@ Bill Otto Lived next to NASA Huntsville;became manned spaceflight SME 的回答
Let’s say you had a very nice flashlight with a 5 degree beamspread and a 1 watt visible beam.
At 60 million km to Mars at close approach, the beam spreads out over 2.5×10^13 sq km or 2.5×10^19 sq meters. That means about 1 photon per second per square meter arrives at Mars.
距离火星近6000万公里，光束扩展到 2.5×10^13 平方公里或 2.5×10^19 平方米。 这意味着每平方米每秒约有1个光子到达火星。
Mars does not move far enough in the time it takes light to get there to need a lead ahead angle with a flashlight.
So yes, a very small amount of the light gets there. Not enough to detect with even advanced sensors, but some gets there.
I get different answers, so maybe I’m missing something. Mars is about 58 million km away at closest approach, so around 200 light-seconds. It orbits at about 24 km/s, and is about 6,800 km across, so in 200 seconds it will move over 70% of a diameter. If you aim at the centre, you’ll miss.
我算出不同的答案，也许是我忽略了一些东西吧。 火星距离最近的位置约有5800万公里，因此光到达那儿大约200秒。 火星以大约24 km / s的速度在轨道上运行，在这段时间内就前进了6,800 km，因此在200秒内它移动的距离超过直径的70％。 如果您之前瞄准火星中心，就会错过火星。
If we assume the earth’s orbital velocity of 30 km/s is parallel to Mars’s motion, and our photons keep that transverse velocity like bullets, you’ll need to aim a third towards the trailing edge of Mars to hit it dead centre.
如果我们假设地球是30 km / s的轨道速度，与火星的运动平行，并且我们的光子像子弹一样保持横向速度，则您需要将三分之一的距离对准火星的前面以使其到达中心。
Sensors can definitely recognize a single photon, and with the largest optical telescopes having an aperture of over 10m, collecting from a 1 m2 area shouldn’t be a problem. The bigger problem will be separating the tiny input from the flashlight from everything else — particularly since at closest approach, the Sun is going to be in the same shot, and bombarding that square metre with about 1,660,000,000,000,000,000,000 photons per second!
The flashlight has a 5 degree spread angle, or a beam diameter about 700 times the diameter of Mars. The beam is so big Mars is not going to move out of the beam in 400 seconds. (Time for light to make a round trip.) If your sensor is a James Webb Telescope, you can detect single photons per square meter, but distinguishing them from other light sources on the Earth would not be possible without some sort of encoding. A flashlight does not have such encoding. That is what I mean by not being able to detect the flashlight.
1623/5000手电筒的散角为5度，或者光束直径约为火星直径的700倍。光束是如此之大，火星不会在400秒内移出光束。 （光传播的时间。）如果您的传感器是James Webb望远镜，则可以每平方米检测单个光子，但是如果不进行某种编码，就无法将它们与地球上的其他光源区分开。手电筒没有这种编码。我的意思是无法检测到手电筒发出的光。
And by the way, you cannot really detect single photons. There are several thousand noise electrons in the detector per second, as well as some signal photo-electrons. Without some way to statistically separate them, like encoding, you cannot detect a single photon per second unambiguously. When your signal to noise is 0.0001, you are not detecting. If I sent pulses at 100 per second, and had a narrowband filter centered on 100 Hz (plus a Doppler offset) then I would have a chance, but that would require integrating for quite a while, since I will only get a photon out of every 100 pulses. But the flashlight is not pulsed.
This is a much more difficult detection problem than you have accounted for. Show me a sensor that can detect a single photon per second without a lot of false detections. It will certainly make my job a lot easier! I do photon-counting ladar. When we say we detect single photons, there is a lot of fine print. What we mean is that we can statistically tell that some of the photo-electrons we detected were signal and not noise, but we can’t really tell which.
@ Marvin Harris Always learning.
My issue here is atmospheric scattering. Between atmospheric dust and particles between us and mars, I’d be pretty impressed if any made it there.
This chart includes the effect of scattering.
I have been involved with sending a lot of laser beams out of the atmosphere. Other people bounce laser beams off the Moon. As long as you don’t use blue light and don’t shoot too close to the horizon, there is no problem.
我参与了向大气中发射大量激光束的工作。 其他人向月球反射激光束。 只要您不使用蓝光且拍摄距离地平线不太近，就没有问题。
This chart is for “one airmass” which means shooting vertically. At 10 degrees elevation angle, the atmosphere is effectively 6 airmasses, which means absorption and scatter will be the 6th power of the absorption at zenith.
该图表适用于“ one airmass ”，即垂直来看。 在仰角为10度时，大气有效为6个 one airmass ，这意味着吸收和散射将是顶部吸收的6次幂。【？？】
There is usually not enough scatter or absorption to worry about with cloudless skies as long as you stay above about 15 degrees elevation..