# Delta-V to accelerate to/decelerate from .5c

I have a reasonably good understanding of delta-V (from playing KSP). However, I don't have a great understanding of how it works once you start running into the limitations of physics. I know that you cannot accelerate past the speed of light. This makes me assume that delta-v gets exponentially harder to increase, with the speed of light as the limit. Am I right so far?

Here's the next wrinkle. Let's say I want to have a ship with future-tech that gives it a terrific ISP, allowing it to accelerate to .6c. Then I want it to decelerate from .6c. Wouldn't the required dV for these maneuvers be 1.2c and therefore impossible?

Or am I thinking about this all wrong? I'm tinkering with a video game exploring mostly-hard science deep-space exploration and I'm curious if interstellar ships can accelerate to .5c on their journeys given a good enough engine ISP (and if such an engine ISP is possible within the bounds of known physics).

Thanks!

• Having 1.2c of delta v (0.6 + 0.6) and accelerating to 1.2 c are two different things – Antzi Feb 15 '19 at 10:24

## 3 Answers

If you can accelerate to a speed you can slow down from that speed, if you accelerate a ship to .6c and then back to 0c the net acceleration is 0, you aren't breaking any laws of physics. If all things are equal (mass of the ship for example) the delta-v to decelerate the ship to 0c is the same it takes to speed it up to .6c in the first place.

As you state there is a relationship between the speed of light and acceleration, what you're looking for is the Lorentz Factor. As you accelerate closer to the speed of light your effective mass increases according to the curve above. At .6c a ship's mass increases by about 25%, which would need to be factored in, either you'd accelerate slower or your engines would need to produce more thrust to compensate.

• So to clarify, my ship CAN have a dV of more than 300 million m/s (speed of light)? It's just that the faster I'm going, the more my mass increases, which makes said dV less effective? So really dV is only relevant to an object standing still, as soon as you start to go faster the amount you can change your speed starts to go down? It just doesn't really matter until you're going super fast? – Bert Haddad Feb 14 '19 at 21:13

Once relativity becomes a factor, you always have to phrase your statements so as to make it clear what point of view you are taking.

From the perspective of the rocket, and assuming you have somehow got unlimited fuel, you can keep accelerating as long as you like. From your perspective, doing so will shorten your journey to a distant star, which can become (again from your perspective) very short. What will appear to be happening is that by accelerating you are reducing the distance to your destination (Lorentz contraction) which is approaching you at almost the speed of light. Adding up the effect of your rocket, you will appear to have used arbitrarily much delta-V.

From the perspective of a non-accelerating observer on the planet you started from, the same events look very different. As you approach the speed of light, the mass of your spaceship appears to increase, and the thrust of the rocket decreases (less fuel is being used per second from this perspective because of time dilation). The duration of your journey will be a little more than the time taken for light to reach your destination and the total delta-V will slowly approach c (for acceleration) and then c more for deceleration.

This article sets out all the equations.

In short:

1. These maneuvers would require 1.2c $$\Delta v$$.
2. Using 0.6c $$\Delta v$$ for linear acceleration would only result $$\frac{1}{\sqrt{1+\frac{v^2}{c^2}}} \approx 0.51c$$ .
3. The $$c$$ limit exists only for the possible speed of objects in reference frames. If you sum speeds, or not related speeds, you can get any big speed num, but it won't have physical meaning (in the sense affected by the Special Relativity).