{"id":371,"date":"2025-05-16T12:25:11","date_gmt":"2025-05-16T12:25:11","guid":{"rendered":"https:\/\/www.cydonis.co.uk\/blog\/?p=371"},"modified":"2025-05-16T12:25:12","modified_gmt":"2025-05-16T12:25:12","slug":"more2-lunar-dreams","status":"publish","type":"post","link":"https:\/\/www.cydonis.co.uk\/blog\/2025\/05\/16\/more2-lunar-dreams\/","title":{"rendered":"(More^2) Lunar Dreams…"},"content":{"rendered":"\n

Fuelling a Lunar Dream: Could Water Launch a Probe from Shetland?<\/h2>\n\n\n\n

Imagine a rocket standing tall on one of the rugged Shetland Islands, ready to embark on an incredible journey. It’s destination? A free return trajectory around the Moon. And it’s fuel? Water, split into its fundamental components, hydrogen and oxygen, using renewable energy from the very winds and sun of the islands.<\/p>\n\n\n\n

It might (to some) sound like science fiction, but the concept of using water as a propellant source for hydrolox (liquid hydrogen and liquid oxygen) engines is very real. The question is: how much water would you actually need to send a 100kg probe on such a mission from a place like Shetland?<\/p>\n\n\n\n

Let’s dive into the fascinating physics and engineering challenges involved!<\/a> ^_^v<\/p>\n\n\n\n

The Rocket Science Behind the Splash.<\/strong><\/h3>\n\n\n\n

Launching anything into space, especially towards the Moon, requires overcoming Earth’s powerful gravity and achieving immense speeds. This is where the concept of Delta-v (<\/strong>\u0394v)<\/strong> comes in. Think of \u0394v as the total “change in velocity” capability your rocket needs to have. For a lunar free return trajectory, starting from Earth’s surface, the required \u0394v is substantial \u2013 thousands of meters per second. Launching from a higher latitude like Shetland means you get slightly less help from the Earth’s spin compared to equatorial launch sites, potentially increasing that \u0394v requirement a little.<\/p>\n\n\n\n

The efficiency of a rocket engine is measured by its Specific Impulse (<\/strong>Isp\u200b)<\/strong>. Hydrolox engines are known for having high Isp\u200b, meaning they get a lot of thrust for the amount of propellant they consume. Our hypothetical engine has a 40% efficiency. This efficiency factor impacts the effective Isp\u200b the engine can achieve in the real world, making it lower than the theoretical maximum.<\/p>\n\n\n\n

The Tyranny of the Rocket Equation!<\/strong><\/h3>\n\n\n\n

The core principle governing how much propellant you need is the Tsiolkovsky Rocket Equation<\/strong>:<\/p>\n\n\n\n

\u0394v=Isp\u200b\u22c5g0\u200b\u22c5ln(mf\u200b\/m0\u200b\u200b)<\/p>\n\n\n\n

Where:<\/p>\n\n\n\n

    \n
  • \u0394v<\/strong> is our required change in velocity.<\/li>\n\n\n\n
  • Isp<\/strong>\u200b is the engine’s effective specific impulse.<\/li>\n\n\n\n
  • g0<\/strong>\u200b is standard gravity.<\/li>\n\n\n\n
  • m0<\/strong>\u200b is the initial mass of the rocket (with propellant).<\/li>\n\n\n\n
  • mf\u200b<\/strong> is the final mass of the rocket (without propellant), also called the dry mass.<\/li>\n<\/ul>\n\n\n\n

    The crucial part here is the mass ratio<\/strong> (mf\u200b:m0\u200b\u200b). This equation tells us that to achieve a certain \u0394v with a given engine efficiency (Isp\u200b), you need a specific mass ratio. The higher the \u0394v or the lower the Isp\u200b, the larger the mass ratio must be. This means the vast majority of your rocket’s initial mass has to be propellant.<\/p>\n\n\n\n

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    Figure 1:<\/p>\n<\/blockquote>\n\n\n\n

    \"\"<\/figure>\n\n\n\n

    This graph illustrates how the required mass ratio (initial mass \/ final mass) escalates rapidly with increasing Delta-v for a fixed engine efficiency (Specific Impulse). Achieving higher speeds requires a disproportionately larger amount of propellant to climb out of Earth’s gravity well, and escape the monstrous ‘homo sapiens singularis’ below.<\/em><\/p>\n\n\n\n

    The Dry Mass Challenge.<\/strong><\/h3>\n\n\n\n

    The dry mass (mf\u200b) isn’t just the 100kg probe. It includes the rocket’s structure, engines, fuel tanks, guidance systems, and importantly, the equipment needed to split the water and power the process using renewables. We’re assuming a structural mass fraction of 1\/8. In rocketry terms, this usually relates the mass of the structure to the total mass or dry mass, and a fraction like 1\/8 suggests a very lightweight structure relative to the total vehicle or dry mass. In our water-splitting scenario, we also need to account for the mass of the electrolysis unit and the power generation\/storage system (solar panels, wind turbine components, batteries).<\/p>\n\n\n\n

    The electrolysis efficiency (~37%) tells us how much of the energy input actually goes into splitting the water. A lower efficiency means you need a more powerful, and likely heavier, power system to produce the required amount of hydrogen and oxygen within a reasonable timeframe for fuelling. This adds to the dry mass.<\/p>\n\n\n\n

    \"\"<\/figure>\n\n\n\n

    Putting Numbers to the Dream (An Illustrative Example).<\/h3>\n\n\n\n

    Let’s try a simplified calculation based on some assumptions, similar to how engineers start to size a rocket:<\/p>\n\n\n\n

      \n
    • Target <\/strong>\u0394v:<\/strong> Let’s assume a challenging but plausible \u0394v requirement of 10,000 m\/s for this mission from Shetland.<\/li>\n\n\n\n
    • Effective <\/strong>Isp\u200b:<\/strong> Using a typical hydrolox vacuum Isp\u200b and considering the 40% engine efficiency (interpreted as an overall efficiency factor applied to the theoretical Isp\u200b potential), let’s work with an effective Isp\u200b of around 400 seconds.<\/li>\n\n\n\n
    • Payload Mass:<\/strong> 100 kg.<\/li>\n\n\n\n
    • Dry Mass Estimate:<\/strong> This is the trickiest part. The structural mass fraction of 1\/8 is very optimistic if applied to the whole vehicle. Let’s instead estimate the combined mass of the structure, engine, tanks, guidance, plus<\/em> the electrolysis and power equipment. For a mission like this, this supporting mass could easily be several times the payload mass. Let’s illustrate<\/em> by assuming this combined mass is 5 times the payload, or 500 kg.\n
        \n
      • So, the estimated dry mass (mf\u200b) = Payload (100 kg) + Structure & Equipment (500 kg) = 600 kg.<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n\n\n\n
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