Is a photon emitted from Earth less energetic that a photon emitted from the Moon Two similar looking photons (locally) are emitted, one from Earth and the other from the Moon, and they are observed at some point X out in space.
On Earth, time is slightly more dilated due to gravity than it is on the Moon, because gravity is stronger. So will the energy measured at point X be similarly reduced?
 A: Yes, the gravitational red shift will be greater for a photon emitted from the Earth than for a similar photon emitted from the Moon.
The energy is given by $E=hf$, and the frequency $f$ is decreased due to the time dilation i.e. the frequency observed far from the Earth or Moon is given by:
$$ f = f_0 \sqrt{1 - \frac{2GM}{c^2 R}} \tag{1} $$
where $M$ is the mass of the Earth/Moon and $R$ is the radius of the surface$^1$. When the time dilation is very small, as it is for both the Earth and the Moon, we can approximate equation (1) using a binomial expansion. In this case we get the expression:
$$ f \approx f_0 \left(1 - \frac{GM}{c^2 R}\right) \tag{2} $$
Note that the Newtonian gravitational potential energy is given by:
$$ \Phi = -\frac{GM}{R} $$
So the frequency and therefore the energy of the photon is simply related to the gravitational potential energy at the surface:
$$ f \approx f_0 \left(1 + \frac{\Phi}{c^2}\right) \tag{3} $$

$^1$ Strictly speaking the $R$ in equation (1) is not the radius of the Earth but rather the circumference of the Earth divided by $2\pi$. In flat spacetime these are of course the same thing, but in a curved spacetime they are different. However for weak gravitational fields like that of the Earth the difference is negligible.
