<blockquote>
  <p>Could the Philae comet lander be recharged with a laser (or possibly many lasers) from earth?</p>
</blockquote>

<p>No.</p>

<p>You could make a laser beam with a pretty low divergence - meaning a beam that remains tight and parallel over a very long distance. However, "a very long distance" is nothing on the cosmic scale.</p>

<p>The Lunar Laser Ranging experiment is using high-power lasers focused by telescopes, shooting a beam into the Moon. The light is reflected back to Earth by mirrors placed there by the Apollo program.</p>

<p><a href="http://en.wikipedia.org/wiki/Lunar_Laser_Ranging_experiment" rel="nofollow">http://en.wikipedia.org/wiki/Lunar_Laser_Ranging_experiment</a></p>

<p>The problem is, even with all the tight focusing at the origin, of the initial 10^17 photons in a pulse, only 1 makes it back to the detector. It's very hard to tell that something makes it back at all, let alone charging any solar panels with it.</p>

<p>The comet and Philae are much, much further away than the Moon. You couldn't possibly project any useful energy density at that distance. Not with the current technology.</p>

CC BY-SA 3.0
2014-12-05T23:19:53.253Z
Perfect. This is just what my answer was missing.
HDE 226868
CC BY-SA 3.0
2014-12-06T03:48:06.327Z
From the Earth to the lunar mirror and back you are basically squaring the factor by which the flux is reduced compared to just from the Earth to the lunar mirror. The beam diverges from here to the Moon and the mirrors capture some fraction of the flux (mirror area divided by beam area at the Moon) and then on the way back, the reflected beam diverges again  and back on Earth the telescope receives a small fraction of what the mirror reflected (telescope aperture divided by beam diameter).
Count Iblis
CC BY-SA 3.0
2014-12-08T22:28:08.210Z
@CountIblis - right. Perhaps the example was not the most apt. In any case, there's no way we could expect to recharge a probe deep into the Solar System when we can't even "put a laser dot" on the dark side of the Moon, that would be visible from Earth. We can't create that much energy density at that distance. And that's just the Moon, basically in our backyard.
Florin Andrei
CC BY-SA 3.0
2014-12-05T19:26:47.103Z
2014-12-06T00:47:30.240Z
2014-12-06T00:47:30.240Z
Florin Andrei
<p>I only have time for a partial answer, so I'll discuss the bit about rotating the comet.</p>

<p>We have to do some calculations with <a href="https://en.wikipedia.org/wiki/Radiation_pressure" rel="nofollow">radiation pressure</a>, the force on an object from light.$^1$ <a href="http://www.webassign.net/question_assets/buelemphys1/chapter22/section22dash3.pdf" rel="nofollow">This pdf</a> has some good starting calculations, although they assume that the source is spherical - the Sun. Our laser clearly is not. However, the beam <em>will</em> diverge over such a long distance, so I'll have to use a substitution (given by CountIblis).</p>

<p>The force due to radiation pressure is
$$F_{\text{rad}}=\frac{2I}{c}A$$
where $I$ is the intensity, $c$ is the ubiquitous <a href="https://en.wikipedia.org/wiki/Speed_of_light" rel="nofollow">speed of light</a>, and $A$ is the area onto which the force is applied. However,
$$I=\frac{\text{power}}{4 \pi r^2}$$
so we get
$$F_{\text{rad}}=\frac{\text{power}\times A}{2 \pi r^2c}$$</p>

<p>The $4 \pi$ has to be replaced (by CountIblis' substitution) with $$\Omega=\pi \left(\frac{\alpha}{2}\right)^2$$
For an ideal laser, $$\alpha=2.44\frac{\lambda}{d}$$
where $\lambda$ is the wavelength and $d$ is the initial beam diameter. Let's assume that our laser has a power output of one megawatt - the <a href="https://en.wikipedia.org/wiki/Chemical_oxygen_iodine_laser" rel="nofollow">COIL</a> laser, from a <a href="https://en.wikipedia.org/wiki/Boeing_YAL-1" rel="nofollow">Boeing YAL-1</a>, as per <a href="https://what-if.xkcd.com/13/" rel="nofollow">Randall Munroe's idea</a>. $1,000,000 \text{ watts}$ is a <em>lot</em> from a laser. Let's also assume that no power is lost.</p>

<p>For COIL, $\lambda=1.315 \times 10^{-6}$ and <a href="http://www.globalsecurity.org/military/systems/aircraft/systems/atl.htm" rel="nofollow">$d=0.1016 \text{ meters}$</a>. This gives an $\alpha$ of $3.158070866 \times 10^{-5}$, and thus an $\Omega$ of $7.83309915 \times 10^{-10}$.</p>

<p>Philae is currently on comet <a href="https://en.wikipedia.org/wiki/67P/Churyumov%E2%80%93Gerasimenko" rel="nofollow">67P/Churyumov-Gerasimenko</a>, which has, on its longest side, a surface area (combined lobes) of about $19,370,000 \text{ meters}$. However, it's about <a href="http://edition.cnn.com/2014/11/12/world/comet-landing-countdown/index.html" rel="nofollow">$310,000,000 \text{ miles}$</a> from Earth, which is $4.9879 \times 10^{11} \text{ meters}$. Knowing that, to rotate the comet, we have to apply force on only half the area of one side, we find that this laser would apply a force of
$$F_{\text{rad}}=\frac{1,000,000 \text{ watts}\times 9,685,000\text{ meters}^2}{7.83309915 \times 10^{-10} \times 2.487914641 \times 10^{23} \text{ meters} \times 299,792,458 \text{ meters/second}}$$
$$=1.65771483 \times 10^{-10} \text{ Newtons}$$</p>

<p>Yep, we're not rotating a comet any time soon. I'll try to add in the bit about recharging Philae by laser sometime later. I do think that will be a lot more plausible.</p>

<hr>

<p>$^1$ Unfortunately, I can't find any calculations on <a href="https://en.wikipedia.org/wiki/Laser_propulsion" rel="nofollow">laser propulsion</a>.</p>

CC BY-SA 3.0
2014-12-05T00:11:02.193Z
The laser will not radiate uniformly in all directions, you should replace the solid angle of 4 pi in the numerator of the formula for the intensity by some Omega which should be calculated from the beam divergence angle alpha as approximately Omega = pi (alpha/2)^2, and in case of an ideal laser whose beam divergence is diffraction limited, we have alpha = 2.44 lambda/d where lambda is the wavelength and d the beam diameter when it exists the laser.
Count Iblis
CC BY-SA 3.0
2014-12-05T00:22:15.197Z
@CountIblis Thank you; I had tried to find an equation for divergence but hadn't been able to.
HDE 226868
CC BY-SA 3.0
2014-12-05T00:31:43.440Z
@CountIblis Thanks again. I made the edit.
HDE 226868
CC BY-SA 3.0
2014-12-04T23:09:20.030Z
2014-12-05T00:31:32.580Z
2014-12-05T00:31:32.580Z
HDE 226868
<p>Could the Philae comet lander be recharged with a laser (or possibly many lasers) from earth?  My thought would be to aim one or more powerful lasers from earth at the Philae lander to recharge it through its solar panels.  </p>

<p>Another thought would be to heat up a small part of the comet with one or more lasers, possibly causing the comet to slowly rotate into a position where the sun light falls onto the lander.  </p>

<p>Would either of these ideas be feasible?</p>

CC BY-SA 3.0
2014-12-07T19:36:32.027Z
Somewhat related and fun to read http://what-if.xkcd.com/13/
toniedzwiedz
<earth><comets><space-probe>
Could the Philae comet lander be recharged by laser?
Jonathan
# Could the Philae comet lander be recharged by laser?

(Asked by: username_0 on 2014-12-04)

Could the Philae comet lander be recharged with a laser (or possibly many lasers) from earth? My thought would be to aim one or more powerful lasers from earth at the Philae lander to recharge it through its solar panels.

Another thought would be to heat up a small part of the comet with one or more lasers, possibly causing the comet to slowly rotate into a position where the sun light falls onto the lander.

Would either of these ideas be feasible?

**Tags:** earth, comets, space-probe
> **Comments:**
> **username_1**: Somewhat related and fun to read http://what-if.xkcd.com/13/  

---

## 2 answers

### Answer by username_2 on 2014-12-05. Score: 4 | Accepted answer

> Could the Philae comet lander be recharged with a laser (or possibly many lasers) from earth?

No.

You could make a laser beam with a pretty low divergence - meaning a beam that remains tight and parallel over a very long distance. However, "a very long distance" is nothing on the cosmic scale.

The Lunar Laser Ranging experiment is using high-power lasers focused by telescopes, shooting a beam into the Moon. The light is reflected back to Earth by mirrors placed there by the Apollo program.

<http://en.wikipedia.org/wiki/Lunar_Laser_Ranging_experiment>

The problem is, even with all the tight focusing at the origin, of the initial 10^17 photons in a pulse, only 1 makes it back to the detector. It's very hard to tell that something makes it back at all, let alone charging any solar panels with it.

The comet and Philae are much, much further away than the Moon. You couldn't possibly project any useful energy density at that distance. Not with the current technology.

> **Comments:**
> **username_3**: Perfect. This is just what my answer was missing.  
> **username_4**: From the Earth to the lunar mirror and back you are basically squaring the factor by which the flux is reduced compared to just from the Earth to the lunar mirror. The beam diverges from here to the Moon and the mirrors capture some fraction of the flux (mirror area divided by beam area at the Moon) and then on the way back, the reflected beam diverges again  and back on Earth the telescope receives a small fraction of what the mirror reflected (telescope aperture divided by beam diameter).  
> **username_2**: @CountIblis - right. Perhaps the example was not the most apt. In any case, there's no way we could expect to recharge a probe deep into the Solar System when we can't even "put a laser dot" on the dark side of the Moon, that would be visible from Earth. We can't create that much energy density at that distance. And that's just the Moon, basically in our backyard.  

---

### Answer by username_3 on 2014-12-04. Score: 3

I only have time for a partial answer, so I'll discuss the bit about rotating the comet.

We have to do some calculations with [radiation pressure](https://en.wikipedia.org/wiki/Radiation_pressure), the force on an object from light.$^1$ [This pdf](http://www.webassign.net/question_assets/buelemphys1/chapter22/section22dash3.pdf) has some good starting calculations, although they assume that the source is spherical - the Sun. Our laser clearly is not. However, the beam *will* diverge over such a long distance, so I'll have to use a substitution (given by CountIblis).

The force due to radiation pressure is
$$F\_{\text{rad}}=\frac{2I}{c}A$$
where $I$ is the intensity, $c$ is the ubiquitous [speed of light](https://en.wikipedia.org/wiki/Speed_of_light), and $A$ is the area onto which the force is applied. However,
$$I=\frac{\text{power}}{4 \pi r^2}$$
so we get
$$F\_{\text{rad}}=\frac{\text{power}\times A}{2 \pi r^2c}$$

The $4 \pi$ has to be replaced (by CountIblis' substitution) with $$\Omega=\pi \left(\frac{\alpha}{2}\right)^2$$
For an ideal laser, $$\alpha=2.44\frac{\lambda}{d}$$
where $\lambda$ is the wavelength and $d$ is the initial beam diameter. Let's assume that our laser has a power output of one megawatt - the [COIL](https://en.wikipedia.org/wiki/Chemical_oxygen_iodine_laser) laser, from a [Boeing YAL-1](https://en.wikipedia.org/wiki/Boeing_YAL-1), as per [Randall Munroe's idea](https://what-if.xkcd.com/13/). $1,000,000 \text{ watts}$ is a *lot* from a laser. Let's also assume that no power is lost.

For COIL, $\lambda=1.315 \times 10^{-6}$ and [$d=0.1016 \text{ meters}$](http://www.globalsecurity.org/military/systems/aircraft/systems/atl.htm). This gives an $\alpha$ of $3.158070866 \times 10^{-5}$, and thus an $\Omega$ of $7.83309915 \times 10^{-10}$.

Philae is currently on comet [67P/Churyumov-Gerasimenko](https://en.wikipedia.org/wiki/67P/Churyumov%E2%80%93Gerasimenko), which has, on its longest side, a surface area (combined lobes) of about $19,370,000 \text{ meters}$. However, it's about [$310,000,000 \text{ miles}$](http://edition.cnn.com/2014/11/12/world/comet-landing-countdown/index.html) from Earth, which is $4.9879 \times 10^{11} \text{ meters}$. Knowing that, to rotate the comet, we have to apply force on only half the area of one side, we find that this laser would apply a force of
$$F\_{\text{rad}}=\frac{1,000,000 \text{ watts}\times 9,685,000\text{ meters}^2}{7.83309915 \times 10^{-10} \times 2.487914641 \times 10^{23} \text{ meters} \times 299,792,458 \text{ meters/second}}$$
$$=1.65771483 \times 10^{-10} \text{ Newtons}$$

Yep, we're not rotating a comet any time soon. I'll try to add in the bit about recharging Philae by laser sometime later. I do think that will be a lot more plausible.

---

$^1$ Unfortunately, I can't find any calculations on [laser propulsion](https://en.wikipedia.org/wiki/Laser_propulsion).

> **Comments:**
> **username_4**: The laser will not radiate uniformly in all directions, you should replace the solid angle of 4 pi in the numerator of the formula for the intensity by some Omega which should be calculated from the beam divergence angle alpha as approximately Omega = pi (alpha/2)^2, and in case of an ideal laser whose beam divergence is diffraction limited, we have alpha = 2.44 lambda/d where lambda is the wavelength and d the beam diameter when it exists the laser.  
> **username_3**: @CountIblis Thank you; I had tried to find an equation for divergence but hadn't been able to.  
> **username_3**: @CountIblis Thanks again. I made the edit.  

---

