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For a collection of sci-fi short stories set in the rather near future I need an approximation of the time it would take a space ship to travel from an Earth orbit to Ceres.

In a previous story of the collection I've simply stated the requirements, doing only a little calculation and research on whether my assumptions generally make sense or are completely off and illogic but now that I've set these values I not only have to make sure they generally make sense but also they fit together, but luckily everything is just an approximation so the physics won't get too complex.

So here's the defined data: the spaceship starts its journey on the moon Triton and flies a slingshot maneuver through the gravity of Neptune. After this point it has reached the peak velocity it may gain from burning its fuel and every additional acceleration experienced afterwards results from the gravitational force of the sun. Luckily, in the year the story is set in the sun, Earth, Ceres and Neptune line up kinda neatly (they really do, I have checked, Space Engine is amazing) so that eliminates a whole lot of difficult problems. Theoretically they might be able to use Saturn for a 2nd slingshot but the planet lacks behind and I'm not versed in interplanetary space navigation so let's just assume they don't. Thus the distance the ship has to cover is:

$$ X_{Neptune\rightarrow Ceres}=27.58AU=4.126*10^{12}m $$

After checking with real spacecraft speeds and assuming technical advances but no wild breakthrough I proposed the travel takes about two years, thus:

$$ V_{Ceres}=\frac{4.126*10^{12}m}{63.072*10^6s}=65.417\frac{km}{s} $$

Considering the fastest spacecraft these days was the New Horizons with $45\frac{km}{s}$ before the Jupiter slingshot - which only accelerated it by $4\frac{km}{s}$ that seems to be an adequate value. Let's assume this is the ship's final velocity before artificially decelerating when reaching Ceres. I know this will extend the travel time but that's much better than decreasing it when assuming it's the starting speed. Also I don't expect that much of a difference, but I do expect the ship from Earth flying to Ceres to be slowed down by a good amount - after all, New Horizons was slowed down to $19\frac{km}{s}$ when it reached Jupiter.

So what I need now is the actual speed the spaceship may reach with its engine and a formula on how much the 2nd ship, the one from Earth, will be slowed down over the time and distance to Ceres - which I assume would be the same formula to calculate the change in velocity of the ship from Neptune. And that is the part I have trouble wrapping my head around as it's been some years since using integral calculations the last time. So, if I want to calculate the actual maximum velocity of the spaceship from burning its fuel I have to subtract the gravitational acceleration. The problem is that the gravitational force increases exponentially the closer the ship gets to the sun. The formula to calculate the gravitational acceleration at a specific distance of the body exerting the force is the following:

$$ g=\frac{G*m}{d^2} $$

In case of the gravitational acceleration exerted on Neptune and Ceres by the sun this is:

$$ g_{Ceres}=\frac{6.673*10^{-11}\frac{N*m^2}{kg^2}*1.9884*10^{30}kg}{(4.099*10^{11}m)^2}=7.8972*10^{-4}\frac{m}{s^2} $$

$$ g_{Neptune}=\frac{6.673*10^{-11}\frac{N*m^2}{kg^2}*1.9884*10^{30}kg}{(4.536*10^{12}m)^2}=6.4493*10^{-6}\frac{m}{s^2} $$

Now while this formula draws a very neat graph (x-axis is in AU)

  • Earth is slightly above $1AU$ with $5.6987*10^{-3}\frac{m}{s^2}$
  • Ceres at $2.7AU$
  • Neptune is at around $30AU$

Graph showing acceleration in relation to distance from sun

It doesn't contain any information about time and the distance travelled by the space ship, so I have no idea how to get a formula that I can derive for the calculation of the velocity in respect to the changing acceleration.

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  • $\begingroup$ Do you know about the en.wikipedia.org/wiki/Hohmann_transfer_orbit ? $\endgroup$ – PM 2Ring Apr 2 '19 at 10:41
  • $\begingroup$ @PM2Ring I didn't. Watching a time lapse of the planets orbiting, they're so far apart that I'd expect the transfer orbit to be very elliptical anyway, or immensely time consuming. Also the article states: "Non-Hohmann transfer paths may have other advantages for a particular mission such as shorter transfer times". A shorter transfer time would certainly be preferable. $\endgroup$ – Otto Abnormalverbraucher Apr 2 '19 at 11:06
  • $\begingroup$ Hohmann transfers aren't fast, but they use minimal fuel. At least, that's the case for the ideal transfer between 2 circular orbits, with no gravitational slingshots. And in the ideal case they're easy to calculate, so they give a useful first approximation of worst transit times. $\endgroup$ – PM 2Ring Apr 2 '19 at 11:11
  • $\begingroup$ Many of the planets are subject to helium-3 mining, which is also used as fuel for the spaceships. There are other stations on various planets and moons, so with little shortage of fuel and long travel times between the various stations cutting short on transfer times is much more important than saving fuel. $\endgroup$ – Otto Abnormalverbraucher Apr 2 '19 at 11:44
  • $\begingroup$ Ok. In that case, I suggest that rather than worrying about the acceleration, look at the energy involved. A body in orbit around the Sun has potential energy & kinetic energy, you can calculate both of those from its mean orbital radius. Determine the energy change required to get your ship from the starting body to the destination, and divide that by the power of the ship to get the time. BTW, rather than using G & the Sun's mass, use the Sun's standard gravitational parameter. $\endgroup$ – PM 2Ring Apr 3 '19 at 5:15
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As an introduction to my recommentation let me tell a related story.

When the writer Andy Weir was doing research for the novel 'the Martian' he assumed that he would have to do all the orbital mechanics calculations by solving the equations. He struggled with that; things were looking hopeless. Then he had an opportunity to ask someone how NASA does that for all the space missions. Weir was surprised to learn that NASA calculates trajectories with numerical analysis.

That is, the engineers set up a numerical simulation, and then they let a computer crunch the numbers. So, starting with a known velocity and direction of motion they calculate where the spacecraft will end up and if that is not at the spot where they want to end up they plan a course correction.

I assume there is some trial and error in the planning stage. To plan a course correction, I imagine, the engineers first evaluate the result of a course correction that should be pretty close. The computer predicts where the spacecraft will end up, and depending on that outcome the plan for the course correction is adjusted. I imagine it will be something of an iterative process, homing in on the sweet spot.

So that is my recommendation to you: set up a simulation, and use an iterative process to home in on the trajectory that fits your requirements.

I'm not a user of Kerbal space program myself, but there are several youtubers with an educational channel who use the Kerbal orbital mechanics capabilities to teach their audience about space travel mechanics. As I understand it, in Kerbal space program you can define a spacecraft, you specify its mass, you specify the thrust of the engines, you specify the amount of fuel supply that the spacecraft carries, and so on, and then you run the simulation.

Or you may decide to use a general purpose simulation platform, one where you enter the equations yourself.

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  • $\begingroup$ You're making this far more complicated than necessary. As said, I only need an approximation. I'm not trying to plot a course, the course is spaceship -> sun, straight up, no complicated simulation setup required. All I need is the straight physics of how to calculate the velocity when the acceleration increases dynamically. Basically what I want to know is whether the ship from Earth to Ceres would take half a year or only two months and the straight distance is enough for this kind of coarse scale. $\endgroup$ – Otto Abnormalverbraucher Apr 2 '19 at 21:13

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