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I was thinking about possibility of cargo transport between, say, earth and mars when a “pusher” accelerates propulsion-less cargo vessel, which is catched and decelerated at the other end of trajectory.

How much energy is needed to accelerate in space a mass of 1 ton from 0 to 11 km/seconds? How much fuel would it take? And how large should be a booster / rocket ?

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    $\begingroup$ You might want to share some context so that it does not look like a homework question asking for someone to do your math. $\endgroup$ – planetmaker May 4 at 8:37
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    $\begingroup$ definitely not a homework. i was thinking about possibility of cargo transport between, say, earth and mars when a “pusher” accelerates propulsion-less cargo vessel, which is catched and decelerated at the other end of trajectory $\endgroup$ – ts. May 4 at 8:47
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    $\begingroup$ More likely you'd attach to a cycler for the journey: en.wikipedia.org/wiki/Mars_cycler $\endgroup$ – GWP May 4 at 18:16
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    $\begingroup$ a pusher could be a space tug that has a lot more fuel and is specialized to meet up with another spacecraft and take it to and from somewhere. I've done that a few times in KSP. $\endgroup$ – James Ervin May 4 at 19:17
  • $\begingroup$ There are some useful concepts in these articles: en.wikipedia.org/wiki/Hohmann_transfer_orbit & en.wikipedia.org/wiki/Specific_orbital_energy $\endgroup$ – PM 2Ring May 4 at 21:34
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How much fuel would it take? And how large should be a booster / rocket ?

That depends on the engine and propellant. Tsiolovksy rocket equation is

$$\Delta V = \ln\frac{m_i}{m_f}I_{sp}g$$

where

  • $\Delta V$ is change in velocity (in this case, 11000 $\frac{m}{s}$)
  • $m_i$ is the initial mass before the burn (payload plus propellant plus dry mass for the engine, tanks, and other hardware)
  • $m_f$ is the final mass after the burn (1 ton, or a little over 907 kg)
  • $I_{sp}$ is the specific impulse of the engine, and
  • $g$ is the force of gravity at the Earth's surface (9.8 $\frac{m}{s^2}$)

To compute our initial mass, we do some rearranging:

$$m_i = m_f e^{\frac{\Delta V}{I_{sp}g}}$$

To keep the math easy, we'll assume magical tanks and engines that have no mass; we are just accounting for propellant mass and payload mass. This gives us the minimum amount of propellant we'd need, but the actual amount will be higher once we factor in the dry mass other than the payload.

The Merlin vacuum engine has a specific impulse of 348 seconds - plugging that into our equation above tells us that we will need at least 22,824 kg of kerolox propellant to accelerate 907 kg of payload to 11 km/s. Again, once we account for the mass of the tanks and the engine itself, the actual amount will be a bit higher.

Hydrolox engines like the SSME have higher specific impulse, on the order of 450 seconds. Plugging that in to our equation gives us at least 10,987 kg of hydrolox propellant.

The ion thrusters on the Dawn spacecraft had a maximum specific impulse of 3200 seconds, meaning we'd only need 318 kg of xenon propellant.

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There are two figures which matter here: one is the energy that ends up in the thing you are accelerating. This is just given by

$$\frac{m_P \Delta v^2}{2}$$

Where $m_P$ is the mass of the thing, and $\Delta v$ is the change in velocity (I am tacitly assuming that this kinetic energy is relative to someone for whom the object is initially at rest).

The second, more frightening figure, is how much energy you need to put into the thing to get this $\Delta v$. If you're using a rocket to do that, then you're subject to the Tsiolkovsky rocket equation (all fashionable rocket scientists have a tattoo of this):

$$\Delta v = v_e \ln \frac{m_0}{m_f}$$

Where $v_e$ is the exhaust velocity of your rocket, $m_0$ is the initial mass and $m_f$ is the final mass. We can make this a bit more complicated and useful:

$$\Delta v = v_e \ln \frac{m_P + m_B + m_F}{m_P + m_B}$$

Here:

  • $m_P$ is the payload mass;
  • $m_B$ is the 'dry', unfueled mass of the booster;
  • $m_F$ is the fuel mass.

You can make things better and yet more complicated by having several boosters some of which you throw away on the way to avoid carting their dry mass around. I won't do that.

You can rearrange this thing to get

$$m_F = (m_P + m_B)\left(e^{\frac{\Delta v}{v_e}} - 1\right)$$

So you can then plug numbers into that to work out what fuel mass you need. Obviously you want to make $m_B$ as tiny as you can and $v_e$ as huge as you can, but there are practical limits.

Once you know $m_F$ you're almost there. From the point of view of the rocket what you're doing is sitting in the back of it and throwing fuel out of the back with $v_e$. I always think of this in terms of a 'pebble rocket' where a person just sits at the back of the rocket, throwing pebbles from a big pile they have to hand: pebble rockets are useful because you can see what the energy is easily, and also because they make clear that the rocket can keep accelerating way beyond $v_e$ which is something people often do not understand. So the kinetic energy you are imparting to the fuel is

$$\frac{m_F v_e^2}{2}$$

But this is only the kinetic energy: you're also making the exhaust very hot usually, which heat is just radiated away into space. I'm not quite sure how to calculate the efficiency of rocket engines, but it's probably not that high.

However, the above formulae should be enough to plug some numbers into to get the things you want. The one thing you have to guess is $v_e$. I believe that $v_e \approx 5000\,\mathrm{m/s}$ is good-but-plausible for chemical rockets: you can do better with ion drieves etc.

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  • $\begingroup$ Awww yeah, it's my favorite symbol, Delta-V $\endgroup$ – James Ervin May 4 at 19:18
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Kinetic energy is $\frac {1}{2}mV^2$ so you're looking at $55$ GigaJoules of energy. That's enough to power ~86 U.S. homes for a year according to this U.S. EIA FAQ

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    $\begingroup$ And that's a small proportion of the energy you need, due to the rocket equation. $\endgroup$ – tfb May 4 at 13:50
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    $\begingroup$ If $v$ is 11,009 m/s and $m$ is 1 metric ton (1000 kg) then $\frac{1}{2} m v^2$ will be 60.5 GJ. $\endgroup$ – uhoh May 4 at 21:39
  • $\begingroup$ @uhoh I had that originally but a different answer used an imperial ton (2000 lbs) and google seems to think a metric ton is a tonne, what should be kept? $\endgroup$ – BrendanLuke15 May 4 at 22:56
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    $\begingroup$ Oh I see that now! And reading further I see what you mean. Hmm... sometimes here in Space SE I see mt used for metric ton, and I think sometimes "ton" is assumed to be a metric ton here, but usually it's contexts where the difference doesn't matter, ballpark approximations etc. I'd say leave it the way you have it now. (here is kilograms, here is metric tons, and metric tons $\endgroup$ – uhoh May 4 at 23:51
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    $\begingroup$ and there's Why did NASA use U.S customary units? Since the US has strangely stuck to imperial-type units for many things and US science-minded folks (like me) are self-conscious of this we probably go out of our way to panic whenever non-metric units wherever they might be seen. The Mars Climate Orbiter mission was lost due to an imperial/metric mixup. $\endgroup$ – uhoh May 4 at 23:52

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