Bush Cancels trip to Pluto.

[Dauphin]

Originally posted by GP
1. How many bounces ar required to transfer energy.

2. What impact does the bounces at the fixed mirror have. (Is any significant energy expended there?)

2) Practically none.

Momentum transfer (making "classical" assumptions) is 2p. Energy transfer per hit will be the increase in kinetic energy ~p²/m_ref where m_ref is the mass of the reflecting object. Therefore the allocation of energy to the fixed item (the moon) and to the spacecraft is the ratio of masses.

Mass of moon = 10²² kg.
Mass of craft = 10³ kg at most.

With regards to 1) I will have to check. I'll get back to you, but it depends on craft velocity and mass. The faster and lighter the less it requires.

Edit - Getting masses of the Earth and Moon mixed up.

[TCO]

OK. I know it's an it dpends answer. I just want a feel it. 4 or 4 million.

Pick some reasonable numbers. If you can't come up with any how about. An MBT at rest with visible light.

MBT = 75tons =80,000 kg. Visible light...say 500 nm.

[MasterBob The Elder]

I guess Bush is afraid of terrorists sabataging his trip to pluto or something.

[Dauphin]

Originally posted by GP
OK. I know it's an it dpends answer. I just want a feel it. 4 or 4 million.

Pick some reasonable numbers. If you can't come up with any how about. An MBT at rest with visible light.

MBT = 75tons =80,000 kg. Visible light...say 500 nm.

I did a quick calculation assuming those figures and two opposable mirrors and came up with 10⁴⁰.

I have serious doubts over my calculation.

[Jack the Bodiless]

For launches in the Ecliptic, just mount the laser on a circumlunar railroad. It wouldn't have to move very fast to counter the Moon's one-month day. And you could do the same with Mercury, using huge fixed or rail-mounted solar cell arrays for power. With solar-powered self-replicating robots, the whole setup could build itself from local materials.

[TCO]

Originally posted by Big Crunch


I did a quick calculation assuming those figures and two opposable mirrors and came up with 10⁴⁰.

I have serious doubts over my calculation.

1. AHHH...who cares about a few tens of orders of magnitude....let the engineers fix it!

2. (more humor) That's one tired little photon...he spends decades shuttling back and forth to make all those bounces...sure hope there isn't an atom out of place on the mirror...or a particle of dust in space anywhere...or our tired friend gets scattered! Poor little guy. Cute little tired photon.

3. (serious question) To check your calcs just see how much energy is transferred in a single bounce. (for instance if the mirror system was impractical...perhaps because the moving craft has an imperfect mirror/optic.) Do we only get ~1/(10^40)* of our laser energy transferred to the craft?


* I realize that the amount of energy will probably change as the photon changes to a lower wavelength...but at this stage orders of ,magnitude seem to be enough to talk about.

[TCO]

Originally posted by Jack the Bodiless
For launches in the Ecliptic, just mount the laser on a circumlunar railroad. It wouldn't have to move very fast to counter the Moon's one-month day. And you could do the same with Mercury, using huge fixed or rail-mounted solar cell arrays for power. With solar-powered self-replicating robots, the whole setup could build itself from local materials.

Or you could just have a few seperate sites and hand off periodically. I agree that geography is not a big issue.

[TCO]

just a check on your calculation, BC. That was for transferring 25% of the energy to the craft? I assume that there are diminishing returns as the photon becomes lower in energy. That's why I picked 25%.

[Dauphin]

Hope you follow this

I assume that there are diminishing returns as the photon becomes lower in energy

The final equation I got for wavelength shift per hit was:

l - l' = h/m_craftc.

Which is identical to the equation for Compton scattering, which raises alarm bells.

So anyway I plugged in the figures:

h = 6.6 x 10^-34
m = 10⁵
c = 3 x 10⁸

Therefore Dl = 2.2 x 10 ^-47, per hit. We want the wavelength to change by 1500nm so divide 1.5 x 10 ⁶ by 2.2 x 10 ^-47 and we get ~ 10 ⁴⁰

Using classical assumptions the energy transfer per hit should be p²_photon/m_craft = (h/l)²/10⁵

Plug in the figures and the energy transfer is ~10^-40 of the photons total energy per hit.

Looks like my calc wasn't out by any serious order of magnitude.

[Dauphin]

Cute little tired photon

Don't worry. Given the relativistic time dilation, photons (from their POV) are destroyed as soon as they are created.

[KrazyHorse]

Originally posted by GP
So anyway...enough of that, let's say you've got an unoccluded path. I grant you that for most geometries/times that will be so.

How many bounces do I need to get 25% of enerrgy expended in making the light converted into spacecraft movement.

Rough idea: 1? 10? 10^6?

GP, I'm disappointed in you. If you'd read my conclusion, you'd see that the energy loss is dependent on "v", the veloctiy of the probe. For classically-dominated regions, the energy transferral is directly proportional to v.

If the probe is moing at, say, 3000 m/s (a reasonable "starting speed" to be achieved using chemical propellants) then 4% of the photon's energy is used up for each bounce. Solving for log_0.960.75, we get ~7 bounces.

[KrazyHorse]

BC, I'm disappointed in you, too.

You forgot to take the probe's motion into acount.

Your calculations are only good for when the probe is "starting out" with 0 velocity wrt the mirror/laser.

[Dauphin]

Originally posted by KrazyHorse
BC, I'm disappointed in you, too.

You forgot to take the probe's motion into acount.

Your calculations are only good for when the probe is "starting out" with 0 velocity wrt the mirror/laser.

Just testing you.

I was too busy looking for a calculation error to spot that assumption error. I overlooked it by saying to myself "from the crafts frame of reference".

[KrazyHorse]

Yeah, I was doing the same thing too, at first, but I was trying to work out the Doppler shift...

Instead, I just ignored it and put in a caveat about non-relativistic speeds.

I love approximations. They make my life much easier.

[Dauphin]

Kicking myself

Why did I not just use the Doppler shift equation?

a reasonable "starting speed" to be achieved using chemical propellants

Or my other favourite, the lunar rail gun, another energy saving device.

As you may have guessed, I am bit of an energy conservationist.