4 ms·

Your "example" seems to predicate on a world without gravity. The reason escape velocity is reduced based on the distance from the surface is that you have alre

Your "example" seems to predicate on a world without gravity. The reason escape velocity is reduced based on the distance from the surface is that you have already provided enough energy to overcome the vertical difference - you can see this reflected in the difference in gravitational potential energy.

nilkn 8y ago
Escape velocity is reduced based on distance because gravity itself is reduced based on distance. This is obvious from the the expression sqrt(2GM/r), which is the escape velocity at a distance r from the center of a spherically symmetric body of mass M. From this expression one can easily calculate the distance I provided. This computation is completely based on Newtonian gravity and would make no sense in a world without gravity.
- garmaine 8y ago
  That’s not how physics works. I suggest reading up on conservation of energy and gravitational potential energy.
  - nilkn 8y ago
    With all due respect, this reply isn’t going to fly. This is a pretty pretentious copout that attempts to end the exchange while saving you from needing to provide any details or engage with the underlying material in any way. I provided a very specific computational counter-example to your statement. You’ll need to start there if you wish to compose a serious reply. I’ll restate the counter-example for your convenience. One can provide just enough continuous thrust to move an object at a fixed unchanging speed of 1mph for a long time (about 283 million years), at which point it will have reached a distance where the escape velocity is only 1mph. Once it has surpassed that distance, one can disable the thrust and the object will have completely escaped Earth’s gravity without ever having moved at more than 1mph. It will then asymptotically slow down to 0 mph over an infinite amount of time.
    - garmaine 8y ago
      Multiple people have pointed out the mistakes you are making, to no avail. I really don’t know what else can be said. It’s like you’re claiming 2+2 is 5. What else can I say? There is ample introductory material on Newtonian physics that you could learn from, only a google search away. It’s disrespectful of my time to assume I must be on the hook to handhold you through it.
      - nilkn 8y ago
        You will have no trouble then pointing out the precise mistake in the calculation.
      - p2t2p 8y ago
        Ok, let me ask a very straight and simple question. Imagine that I have a spacecraft with imaginable engine which is able to indefinitely provide thrust enough for this spacecraft to move straight up from the planet’s surface at the constant speed of 1 mph. Will this spacecraft leave the planet in indefinite amount of time or not? If not - what is going to stop it?
      - deleted 8y ago
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      - deleted 8y ago
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      - zpconn 8y ago
        I found a number of sources online that disagree with you. https://www.quora.com/Escape-velocity-is-supposed-to-be-24-000-mph-but-our-rockets-never-achieve-this-speed-How-does-that-work?share=1 https://www.quora.com/Escape-velocity-is-supposed-to-be-24-0... Multiple answers explain that there isn't a fixed minimum change in velocity required for an object to escape Earth's gravity. https://www.quora.com/Does-rocket-always-need-the-escape-speed-11-2km-sec-to-escape-the-earth?share=1 https://www.quora.com/Does-rocket-always-need-the-escape-spe... The first answer here explains the same thing, using basically the same example (1 m/s instead of 1 mph). https://space.stackexchange.com/questions/4688/couldnt-i-escape-earths-gravity-traveling-only-1-mph-0-45-m-s https://space.stackexchange.com/questions/4688/couldnt-i-esc... The answers here overwhelmingly reiterate and support the same example in this thread.

nilkn 8y ago

Escape velocity is reduced based on distance because gravity itself is reduced based on distance. This is obvious from the the expression sqrt(2GM/r), which is the escape velocity at a distance r from the center of a spherically symmetric body of mass M. From this expression one can easily calculate the distance I provided. This computation is completely based on Newtonian gravity and would make no sense in a world without gravity.

garmaine 8y ago
That’s not how physics works. I suggest reading up on conservation of energy and gravitational potential energy.
- nilkn 8y ago
  With all due respect, this reply isn’t going to fly. This is a pretty pretentious copout that attempts to end the exchange while saving you from needing to provide any details or engage with the underlying material in any way. I provided a very specific computational counter-example to your statement. You’ll need to start there if you wish to compose a serious reply. I’ll restate the counter-example for your convenience. One can provide just enough continuous thrust to move an object at a fixed unchanging speed of 1mph for a long time (about 283 million years), at which point it will have reached a distance where the escape velocity is only 1mph. Once it has surpassed that distance, one can disable the thrust and the object will have completely escaped Earth’s gravity without ever having moved at more than 1mph. It will then asymptotically slow down to 0 mph over an infinite amount of time.
  - garmaine 8y ago
    Multiple people have pointed out the mistakes you are making, to no avail. I really don’t know what else can be said. It’s like you’re claiming 2+2 is 5. What else can I say? There is ample introductory material on Newtonian physics that you could learn from, only a google search away. It’s disrespectful of my time to assume I must be on the hook to handhold you through it.
    - nilkn 8y ago
      You will have no trouble then pointing out the precise mistake in the calculation.
    - p2t2p 8y ago
      Ok, let me ask a very straight and simple question. Imagine that I have a spacecraft with imaginable engine which is able to indefinitely provide thrust enough for this spacecraft to move straight up from the planet’s surface at the constant speed of 1 mph. Will this spacecraft leave the planet in indefinite amount of time or not? If not - what is going to stop it?
    - deleted 8y ago
      [deleted]
    - deleted 8y ago
      [deleted]
    - zpconn 8y ago
      I found a number of sources online that disagree with you. https://www.quora.com/Escape-velocity-is-supposed-to-be-24-000-mph-but-our-rockets-never-achieve-this-speed-How-does-that-work?share=1 https://www.quora.com/Escape-velocity-is-supposed-to-be-24-0... Multiple answers explain that there isn't a fixed minimum change in velocity required for an object to escape Earth's gravity. https://www.quora.com/Does-rocket-always-need-the-escape-speed-11-2km-sec-to-escape-the-earth?share=1 https://www.quora.com/Does-rocket-always-need-the-escape-spe... The first answer here explains the same thing, using basically the same example (1 m/s instead of 1 mph). https://space.stackexchange.com/questions/4688/couldnt-i-escape-earths-gravity-traveling-only-1-mph-0-45-m-s https://space.stackexchange.com/questions/4688/couldnt-i-esc... The answers here overwhelmingly reiterate and support the same example in this thread.