# What speed could a Nicoll-Dyson beam propel a solar sail ship to?

So far In my research I have not found any estimates for the speed that a solar sail spaceship could be propelled to using a Dyson beam.

I have not decided what mass the ships will be that I want propelled as although a Dyson beam would be very powerful, the mass of the ship may make other propulsion methods a better option.

I would assume a blue giant type star would be the best option for the highest speeds but if this method was used by an old civilization, longer living stars may be the type used to propel ships.

Taking into account the possible star type and mass of the ship, what speeds could a Nicoll-Dyson beam propel a sail spaceship?

• Not an answer, but: have you considered how you might stop? Without an equal amount of deceleration you'll pass through your destination's entire solar system in a matter of hours. OTOH once you get up to speed, there will be some drag from interstellar gas, but any dust particles will do huge amounts of damage Jan 10, 2022 at 14:06
• @ChrisH It needs more thought but possible options are spinning the ship and sail around to be slowed down by a beam from the destination star, Using the gravity well and solar winds of the destination star, I read about a secondary sail arrangement, or with the ship masses given in the answers a large amount of fuel could be carried for this purpose. I was originally thinking of using the first idea within an empire but other options would be needed for new destinations. Jan 10, 2022 at 18:32
• @ChrisH True, the maintenance of the sail could be necessary even with the strongest light materials. Jan 10, 2022 at 18:35

Omitting the technical details:

• An entire Nicoll-Dyson beam is so overwhelmingly powerful that it would probably just incinerate any sail it was pointed at in an instant. This means that your lightsail ships can accelerate at their maximum possible rate given their ability to radiate away absorbed energy, whilst running your laser at far below maximum power. You'd probably drive many sails at the same time, probably at different targets.
• An ND beam can push a lightsail that's as far away than the Andromeda galaxy, assuming it can get a firing solution on something that far away. This effectively means you can accelerate your lightsail at its maximum rate for as long as anyone might reasonably want, but given the sheer mindboggling amount of power available you won't need to push any reasonably sized sail for very long before it hits silly speeds.

Do you want high relativistic speeds? Sure, you can have that. You could accelerate a 4 million tonne ship at 1G with a beam power of less than 6 exawatts, which is less than a millionth of the output of the Sun. You'd hit .9c in a bit over 2 years. If the sail were 1000km in diameter, it would be illuminated with ~6.7MW of energy per square metre. It would need to be very reflective indeed. Arranging such a material is left as an exercise to the reader, though if you can consider building a Nicoll-Dyson array around a blue giant star I'm sure you'll do just fine.

The slightly unhelpful but obvious answer is "asymptotically close to the speed of light", limited only by the expansion of space. The acceleration of a lightsail will eventually become negligible when it is far enough away from the emitter so there's a practical speed limit associated with your patience.

Your laser will have a distance beyond which it can no longer focus all of its energy onto the lightsail. You can't get any better than a diffraction limited beam. According to Rayleigh's criterion, a laser of wavelength $$\lambda$$ fired from an emitting element with diameter $$d_l$$ will fill a sail of diameter $$d_s$$ at range $$r = {d_sd_l \over 2.44\lambda}$$

Lets say you have a sail that's 1000km across, and a laser firing 500nm yellow light, then to have your diffraction-limited range be 10 lightyears then your emitting element would need to be 115km across. A Nicoll-Dyson array with a radius of 1AU could hit that same sail at 25 million lightyears away. One with an orbital radius equivalent to Mercury could still focus at 5 million lightyears. That's intergalatic range. Aiming and focussing is also left as an exercise to the reader ;-)

The velocity $$v$$ reached after a certain period $$t$$ of constant acceleration $$a$$, accounting for relativistic effects, is $$at \over \sqrt{1 + \left ( \frac{at}{c}^2 \right )}$$ (where $$c$$ is the speed of light). If you're prepared to keep your laser running for 25 million years, then even if it only imparted an acceleration of a microgee, you'd still hit .999c.

Lightsails accelerate because photons have momentum. From The Starflight Handbook, if a beam of light with energy $$E_b$$ is perfectly reflected by a sail of mass $$M_s$$, then that sail will experience a change in velocity $$\dot{V_s} = {2E_b \over M_s c}$$.

If your 1000km diameter sail starship had an areal mass of 5g per square metre all in, it would weigh about 4 million tonnes. Upsilon Orionis is a class B0V star with a luminosity about 60000 times that of the sun. If your ND laser could convert 10% of that into beam power, you'd be shooting out about 2 x 1030W, which is enough to push the sail at about 340 billion gravities... that's probably a bit impractical to reach in practise, to say the least. With a merely sun-like star and the same 10% efficiency, you'd still get 6 million gravities.

• "Do you want high relativistic speeds? Sure, you can have that. You could accelerate a 4 million tonne ship at 1G with a beam power of less than 6 exawatts, which is less than a millionth of the output of the Sun. You'd hit .9c in a bit over 2 years." remarkable.. so bottom line you actually won't need this giant beam to go sun sailing with ships that are much smaller than that.. +1, great explanation Jan 9, 2022 at 20:51

As Big as This Guy

You can accelerate something as big as the asteroid that smooshed all the dinosaurs to relativistic speeds in a matter of years.

The total energy output of Earth's sun is about $$4 \times 10^{26}$$ Watts or Joules per second. Let's ignore the $$4$$ for simplicity.

The space shuttle is about 2000 tons or $$2 \times 10^6$$ kg. Let's ignore the $$2$$ for simplicity.

The relativistic kinetic evergy of the shuttle is:

$$E = \left( \frac{1}{\sqrt{1-v^2/c^2}}-1\right)mc^2$$

for $$v$$ the velocity measured in metres per second and $$c$$ the speed of light in metres per second.

Note: Here $$m$$ is the rest mass. The following can me reformulated in terms of the so-called relativistic mass

$$M = \frac{m}{\sqrt{1-v^2/c^2}}$$

Notice $$M \simeq m$$ for small $$v$$ and $$M \to \infty$$ as $$v \to c$$. If we write things like $$E = \frac{1}{2} Mv^2$$ we can see that as $$v \to c$$ an infinite amount of energy is needed.

We know $$c \simeq 3 \times 10^8 \simeq 10^8$$. The $$-1$$ in the formula is unimportant when $$v$$ is close to $$c$$. So we get

$$E \simeq \frac{mc^2}{\sqrt{1-v^2/c^2}}$$

Flip this around to get

$$v =c \sqrt{1 - \frac{m^2 c^4}{ E^2}}$$

Not this goes to $$c$$ as $$E \to \infty$$.

We can graph the above in terms of exposure time to the beam. After $$t$$ seconds of exposure we have transferred $$t \times 10^{26}$$ Joules of energy. So we get

$$v \simeq c \sqrt{1 - \frac{10^{12} 10^{32}}{ t^2 10^{52}}} = c \sqrt{1 - \frac{10^{44}}{ t^2 10^{52}}} = c \sqrt{1 - \frac{1}{10^8} \frac{1}{t^2}}$$

We could graph this but it's easy to see that even for 1 second of exposure the thing under the square root is very close to 1 and so $$v$$ is almost the speed of light.

To see how big a spaceship we can accelerate consider instead the moon which weight about $$10^{24}$$ kg. Now the velocity becomes

$$v \simeq c \sqrt{1 - \frac{10^{48} 10^{32}}{ t^2 10^{52}}} = c \sqrt{1 - \frac{10^{28}}{t^2}}$$

Again we could graph this but even after a year or $$10^7$$ seconds we still only have $$c \sqrt{1 - \frac{10^{28}}{10^{14}}} = c \sqrt{1 - {10^{14}}}$$ which doesn't make sense. This comes from our ignoring the $$-1$$ earlier and means we are nowhere close to the speed of light.

Let's instead consider a $$10^{17}$$ kg spaceship. That's about the weight of the asteroid that smooshed all the dinosaurs. Then the exponents cancel and the speed becomes

$$v \simeq c \sqrt{1 - \frac{10^{14}}{t^2}}$$

If we measure time $$T$$ in years rather than seconds $$t$$ this becomes

$$v \simeq c \sqrt{1 - \frac{1}{T^2}}$$

which increases like this:

Note the graph is only accurate as we approach the speed of light (1 on the y-axis) and even then it only gives an indication of the order of magnitude we can expect. So anything from $$10^{15}$$ kg or so can be brought to the speed of light in a few years.

• Does this take into account the increasing amount of energy needed to take a ship to c? I thought that the mass of something increased as it reaches speed of light but I just read it is not described like that now but the principle of needing vastly more energy is the same. Jan 9, 2022 at 18:21
• It does indeed! That is why the speed never goes past $c$. You can rephrase all this in terms of the quantity $M = m/ \sqrt{1-v^2/c^2}$ which is called the relativistic mass. Notice it starts at $M = m$ for $v=0$ and becomes infinite as the speed approaches lightspeed. Then you can write things like $E = \frac{1}{2} M v^2$. Jan 9, 2022 at 18:29
• @AlanDavies See the note. Jan 9, 2022 at 18:35
• @StarfishPrime It's the crater caused by the asteroid that smooshed all the dinosaurs. Jan 9, 2022 at 20:05
• Great read & explanation +1 Jan 9, 2022 at 20:47