There's enough energy in a Tesla battery to for the Tesla to reach escape velocity. If you could simply drive at max acceleration (and the car didn't fall apart, and the tires continued to have grip, and a million other reasons why this is impossible) eventually you'd reach escape velocity and still have some percentage left.
In a more realistic sense, a long railgun type system would be very practical in a no-atmosphere environment, and then not being subject to the tyranny of the rocket equation, you could launch whatever you wanted. Enough fuel to decelerate is no problem.
No, it's not even remotely close. A Model S weighs around 2000kg and has a battery of 100 kWh. That's √((100 kWh)/(1/2*2000kg)) = 600m/s of delta-v. Escape velocity for Earth is 11.2km/s, almost a factor 20 more.
This nerd sniped me a LOT, I’m wondering if it’s possible for a chemical battery to reach orbital velocity (not escape).
An idealized Tesla would just be its battery (500kg) perfectly dumping energy to mechanical forward speed. Cutting 3/4 of the weight gets you closer to the delta-v you need, but youre still off by a factor of 5. Though orbital velocity, and leaving from the equator and gaining that speed, means you only need to get up to ~7.2 km/s. Still only a third of the way there.
Maybe you could split your battery into chunks, and expel them once they’re expended?
Once you hit an orbit intersecting the ground, you have to scrub all your speed in whatever that amount of time is, which is gonna be short. It's basically an orbital launch in reverse.
https://en.wikipedia.org/wiki/Tsiolkovsky_rocket_equation
double the Δv means you square the mass ratio. The space shuttle had a mass ratio of about 16, a mass ratio of 256 would be absolutely insane.
You get this velocity change at the cost of dealing with the heat and all but a tiny fraction of that heat ends up immediately in the atmosphere.