Comment on What on Earth can we still not measure speed by or at least have a transfer rate?
FiniteBanjo@feddit.online 3 days ago
I don’t think there is anything we can’t use as a measurement but it would be super inefficient or inaccurate to measure most speeds in particle width per hour.
Are you sure you meant to write “measure speed by”?
Patnou@lemmy.world 3 days ago
Now that you bring it up I guess…Good catch btw. …How about I rephrase. Is there anything on Earth that we know about but still have no measurement of how fast or slow it goes? And if we do would that measurement have to include a transfer of speed, hence applying a rate or acceleration between two points? Sorry still the morning.
FiniteBanjo@feddit.online 3 days ago
No, I don’t think anything like that exists.
The minimum speed is 0 for everything and the maximum ever observed is the speed of light.
With the ecception of light everything has a mass and depending on its mass the amount of energy put in determines its speed.
If you put in enough energy anything can in theory go the speed of light.
The Formula for Kinetic Energy is equal to half of Mass times velocity squared.
E = 1/2 ∗ M ∗ v^2
Where v^2 is velocity which is
(Distance / Time )^2
and that means that v without being squared is speed on its own.
So you can rewrite the kinetic energy formula as
Speed = square root(2E / M)
And this works for literally everything in the universe because everything except light has mass. Light might also have mass but we’ve never observed it, which is probably why it goes the speed of light: any amount of energy is enough to send it the speed of light.
The only way to use less energy to make the same object go faster is to reduce resisting forces: remove friction such as a smooth surface, go far away from any sources of gravity, etc.
Blue_Morpho@lemmy.world 3 days ago
You can approach the speed of light but you cannot reach it. Your equations are missing the relativistic corrections.
FiniteBanjo@feddit.online 3 days ago
Can we agree that 99.9999999% of C is effectively C?