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- Thread starter Cpt. Bob
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quantumdude

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Originally posted by Cpt. Bob

Is this correct, and if so why?

It is correct, and it is due to the conservation of energy. The escape velocity is calculated by setting the total energy equal to zero. That is, if you start at r=∞ from rest and let go, you would return to the Earth at the escape velocity.

Would it be incorrect in the example above, but correct in the real world because of other more powerful gravitational forces, such as the suns, countering acceleration due to earths gravity till the spacecraft is very near earth, or near enough for earths gravity to become dominant? Thx in advance for any clarification.

I think you've got it backwards: It is correct in the (idealized) example above, but incorrect in the real world.

When calculating the escape velocity of the Earth, the influence of other bodies is neglected. In the real world, that effect must be taken into account to be accurate.

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arcnets

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Yes.Originally posted by Cpt. Bob

if the spacecraft was placed stationary any distance from the earth, the maximum velocity it could ever achieve is 11 km/sec. Is this correct

Because in this scenario, the ship can only gain kinetic energy from its potential energy in the earth's gravitational field. And, as the distance earth-ship approaches infinity, the potential energy approaches a certain limit.and if so why?

Yet we know that meteorites can have much higher velocities than 11 km/s. Which means they carry surplus energy that does not stem from earth's gravity alone.

I'd put it the other way round:Would it be incorrect in the example above, but correct in the real world

Yes, a celestial body that is much more massive than earth, but not much farther away (at a certain time), could (have) provide(d) the surplus energy.because of other more powerful gravitational forces, such as the suns

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Great, thx for the help. I hadnt even thought about conservation of energy as it applies here.

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HallsofIvy

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Pretty good answers, guys!

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