*No way*, I thought to myself.

*Satellites are too heavy*. However, his question has been hovering around in my mind, so I've decided to try and answer it: can we use a laser to launch a satellite into orbit?

Let us begin with a few simplifying assumptions. I'm going to assume that we want our launched satellite to reach the escape velocity of the earth. Of course, we don't want a satellite to escape orbit, but escape velocity is calculated without considering any kind of drag forces on our launched object. To first order then, I expect achieving escape velocity will get our satellite into a relatively high orbit without actually escaping earth's gravity well. Second, I'm going to assume our laser is on the ground (reusable launching device!), and that our satellite is perfectly reflective, so we're not going to be melting it. Finally, I'm going to assume that the laser will remain effective at targeting our object up to 15 km above the surface, around the effective range for lasers used as guides for adaptive optics in telescopes.

Now, to the meat of the problem. First, we need to find the escape velocity from the earth. This is defined as the velocity we would need to give an object to overcome the gravitational energy of the earth upon the object. This is found by setting the kinetic energy of the object equal to the change in gravitational potential energy from the earth's surface to infinity:

\[ KE=\Delta PE \]

so

so

\[\tfrac{1}{2}mv_{esc}^2 = \frac{GM_{earth}m}{R_{earth}}\]

(for those interested, the gravitational potential energy at infinity is defined to be zero). We can easily solve this for v,

\[ v_{esc}=\sqrt{\frac{2GM_{earth}}{R_{earth}}\]

To me, the most remarkable feature of the escape velocity is that it is independent of the mass of the object! We can now plug in numbers, and find an escape velocity if ~11.2 km/s.

We now make use of our constraint, that we have to achieve this escape velocity over 15 km. We assume the laser outputs a constant number of photons per second, n. The force the laser provides the satellite is the change in momentum with time,

\[F=\frac{dp}{dt}\]

How is our laser imparting momentum to our system? Well, light is composed of photons, tiny packets of light. Each photon has momentum. Assume each photon reflects perfectly, that is, straight backwards and with no energy loss. Then the photon has reversed its momentum. Since momentum is conserved, the satellite has gained a momentum of 2p. The momentum of a photon is

\[p=\frac{h}{\lambda}\]

where h is Planck's constant. This gives

\[F=\frac{2nh}{\lambda}\]

Given a constant force, the time taken to reach some velocity is just

\[t=mv/F\]

In our case the v is the escape velocity. Now, given a constant force, and no initial velocity, an object of mass m travels a distance d in a time t given by

\[d=\frac{F}{2m}t^2\]

We can substitute our expression for t into this expression, giving

\[d=\left(\frac{F}{2m}\right)\left(\frac{mv}{F}\right)^2\]

\[d=\frac{mv^2}{2F}\]

What we are really interested in is n, the number of photons per second this process takes, so we substitute our expression for the force, and solve for n:

\[d=\frac{mv^2\lambda}{4nh}\]

s0

\[n=\frac{mv^2\lambda}{4hd}\]

Now that we have the number of photons per second our laser supplies, all that is left is to find the power of laser this would take. The power of a laser will be given by the number of photons per second times the energy per photon. The energy per photon is given by E=hf where f is the frequency of our light. So, the power, P, of our laser is

\[P=nE\]

or

or

\[P=\frac{mv_{esc}^2\lambda f}{4d}\]

so

so

\[P=\frac{mv_{esc}^2 c}{4d}\]

Where we have recognized that the frequency times the wavelength of a photon is just the speed of light, c.

All that remains is to plug in numbers. A medium sized satellite might be about 750kg, giving a laser power of 4.7*10^14W. This is .5 PW. According to wikipedia, the greatest power output of a continuous operation laser is on the order of 1 kW, or ~10^12 times less than our necessary power! There's no way we're getting a normal sized satellite into orbit with a laser. What about a smaller satellite? In recent years picosatellites have proposed, with masses of ~.1kg. This gives a necessary power of 6.2*10^10, or 620 TW. This is still ~10^8 times greater than our most powerful continuous laser. In fact, reversing the calculation, our laser could launch a satellite with a mass of ~1.6 mg. A little googling reveals this is the same order of magnitude as a grain of rice! Simply put, we're not going to be launching satellites with lasers anytime soon.

I heard an interesting talk a while back on a variant of the laser launching idea, which was to use a high-powered ground-based laser as a kind of heating element. The idea is that you blast your "rocket" with a laser from the ground, heating a store of inert gas with the light, and shooting it out the back end at high speed, providing propulsion.

ReplyDeleteThis is a net win because it eliminates the need to carry two components of fuel (both a fuel and an oxidizer), and a lot of the safety problems that go along with that. And it cuts down on the weight needed.

Of course, it means that any space launch is accompanied by a giant laser beam tracking the launch vehicle across the sky, which I'm sure would be a huge hit with the FAA...

I am very much not a math guy (to my chagrin), but it seems all of this only takes into account light pressure.

ReplyDeleteHowever, there is another form of light propulsion called Ablative Laser Propulsion, and I wonder what the numbers on that would reveal.

The Wikipedia link: http://en.wikipedia.org/wiki/Laser_propulsion.

This method involves using a ground-based laser to convert material on the launch vehicle into plasma which is then vented to create thrust. I've seen footage of scale model tests, and it worked reasonably well.

Might the numbers be more favorable using this methodology, and something that we could hope to see as a usable launch method in the near future?

I would certainly be interested to see how the math works out...

I did actually see that brief on wikipedia when I was doing a little background reading on laser propulsion systems. I've just grabbed a few recent articles on ablative laser propulsion, and I'll see if I can come up with a review of the physics and the practicality in the next day or two.

ReplyDelete"To me, the most remarkable feature of the escape velocity is that it is independent of the mass of the object!"

ReplyDeleteAh, my friend, that's what the equivalence principle is all about!

also, I have some issues with your derivations; namely, escape velocity is only relevant when you are supplying some initial velocity and then letting go. Here what we're doing is constantly supplying force - so what we really need is just to push harder than g, the earth's pull.

ReplyDeleteWe want

\[F > mg = m\frac{MG}{R^2}\]

which leads to a result similar to yours but slightly different,

\[n > \frac{MGm\lambda}{2R^2 h} \]

or

\[ P > \frac{MGm c}{2R^2}\]

which comes to about 1.1 TW. If we want to push it fast we'll probably need double that - so let's say 2 terawatts. A couple of orders of magnitude easier, but still not currently achievable.

Well, just pushing harder than the earth's gravitational force doesn't guarantee that something won't fall back to the ground, unless you keep on pushing. The essence of my claim is that I want to supply enough force to get this thing to escape velocity when it is at 15 km up, as that is when I'm 'letting go'. I figured that would be a good estimate of the amount of energy to get into a high orbit, because we've been neglecting drag forces, so it really won't hit that escape velocity, it'll have an energy somewhat less than this.

ReplyDeleteSo, I did play a little fast and loose with the numbers. What I really wanted was to achieve the escape velocity of 15 km above the surface of the earth at 15 km above the surface of the earth. However, if you look at the escape velocity at 15 km up, that's given by

\[v_{esc}=\sqrt{\frac{2GM_{earth}}{R_{earth}+15km}} \approx 11.2 km/s\]

This is because the radius of the earth is ~6.3 Mm. So a 15 km difference doesn't really change the escape velocity.

Hopefully that clears up my calculation a little bit.

Ah, yes, I missed the part about only pushing it for 15 km. In that case your derivation is correct. I suppose another way to say it is that you can either supply a little energy over a long time or a lot of energy over a short time. I suspect that totaling the energy from both methods will give you the potential difference you need to overcome to break free of the Earth.

ReplyDelete