Why would interstellar space travelers need to decelerate their whole ship?

Question

In many sci-fi novels I have read, interstellar travelers had their ship accelerate half the way, then decelerate half the way. I wonder why this is the chosen approach. Would not it be better to accelerate all the way, then leave the ship for small landing module(s) and decelerate them while allowing the interstellar ship to go away from the star system? Maybe, even eject the landing modules from the mother ship in opposite direction.

Answer 1

In general, you want travel as light as possible, as this will enhance your delta-v budget.

With current technology, the mass of your spacecraft will vastly decrease towards end of mission. There will be some consumables that can not be recycled. Most notably, propellant will be vented (as exhaust gas). Thats even true for ion-thrusters. The less consumables/propellant you carry, the less structure your craft needs.

The ratio of mass reduction vs remaining mission time with current technology is relatively high, so it makes sense to get rid of structure that is not needed any more. Think of ascent stages that get jettisoned as soon as their fuel is depleted.

Sci-Fi is about fictitious technology. If you manage to travel and sustain life on board with less consumables, that may be a completely different story: what you need in the end may not be much less than what you started with.

Regarding "eject in opposite direction": This is what propulsion is about, eject something heavy with the highest possible velocity. The main ship has much more mass than your thrusters exhaust gases, thats a good thing. But it may be difficult to achieve a decent delta-v on ejection without disintegrating the whole thing. Maybe a good idea would be to have a rail-gun and shoot little pieces of your mother ship in a prograde direction one by a time, thereby spreading the delta-v over a longer time.

Answer 2

You need to decelerate because you only brought what you needed.

Space travel is expensive. Interstellar space travel, doubly so. Lots of effort is going to be focused on efficiency. How little can you send to another planet and still have a viable mission?

If you think about it, you can take inventory of what you wont want to bring with you for your new colony. It should be a very minimal list. Everything you brought with you should be useful. Heck, even the hull of your spacecraft is surely highly valuable. Waste not want not. When you really get down to it, there's not much mass that you would want to let keep flying into the darkness of space.

Also, consider the strength of the acceleration. If we're talking about most sci-fi interstellar travel, you spend a lot of time accelerating at a very slow rate. Ideally, you do so in a fuel efficient manner. If you jettison close to the planet, you're going to need very powerful retro-rockets to quickly slow your smaller vessels. That means lots of gee forces. Too close, and your humans may not survive the deceleration.

On the other hand, if you have to decelerate your small craft for a long time, then the distinction between your small craft and your larger craft becomes blurry. Your small craft suddenly needs all of the structure and life support needed to keep your passengers alive during this deceleration period. If you're not careful, suddenly you find yourself needing so much stuff that you might as well have just decelerated the whole ship!

Answer 3

Reason 1: Leaving the system again

The problem with your proposal is that there's likely no way for the small landing craft to leave the planetary system. In most sci-fi stories/movies, in my experience, the landing craft are small, sleek, and designed to go down to the planet and then return. They have engines powerful enough only to get back to the mothership, and not a lot of fuel. The designs are minimalist.

Here's an example. Take Obi-Wan's Starfighter from Attack of the Clones:

It has a separate ring to travel through hyperspace; the engines on the ship itself just aren't powerful enough. At the same time, the ring is not, I would assume, very helpful during re-entry, or when landing, for that matter. You need some other way of catching up with the mothership, or a way to get off the system.

If you're writing hard science fiction, then a hyperspace ring is probably not going to feature in your story. Odds are good, then, that your landing craft will need a way to get to the star system and leave again. The mothership needs to stay there so the craft can leave again, unless the crew are on a one-way journey.

Reason 2: Entering the system

At any rate, though, the craft will need to decelerate. If you're traveling at any speed that makes interstellar travel plausible, you're going pretty damn fast. Chances are good that you're going at a speed faster than the escape velocity for the planet, if not faster than the escape velocity for the system. If you intend to have any control whatsoever and land on the planet without crashing, then you should probably decelerate. A lot. Again, the landing craft likely don't have powerful enough engines for this.

Answer 4

Would not it be better to accelerate all the way, then leave the ship for small landing module(s) and decelerate them while allowing the interstellar ship to go away from the star system?

This implies that the landing is a one-way trip - if the small modules slow down to land while the mothership keeps accelerating, then they'll never catch up to the mothership again. If that's the intent, fine. If not, you've just stranded people on a strange planet forever.

The first issue that rears its ugly head is acceleration. Suppose your mothership has been accelerating at 1 g all the way from the source to the destination. If you want the load on your landing module to also be 1 g, then you will have to dispatch it at the halfway point. Dispatch it any later, and the g load will be higher. For equipment designed to withstand high g loads, that's not so much of a problem. For squishy payloads (a.k.a. meat sacks, a.k.a. people), you're going to have to keep it around 1 g or so.

Obviously, if your mothership is accelerating a slower rate, you can dispatch the landing module later in the trip to keep the acceleration around 1 g. But no matter what, you are going to have to account for that acceleration.

This brings us to a second issue - the amount of time spent slowing down. How long will it take your landing module to slow down at 1 g or so? A few hours? A few days? A few weeks? A few years? How much life support does it carry (if any)?

Let's say your mothership is booking along at 10% c (~30000000 m/s). At 1 g, it would take your lander on the order of a month to slow down (not accounting for any relativistic effects).

But, okay, we can deal with acceleration, we can deal with time to slow down. This brings us to the last issue.

If you're using any kind of a reaction drive (chemical rocket, ion engine, nuclear-pulse drive, etc.), you need reaction mass (propellant), and the amount of reaction mass you need rises exponentially with the change in velocity (ΔV). This comes from the Tsiolkovsky rocket equation:

$$\Delta V = ln\left(\frac{m_i}{m_f}\right)v_e$$

where

• $\Delta V$ is the total change in velocity;
  
• $m_i$ is the mass before the burn (payload plus propellant);
  
• $m_f$ is the mass after the burn (payload only); and
  
• $v_e$ is the effective exhaust velocity of your drive (which can also be expressed as $g * I_{sp}$, where $g$ is one standard Earth gravity (9.8 m/s²) and $I_{sp}$ is the specific impulse of your drive).

You may have heard the phrase "the tyranny of the rocket equation". That logarithmic term is what makes it so tyrannical. If you flip it around like so:

$$\frac{m_i}{m_f} = e^{\left(\frac{\Delta V}{v_e}\right)}$$

you see that your mass ratio $\frac{m_i}{m_f}$ grows exponentially with ΔV.

The higher the mass ratio, the more kg of propellant you need to accelerate each kg of spacecraft mass. A MR of 2 means you need 1 kg of propellant for each kg of spacecraft mass. A MR of 3 means you need 2 kg of propellant for each kg of spacecraft mass.

So let's assume you're using an engine with a specific impulse ($I_{sp}$) of 1500 seconds (sort of like the ion engines on the Dawn spacecraft). A ΔV of 30000 m/s (~0.01% c) gives us a mass ratio of e^(30000/(9.8*1500)) =~ 7.7. For every kg of spacecraft you want to accelerate, you have to expend 6.7 kg of propellant. A ΔV of 60000 m/s (~0.02% c) gives us a mass ratio of 59.24. A ΔV of 120000 m/s (~0.04% c) gives us a mass ratio of 3510.

For this type of engine, to reach 0.28% of light speed (a little over one quarter of one percent), you would need on the order of almost one Earth's mass of propellant (2.36e24 kg) for each kilogram of spacecraft mass you want to accelerate.

Obviously, we're not flying to another star system using ion engines, or frankly any other reaction drive we have the technology to build. You'd need drives with specific impulse values on the order of 100000 seconds or better, which we are nowhere near producing, or a true reactionless drive (one that doesn't require exchanging momentum), which are are also nowhere near producing.

Answer 5

Basically, the answer comes down to the amount of fuel you need to accelerate and decelerate. When planning an interstellar mission, your payload space is limited by how much fuel you can have onboard, so every available space is filled with essentials. You don't really have room for luxuries. If you want to use this escape pod method, those escape pods take up weight and space, contributing to the amount of fuel you need to accelerate the ship in the first place.

The escape pods themselves will also need a very large amount of fuel in order to perform the deceleration necessary. As we know from the Tsiolkovsky Rocket Equation, $\Delta v = v_e ln \frac{m_0}{m_f}$, meaning that as delta-v grows, the fuel needed grows at an exponential rate. Even if you fire them from the ship in the opposite direction, the ship itself will have to be carrying the fuel necessary to do that.

Add to this the fact that, since you've decided to let your ship accelerate the ENTIRE way, it's going to be traveling insanely fast, so those little escape pods will need to decelerate a huge amount.

In other words, while it is possible to do as you suggest, the fuel requirements would be tremendous. It would almost always be easier just to decelerate the whole ship. That way you can use all of the space onboard for necessary equipment and personnel. Plus once it arrives at the destination, you can reuse it, or park it in orbit to act like a space station.

Answer 6

This would only make sense if there was a substantial amount of the ship that could be disposed of that was only needed for launch or transit, and not for the final stretch. First of all, let's take a look at some assumptions.

• The nearest star- 4.3 light years.
• Assuming 10% of the speed of light, that's about 43 years, give or take, to get there.
• Assuming an acceleration of 0.1g, that's about a year to accelerate and decelerate.

Okay, so any mass that you would have to get rid of would have to be during the decelerate portion. That is still a sizable portion of the trip. You could safely dump at that time any empty containers for food, but for that long of a trip, you are probably having to grow food somehow. You could dump anything that was only needed for the launch phase. But I'd be pretty willing to bet that you need most of the ship to sustain life for the year required to decelerate.

Answer 7

There are basic laws in physics, one of them the conservation of momentum and energy, also known to be one of newtons laws. It says: "As long as no force changes the momentum of an object, its momentum and kinetic energy will not change."

Case A: constant acceleration

Now, let's take a spaceship, that accelerates all the way from A to B with constant acceleration $\vec a$. As long as it doesn't go relativistic speeds, it will achieve the velocity $\vec v = t \times \vec a$, wherin t is the travel time from A to B. As t is really large, this will be a very very huge speed.

Case B: deceleration at halfway point.

Now, we have our rocket accelerate with $\vec a$ for half the flight time, and then decelerate with a for the other half. Decelerating is accelerating with $-\vec a$ (we use vectors here!). The velocity at the end will be: $\vec v = \frac 1 2 t \times \vec a + \frac 1 2 t \times \vec -a = \frac 1 2 t \times (\vec a - \vec a) = 0$

But why should we want B We'll need much more time?!

Discussion

Our Energy is dependent on the speed: $E_{kin}=\frac 1 2 m v^2$. This kinetic energy is stored in the whole vessel and will continue to keep the ship on the path, as it is directly proportional to its momentum and inertia.

Now, we want to drop our colonist capsule. Let's say the capsule is 1/10th of the weight of the carrier ship. Then the capsule will carry 1/10th of the kinetic energy while still travelling almost at its speed before being launched.

Let's look at our cases: In case A, $\vec v$ is still that very huge number, in case B it is 0, or close to, if the deceleration was not for the full second half of the travel but some less. To get into a nice orbit around the destination's sun, the ship should do a bit less deceleration than acceleration.

Solution

Now, our colonist capsule needs to achieve several steps to actually land, and each one requires it to be slower (in relative terms) than in the step before:

• get an orbit with roughly the same speed and radius around the star as the planet it wants to land on
• then decelerate into an orbit around that planet
• then decelerate into a decaying orbit to land on the planet

So we need to already pack a huge amount of fuel to get through these maneuvers, and if we want to decelerate just the capsule in the solar system, that will not only demand packing even more fuel, it will demand extremely powerful engines lest the ship exits the system before it is out of it again.

We can however reduce the fuel cost for the deceleration: Whenever a fuel tank is leeched empty, we abandon it to continue on the trajectory to our destination, reducing our mass by a bit and thus the reducing time needed for the next phase of acceleration or deceleration.

Answer 8

The distinction that must be made here is between comfort and speed. A spacecraft accelerates half-way to gain the fastest velocity to reach its destination in the shortest possible time, but at a rate of acceleration that normal humans can go about their business and lives on board that is sufficiently comfortable.

If the spacecraft has to decelerate it will do so at the same rate of deceleration as its acceleration phase. This would involve decelerating the whole spacecraft. However, if instead of decelerating and stopping the spacecraft, once it had accelerated to the halfway mark, dropped off landing craft those landing craft would have to decelerate all the way until they reached the target system for colonization.

The reason that the landing craft have to decelerate from the halfway point is they have to lose the same amount of velocity as the main spacecraft and at a rate of deceleration that is comfortable and convenient for the human colonists it is carrying.

If the landing craft were dropped off closer to the target system the forces of deceleration will increase proportionally. The shorter the distance and the shorter the time to decelerate the higher the g forces will be.

Unless you want your colonists to arrive resembling raspberry jam, a longer and slower deceleration will be the way to go. Almost invariably this will mean if your spacecraft accelerated to halfway mark, it will have to decelerate the rest of the way to arrive safely and in one piece.

Answer 9

Short answer: Presumably "the whole ship" and all of it's weight is the very mechanism which provides the ability to dramatically change momentum (whether accelerating or decelerating), so decelerating without the whole ship begs the question: How?

Also, as Michael Kjörling has pointed out, the primary limiting factor of the acceleration/deceleration rate is probably going to be the force it exerts on the travelers, unless the story invokes some sort of "inertial dampener", which according to modern physics is impossible.

Andreas also makes a good point, that the total useful mass of the ship should be somewhat less by the midpoint of the trip, but if you have already maximized the force you are willing to subject your travelers to for the first half of the trip, then to achieve deceleration without exceeding that same maximum force will still require the deceleration to begin right around that midpoint. What changes is the expenditure of fuel required to achieve that deceleration rate.

I think you have a good idea at the heart of your question though. At the time of arrival, assuming you have no intention of reusing the vehicle, you have this whole big hunk of interstellar spaceship that is no use to anyone. It strikes me as a fine innovation to find ways to use parts of that hulking machinery as propellant mass. For example, let's say the fuel used for much of the first part of your journey required some sort of containment tanks, most of which are empty near the end of the journey. If these tanks could be ground into fine particles which are somehow injected into the propellant to add mass to the exhaust without completely wrecking the fine balance in the fuel, perhaps it could add to the efficiency of the deceleration and reduce the overall fuel payload. The engine that could make use of this dirty fuel without exploding and the grinder that is itself of some relatively negligible mass probably remain still in the realm of science fiction, but if such a thing were possible, decreasing the fuel payload would be a major gain. Fuel mass is a critical factor in any long range space flight, because adding fuel also adds weight, which requires even more fuel... you see where this is going.