How fast does an object have to be going to turn into energy

Question

Albert Einstein's equation, $E=mc^2$, says that an object has to be going at the speed of light squared to turn into energy. How fast would this be in miles per hour?

Answer 1

$c$ is not only an invariant speed, $c$ is also a physical constant that factors in many well known formula, e.g., the electromagnetic fine structure constant

$$\alpha = \frac{e^2}{4 \pi \epsilon_0 \hbar c}$$

In the case of the famous

$$E = mc^2$$

the particle with mass $m$ has zero speed (in this frame of reference). If the particle has a speed $v$ in this frame, then the equation for the particle's energy is

$$E = \gamma_v m c^2 = \frac{mc^2}{\sqrt{1 - \frac{v^2}{c^2}}}$$

says that an object has to be going at the speed of light squared to turn into energy.

I'm afraid this is a misunderstanding of the mass-energy relationship which gives the equivalent energy of the (invariant) mass of a particle.

As I showed above, the energy of a massive particle goes to infinity as it approaches the speed of light and, thus, cannot ever travel at the speed of light in any reference frame.

Answer 2

$E=mc^2$ is actually not the full equation, although it is clearly the most famous form of it. The full equation is

$E^2 = p^2c^2 + m^2c^4.$

This formula says that a particle's energy squared is equal to the sum of the momentum squared and the mass squared (with factors of the speed of light thrown in to make it dimensionally correct).

The simple $E=mc^2$ follows for $p=0$, which is to say, for an object at rest. All massive objects are at rest in their own reference frame, and in contrast, light can never be still--it has no rest frame. So the equation $E=mc^2$ represents the latent energy possessed by a massive particle due to it's mass alone. Giving it a non-zero momentum then adds kinetic energy onto this rest mass contribution. This is the energy that atomic processes like fission can tap into, which is why the equation is so famous in this form.