Bush Cancels trip to Pluto.

[KrazyHorse]

GP....hmmm.

Let's see.

Again assuming non-relativistic non-zero motion (v/c < 0.01, mv >> hf/c) and to first order:

We have the impetus delivered by the photon as 2hf/c, so original momentum is mv and the final momentum is mv + 2hf/c. Energy transfer is elastic (or so) so the change in energy of the probe is (p_f^2)/2m - (p_i^2)/2m

To first order again, the energy increment is 4hfv/c, therefore this is how much of the original energy the photon loses. Note that we can see why the constraints are necessary, since for v/c > 0.25 we have that the photon descends into a negative energy state with this approximation, and for v=0 we get "something for nothing" as the photon does not lose energy in the process.

[TCO]

Ok so how much does the photon change wavelength? (What percent of the energy is transferred in each bounce? How much does the wavelenght change. Let's say we start with an MBT sized body (75 tons) at rest and the initial light wavelength is 500nm.

I guess I should be able to crunch this myself but I get confused with nu (funny looking v) and f and all that.

[TCO]

The mirror bounce system onbiously seems difficult (optical alignment is tough enough on an optical table...but you're gonna do it over millions of miles?) but let's say we can do it. The less energy is transferred by a bounce, the more dependant you are on precision alignment...) Or looking at it another way, given a set ability in precision alignment...(let's say an average of 4 bounces before the photon is scattered) than the lower the energy transfer the more significant the wasted energy becomes.

*GP as always...bowing to the high poobas of physics...but attempting to approach engineering problems with an engineering mindset*

[Roman]

Actually, I just realised something. Even with the mirror-bounce system with no light absorbtion, the energy trasfer could not be 100% efficient, because energy is also transfered to the other mirror.

[TCO]

yeah. I was thinking about this too. I think that the lighter the "struck body" the more energy is transferred. So that you would expect more energy to go to the spacecraft. But I'm not sure if you would need a very rigid mounting on the Moon mirror for instance to stop loss of energy through vibrations. Oh and if it were on the moon and the spaceship was being sent to very far away from the moon, you'd obviously have frequent lost contact because of the rotation of the moon around the Earth. (The Moon has a fixed face, right?)

[Roman]

Well, yes, if the mirror was properly fixed on the Moon the energy loss would be negligible.

Tracking the probe would indeed present a problem. Maybe it could be boosted intermittently - only when the beam can reach it.

[TCO]

If the mirror/laser were mounted on the Moon and the probe was sent away in the direction of the ecliptic (say to an outer planet) than it would be untrackable about 50% of the time, because of the Moon's rotation around the Earth.

[sulla]

Why did we do something so stupid?

[Dauphin]

. Oh and if it were on the moon and the spaceship was being sent to very far away from the moon, you'd obviously have frequent lost contact because of the rotation of the moon around the Earth. (The Moon has a fixed face, right?)

The Moon rotates once every 28 days, however the solution to this problem is to place the system at either of the the lunar poles. These will maintain a fixed orientation all year round.

[TCO]

BC, that would work ok for flights out of the plane of the ecliptic. (I said that.)

[Dauphin]

If the mirror/laser were mounted on the Moon and the probe was sent away in the direction of the ecliptic (say to an outer planet) than it would be untrackable about 50% of the time, because of the Moon's rotation around the Earth.

The Earth does not block the path between the Moon very often. What's the problem?

The moon follows a plane that is inclined ~5 degrees to the Earths rotation plane around the sun. Which is why we don't get eclipses every month.

[TCO]

Originally posted by Big Crunch

If the mirror/laser were mounted on the Moon and the probe was sent away in the direction of the ecliptic (say to an outer planet) than it would be untrackable about 50% of the time, because of the Moon's rotation around the Earth.

The Earth does not block the path between the Moon very often. What's the problem?

The moon follows a plane that is inclined ~5 degrees to the Earths rotation plane around the sun. Which is why we don't get eclipses every month.

I'm wasn't talking about the "blocking problem" but about the moving around the earth issue. Imagine a dime on a phonograph.

Having a polar laser/mirror takes car of this problem excepts for louanch in the ecliptic. (unless you assume that the Moon is a perfect sphere...you need to have some elavation above 0 degrees or else the laser will be occluded by hills and stuff.)

[TCO]

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?

[Dauphin]

OK, so you mean because you are firing towards the horizon.

I don't believe that to be a serious problem. As I say the Moon is inclined about 5 degrees to the plane of rotation of the planets, so only if the poles are located in a valley, and surrounded by mountains are you going to have a problem.

Any planetary craft would be 5 degrees above the horizon, which means the line of sight would only be blocked by hills above 80m high within a 1km radius, 800m within 10km.

[TCO]

Originally posted by Big Crunch
OK, so you mean because you are firing towards the horizon.

I don't believe that to be a serious problem. As I say the Moon is inclined about 5 degrees to the plane of rotation of the planets, so only if the poles are located in a valley, and surrounded by mountains are you going to have a problem.

Any planetary craft would be 5 degrees above the horizon, which means the line of sight would only be blocked by hills above 80m high within a 1km radius, 800m within 10km.

OK, cool. You're right. Geometery will be fine almost all the time.

Could you address some of the other questions?

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?)