Does the photons of light have an effect the movement of solar planets In space, solar sails are used in order to move some space exploration crafts.  Since in space there is vacuum the slightest force could be enough to propel and move the craft.  Although the earth is much larger and much heavier, does these photons have any effect on Earth and other planets  ?  It has a very large surface area and it has been hit by light for billions of years so it could be that these photons affected its movement?  
 A: At Earth's distance from the Sun (which by far dominates the flux on Earth), the incoming flux density is on average $J = 1362\,\mathrm{W}\,\mathrm{m}^{-2}$ (the "Solar constant").
Earth has a radius of $6\,371\,\mathrm{km}$, and hence a cross-sectional area of $A = 1.3\times10^{14}\,\mathrm{m}^2$.
If the photons are all absorbed$^\dagger$, they exert a pressure $P = J/c = 4.5\times10^{-6}\,\mathrm{Pa}$, where $c$ is the speed of light. The force on Earth is then $F = PA = 5.8\times10^8\,\mathrm{N}$. Since the mass of the Earth is $M = 5.97\times10^{24}\,\mathrm{kg}$, it will be experience an acceleration of $a = F/M \simeq 10^{-16}\,\mathrm{m}\,\mathrm{s}^{-2}$.
The mass of the Sun is $M_\odot = 2\times10^{30}\,\mathrm{kg}$, and it's $d = 1.5\times10^8\,\mathrm{km}$ away, so it causes Earth to accelerate at $G M_\odot / d^2 = 5.9\times10^{-3}\,\mathrm{m}\,\mathrm{s}^{-2}$, roughly 14 orders of magnitude higher. The effect is a correspondingly slightly larger radius of the orbit.
Other planets
Since the transferred momentum scales with the area of the planet and hence its radius squared, but the acceleration is inversely proportional to the mass and hence radius cubed, smaller planets are affected more that larger ones, the acceleration being proportional the radius.
On the other hand, the distance of the planet doesn't influence the effect relative to the acceleration from the Sun's mass, since both the flux and the Sun's gravitational force decrease with distance squared.

$^\dagger$Roughly 30% of the photons are reflected, thereby transferring more momentum to Earth (since they are not only "stopped" by Earth, but re-emitted back into the direction it came from, and hence — by conservation of momentum — must deliver momentum to Earth). This increases the final result by roughly 10–20%.
A: We can do a quick "back of the envelop calculation" to gauge the magnitude of this effect. For this I will make many simplifications, but it should give a good enough estimate.
Let's suppose that the Earth is a perfect mirror, (in reality it reflects roughly 30% of received light, so it would be better to assume it absorbs everything, but let's take the upper bound here). Let us also suppose that the Earth is a disc, so that every photon of the sun hits her perpendicularly.
According to Wikipedia, the flux emitted by the sun and received by Earth is $1362$ $W/m^2$, let's round that down to $1000 W/m^2$. Taking $6000$ $km$ for Earth's radius, the surface of our disc is roughly $A = \pi (6\cdot10^{6})^2m^2\approx 10^{14}m^2$. So the power radiated on Earth is $\approx 10^{17} W$.
How do we translate this information into a force exerted by the photons ? Each photon will have energy $E = h\nu$, where $\nu$ is it's frequency. If we suppose all photons have the same frequency, then there are $\frac{10^{11}}{h\nu}$ photons$/s$ hitting the Earth. Their momentum is simply $p = \frac{E}{c}$. 
When they are reflected on the surface of the Earth, their momentum is flipped sign, so they undergo a change in momentum of $\delta p = 2\frac{E}{c}$. We know by Newton's law that they must have been acted on by a force $F$ satisfying $F=\frac{\delta p}{\delta t} = \frac{2E}{c\delta t}$. But by Newton's third law, that means they acted on Earth by a force of the same magnitude ! 
We now have all the pieces. In a time $\delta t$, $N = \frac{10^{11}}{h\nu}\delta t$ photons will hit the Earth, each exerting a force $F = \frac{2E}{c\delta t}$ on it. The total force applied to the earth during that time is thus $F\times N = 10^{17}\frac{2E\not{\delta t}}{h\nu\not{\delta t}}=10^{17}\frac{2(\not{h\nu})}{c\not{h\nu}}=\frac{2}{c}10^{17} \approx 10^9 N$
As we can see, $\nu$ drops out conveniently so we don't have to assume anything about the frequency of the photon emitted by the sun. So there we have it, the force of the solar radiation is $10^9$ $N$, which is by no means a "weak" force, at least from a human's point of view. But let's compare it to something more comparable, as for example the force exerted by the Moon on the earth.
$ F_{moon} = G \frac{M_{moon}M_{Earth}}{r_{moon-earth}^2} \approx 10^{20} N$
So the force exerted by the Sun's photon are $0.000000001$% of the force exerted by the moon. As you can see, the effect is ridiculously small, and that is compared to the force exerted by the moon, which is in turn ridiculously small compared to the force exerted by the Sun on Earth.
In other words, I don't think we would be able to estimate the effects on Earth's orbit anytime soon !
(although it would be interesting to estimate what kind of difference in the orbit radius we are talking about, I will probably do the computation if I get some time !)
A: *

*the answer is yes, photons can excerpt pressure on what they interact with. 

*there is elastic and inelastic scattering, when photons do not get absorbed, but instead they give some or non of their energy to the atom they interact with.

*in this case, the photons do excerpt pressure on earth, at least some of the photons. Since they only give part of their energy to the system they interact with, this effect is very little.

*yes, if we had sensitive enough tools, we could show that the Earth's orbit is changed due to the photons pressure
A: The pressure is 4.5 micro Pascal at 1 astronomical unit away from the sun. For Earth this gives a force of pi*36e12*4.5e-6=4.5e8 N that is an extremely small acceleration of ~10-16 m/s2. For a hydrogen atom however it gives a force of order pi*0.25e-20*4.5e-6/1.7e-27=20 m/s2 but only a small fraction of the photons contribute so you end up with something like 1 cm/s2 acceleration. This is comparable to the solar acceleration of $v^2/r= 1e9/1.5e11 \sim 1cm/s2$. I would expect that hydrogen will is blown out of the solar system unless captured by a planet.
