Force of photons from the Sun hitting a football field = weight of 1 dime? I read, I think, some time ago that the "weight" of photons from the Sun hitting an area the size of a football field at noon on a sunny day would be about the "weight" of a dime?
Would appreciate it someone could flesh that out, verify if correct or false?
 A: Photons are massless so their weight is 0. However, photons do have momentum so they can exert force. This force is due to their momentum and would occur even in the absence of gravity, so it is not a weight.
The solar irradiance during peak hours is approximately $1000 \mathrm{ \ W \ m^{-2}}$ and the size of a football field is about $7200 \mathrm{ \ m^2}$ for a total radiant power of $7.2 \mathrm{ \ MW}$. Since $p=E/c$ and $F=\frac{dp}{dt}$ we get that the force from this energy is $(7.2 \mathrm{\ MW})/c = 0.024 \mathrm{\ N}$.
In comparison, a dime has a mass of $2.268 \mathrm{\ g}$ which on the earth turns into a gravitational force, or weight, of $0.022 \mathrm{\ N}$.
So the force of the sunlight on a football field during peak solar hours is close to the weight of a dime.
A: "Weight" can be understood as a type of force - standing on the floor, you impart a force on the floor.
Light can impart force on a surface due to the transfer of momentum involved. In other words, if a photon with momentum $p$ strikes a surface and is reflected in the opposite direction, a total momentum of $2p$ is imparted on the surface. This is called radiation pressure. In Newtonian mechanics, you need mass to have momentum, but in relativistic mechanics, you only need energy to have momentum. And photons certainly carry energy.
So how do we compare the radiation pressure on a football field to the weight of a dime? Consider the units: Pressure is force per area, as measured in newton per square meter, N/m$^2$. You would typically think about weight as measured in kilograms, but that is actually mass. The weight is the force on the object due to gravity, so if the dime has mass $m$ and we have gravitational acceleration $g$, the weight is $F = m g$.
Force is also defined as change in momentum. So if we say $N$ photons with momentum $p$ are being reflected off a football field with area $A$ per time $t$, in total their momentum is changed by $2 N p / t$. The pressure on the football field is $2 N p/(t A)$. If we imagine a dime spread over the football field, the pressure from this dime would be $m g/A$. So, by saying that the weight of a football field of photons is the same as that of a dime, we are saying
$$\frac{2 N p}{t} = m g.$$
As for whether it is true or false, that is a simple question of estimating the parameters in this equation, which I leave as an exercise for the reader.
A: Adding to Dale's answer, the dimensions of a dime are $17.91$ mm diameter and $1.35$ mm thick. So the volume is $341.1$ mm$^3$.
If you spread that out over a football field with an area of $7200 \cdot 10^6$ mm$^2$, the thickness is $4.72 \cdot 10^{-8}$ mm. Or about $1/2$ an atom thick.
