While studying ##E=mc^2## I keep noticing that kinetic energy is not ##\frac{1}{2}mv^2## as we were always taught in school. The relativistic formula is, in various forms, $$K_{rel}=\frac{c^3}{\sqrt{c^2-v^2}}-c^2=c^2\left(\frac{1}{\sqrt{1-\beta^2}}-1\right)=c^2\left(\frac{1}{2}\beta^2+\frac{3}{8}\beta^4+\ldots\right)$$where ##\beta=v/c##.
On May 2nd 2021 the Parker Solar Probe achieved a velocity of 532,000 km/h which is about ##0.0005c##. That gives
\begin{align}\frac{K_{rel}}{c^2}=1.25000023\times{10}^{-7}\ ,\ \frac{K_{cl}}{c^2}=1.25\times{10}^{-7}&\phantom {10000}(8)\nonumber\end{align}Not much difference. The mass of the probe is about 600kg so the difference is ##2.3\times{10}^3\ Joules## which might keep a phone going for an hour.
The difference is more noticeable at much higher velocities. At ##0.1c## the difference is about 1%. It goes wild as you approach the speed of light.
A bit more here: 2.2 Kinetic energy.pdf (1 page)
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