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?
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.
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, the force on an object from light.$^1$ This 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, 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 laser, from a Boeing YAL-1, as per Randall Munroe's idea. $1,000,000 \text{ watts}$ is a lot from a laser. Let's also assume that no power is lost.
For COIL, $\lambda=1.315 \times 10^{-6}$ and $d=0.1016 \text{ meters}$. 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, which has, on its longest side, a surface area (combined lobes) of about $19,370,000 \text{ meters}$. However, it's about $310,000,000 \text{ miles}$ 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.
