# Mass of the Light

*Most recent answer: 02/28/2015*

- Zoey (age 24)

Canada

First of all, force is not something that an object has, unlike say energy or momentum. The force between two objects is just the rate at which momentum goes from one to the other, keeping the same total momentum. Since light has momentum, it can exert forces on other things. We've discussed that before: ,

Only particles with zero rest mass (like light's photons) can travel at the speed of light. It would take infinite energy to get that car to the speed of light, so it just can't happen.

You ask if launching a car could affect the sort of gravity we feel on Earth. The mass loss from the car leaving would be far too small to notice a reduction of the earth's gravity. Our sense of gravity depends also on a small acceleration due to the Earth's spin. That spin creates an effective centrifugal force, reducing how much things weighh on scales by a very small amount. If the car were launched horizontally, could it change the earth's spin enough to have a noticeable effect on that? Here's a calculation about that.

The gravitational acceleration (g) on the Earth's surface have three sources. First of all, there is the gravitation itself. Counteracting this is a cetrifugal force due to spinning, proportional to the square of Earth's angular velocity (ω^{2}). Third is the Corriolis force on moving objects also proportional to ω. The total angular momentum of the Earth+flying car system should be conserved, so the Earth will react by rotating in the opposite direction to that of the car's movement. This is similar to the reason behind how ice skaters can spin. Special theory of relativity is applicable in the absence of accelerations only, so let us make some crude semi-quantitative arguments. When you accelerate your car to enormous speeds such as 0.6c, it will gain a momentum p= 0.75*mc ~ 3. 10^{11} Nm. Treating the system classically,

L = pR = ω I = 0.4 M R^{2} δω

δω = 2.5 p / MR where M is the mass of the Earth and R is it radius. δω is the induced angular velocity of the Earth. Substituting everything,

δω = 2.5 3 10^{11}/ (6 10^{6}m 6 10^{24} kg) ~ 10^{-20} rad/s, which is indeed very small compared to that due to daily rotation ~10^{-4 }rad/s. The Earth is so huge! Because the change in ω is negligible, so is the change in g.

Tunc + mw

*(published on 02/28/2015)*