch mass. When you move from LEO to GEO you're moving a much smaller mass so even though you're moving a lot farther it takes less energy.

Fixed mass ship

The gravitational potential energy at a distance $r$ from the Earth's centre is:

$$ V(r) = -\frac{GMm}{r} $$

so the difference in energy between two distances $r_a$ and $r_b$ is simply:

$$ \Delta V = -GMm \left( \frac{1}{r_a} - \frac{1}{r_b} \right)$$

The term low Earth orbit seems to cover a large range of altitudes, so let's take it to be the altitude of the ISS i.e. about 350km. Geostationary orbit is 35,786km and the Earth's radius is 6378km, so the potential energy change per kg is:

$$ \Delta V(\text{surface to LEO}) = 3.25MJ/kg $$

$$ \Delta V(\text{LEO to GEO}) = 49.8MJMJ/kg $$

So it takes 15 times more energy to get a fixed mass from LEO to GEO than from the surface to LEO.

This is unlikely to be relevant to any spaceship, since I know of no way for a spaceship to get to GEO without changing it's mass. However it is relevant to a space elevator if we ever manage to buld one.

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I'm somewhat confused then. Does the delta-v take the mass-reduction of spent fuel into account? What is the comparison between the energy requirements of a theoretical fixed mass ship (say, an ion driven one) from Surface -> LEO vs LEO -> GEO? Your first paragraph seems like it's less for LEO -> GEO, but then your last one seems to say that it's more. –  Ehryk Sep 12 '12 at 10:51
    
Yes, the delta-v figure does take mass reduction of spent fuel into account. The delta-v figures quoted assume a current design of rocket. That's why the delta-v for LEO to GEO is less than to get from the Earth to LEO. A theoretical fixed mass ship would expend less energy getting from Earth to LEO than from LEO to GEO, but no ship of this form is ever likely to launch from Earth's surface. Current ion drive ships are launched into orbit using a conventional booster, and only use the ion drive when well away from the Earth. –  John Rennie Sep 12 '12 at 12:12
    
What's the ratio for the fixed mass ship? –  Ehryk Sep 12 '12 at 17:50
    
@Ehryk: I've edited my answer to discuss a fixed mass ship. –  John Rennie Sep 13 '12 at 7:38
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@JohnRennie +1, nice answer. But your energy calculations for the fixed mass ship are misleading: most of the energy required to get into LEO is kinetic energy. This adds about 60 MJ/kg to the LEO energy and 9 MJ/kg to the GEO energy. So we have $\Delta E(\text{surface to LEO}) \approx 65\,{\rm MJ/kg}$ and $\Delta E(\text{LEO to GEO}) \approx 25\,{\rm MJ/kg}$ –  mmc Sep 13 '12 at 17:05
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