4 ms·
This is very confused. To leave the Earth’s gravitational grip you need to impart a certain minimum change in velocity. You can do this all at once or or we tim
by garmaine 9y ago
This is very confused. To leave the Earth’s gravitational grip you need to impart a certain minimum change in velocity. You can do this all at once or or we time but it doesn’t matter. The delta-V is the same.
- nilkn 9y agoThis is simply false. The needed escape velocity decreases with distance from the planet, so applying a change over a long period of time actually can very materially change the requirements. To give an example, if you provided propulsion to move an object at 1 mile per hour, you could turn off all propulsion and the object would have escaped Earth's gravity once it was 4 * 10^12 km away. This is, of course, an extraordinary waste and probably practically impossible.
- larkeith 9y agoYour "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 9y agoEscape 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 9y agoThat’s not how physics works. I suggest reading up on conservation of energy and gravitational potential energy.
- nilkn 9y agoWith 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 9y agoMultiple 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 9y agoYou will have no trouble then pointing out the precise mistake in the calculation.
- p2t2p 9y agoOk, 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 9y ago[deleted]
- marcosdumay 9y agoYou can reduce delta-v it a little by floating to the edge of the atmosphere. Beyond that, there is no method of getting higher that does not involve gaining some speed.
- muyyatin 9y agoBut, ignoring things like air resistance, the lower the thrust is, the more delta-V/energy it takes to escape? For instance, if you had a rocket with a fixed delta-V and it used all of its energy almost instantaneously, say it escapes from Earth. But if you lower its thrust so that it is not accelerating upward (lower than the force of gravity), it could use the same delta-V without leaving the ground.
- garmaine 9y agoNo, it’s the things you’re “ignoring” which keep it on the ground. A solar electric thruster provides less thrust than the surface weight of the spacecraft it is pushing, but it is nevertheless able to push a probe to escape velocity and beyond over months or years of continuous thrust.