I Escape velocity and kinetic energy

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1. Mar 4, 2016

nibbel11

is it right to say, "when all the potential energy is converted in kinetic energy the object is moving at the escapevelocity.
and "when the change in potential energy and kinetic energy is constant at the same time it is laying still on the ground or in a perfect circulair orbit.
and the last one "if the planet is rotating faster than the escape velocity you are going to get flung of the planet"
are these statements right?

2. Mar 4, 2016

Staff: Mentor

No. That's impact velocity. It varies with the altitude you drop something from.
That's a grammatically cumbersome sentence. But "change...is constant" implies to me a continuous addition of energy. So no, that wouldn't be an orbit or sitting still on the ground. Maybe a rocket could have a continuously increasing energy as it is launched.

An orbit or sitting on the ground has zero change in total mechanical energy, or rather, that the total energy is constant.

Of course, so does an object plunging to Earth...
That one is true.

3. Mar 4, 2016

Simon Bridge

No. Russ beat me to it ... I'll just add:
Say I drop a stone from 1m ... just before it hits the ground, it is not moving at escape velocity.
Technically it has not converted all it's potential energy either ... that would happen at the center of the Earth perhaps. In that case, it will have exactly enough kinetic energy to rise to 1m hight ... still not escape velocity.

An object at escape velocity has the same kinetic energy as the potential energy difference between where it is and infinity.
An object falling from infinity, with no initial velocity, converting all potential energy to kinetic, the the impact velocity has the same magnitude and opposite direction to the escape velocity at the impact site.

4. Mar 4, 2016

Staff: Mentor

Kinetic and potential energy are reference frame dependent and a spot on the surface of the earth is often chosen as the origin of the reference frame.

5. Mar 4, 2016

Simon Bridge

Even so: the stone still loses potential energy on it's way to the centre - so, on the surface, I can argue it still has potential energy to lose even though the "tank level" reads zero, that's just a number chalked on the side of the tank. I can probably get more wiggle room in there if I really tried but it's 23:16 here and I'm not that committed.

Are escape velocity calculations are usually done by taking a spot on the surface as zero?