Release speed

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dede2002
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Re: Release speed




by dede2002 » 21/05/21, 12:19

ABC2019 wrote:the direction of the speed does not matter (good on condition of not going to crash on the ground anyway).

I made the calculation of the mechanical energy, the direction of the speed does not intervene in it.


Ok in relation to the earth, but the earth revolves around the sun with a circumferential speed of 100 km / h, as soon as you leave the earth, you have a speed of 000 + - 100 km / h . So the direction of speed is important, so as not to crash into the sun ... :)
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Re: Release speed




by dede2002 » 21/05/21, 12:54

There is something I would like to know: can, following a possible collision between satellites, pieces slowed down by the shock fall to earth slowly enough not to be completely consumed?
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Re: Release speed




by Obamot » 21/05/21, 16:04

Yes, but not a big risk. Ladle:
There would be a very low probability of 1 in 195 million that space debris would fall near an inhabited area on Earth.
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Re: Release speed




by Exnihiloest » 21/05/21, 18:17

dede2002 wrote:Thank you I got it, or almost ... :)

The direction of the speed may not matter. but it is easier to reach a horizontal speed than uphill.


When it is to turn in orbit, it is above all that it is necessary to arrive with a tangential speed to the planned orbit.
From an energy point of view, it theoretically doesn't make a difference.
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dede2002
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Re: Release speed




by dede2002 » 22/05/21, 12:22

Ok, theoretically. but in practice it takes less power (and more time) to reach speed on a gentle slope than vertically, so a less heavy engine. Less mass therefore more height with the same energy. :)
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Re: Release speed




by dede2002 » 27/05/21, 11:18

ABC2019 wrote:
dede2002 wrote:Thank you I got it, or almost ... :)

The direction of the speed may not matter. but it is easier to reach a horizontal speed than uphill.

with a very short pulse time, no, it's the same. A rifle bullet comes out at the same speed no matter which direction you shoot. You have in the head the speed of a car, but precisely we are in the case of a gradual push, and the difference comes from the fact that you have to overcome the potential energy during the acceleration time when you go up, while you don't have to do it horizontally. with the same power expended at the start, the speed achieved includes the fact that you expended part of the energy to go up in one case, and not in the other.

But without friction, for example on the Moon, a bullet which would reach the speed of release will escape the attraction of the Moon whether it is fired vertically or horizontally ...
.


This centrifugal force is spinning in my head ...

If it is calculated from the center of the planet, horizontally it starts from the axis, while vertically it starts at a distance from the center equal to the radius, it could leave less quickly to be released?

Trying to do calculations (imagining starting from the center at the release speed with a deceleration of 1g, I come to a stop at the surface *) I find that the release speed is equal to the acceleration (g ) over the distance of the radius.
Same result on earth or on the moon ...

* I know this is wrong because in the center we must be in a situation of weightlessness, but this consistency intrigues me ...?

ps: the bullet, upwards it has its acceleration due to the thrust minus g, downwards more g, little difference indeed ... :)
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Re: Release speed




by ABC2019 » 27/05/21, 11:54

dede2002 wrote:
ABC2019 wrote:
dede2002 wrote:Thank you I got it, or almost ... :)

The direction of the speed may not matter. but it is easier to reach a horizontal speed than uphill.

with a very short pulse time, no, it's the same. A rifle bullet comes out at the same speed no matter which direction you shoot. You have in the head the speed of a car, but precisely we are in the case of a gradual push, and the difference comes from the fact that you have to overcome the potential energy during the acceleration time when you go up, while you don't have to do it horizontally. with the same power expended at the start, the speed achieved includes the fact that you expended part of the energy to go up in one case, and not in the other.

But without friction, for example on the Moon, a bullet which would reach the speed of release will escape the attraction of the Moon whether it is fired vertically or horizontally ...
.


This centrifugal force is spinning in my head ...

If it is calculated from the center of the planet, horizontally it starts from the axis, while vertically it starts at a distance from the center equal to the radius, it could leave less quickly to be released?

Trying to do calculations (imagining starting from the center at the release speed with a deceleration of 1g, I come to a stop at the surface *) I find that the release speed is equal to the acceleration (g ) over the distance of the radius.
Same result on earth or on the moon ...

* I know this is wrong because in the center we must be in a situation of weightlessness, but this consistency intrigues me ...?

ps: the bullet, upwards it has its acceleration due to the thrust minus g, downwards more g, little difference indeed ... :)


centrifugal force is not a "real" force, it depends on the frame of reference. It only exists when you locate the movement in relation to a frame of reference whose axes are rotating.

If you look at the movement of a satellite around the Earth in a frame of reference whose axes are fixed (pointing towards the stars), there is no centrifugal force. There is a centripetal force (gravitation) and a rotational movement around the Earth due to this force.

If you look at it in a frame of reference whose axes rotate with the satellite (which is the case when you place a camera in the space station, because the "frame" is the station itself and the axes rotate with it), then it there is a centrifugal force which balances gravity .. and everything seems still!

both points of view are valid, but they should not be mixed. This is why in Mechanics the first thing to do is to define the frame of reference against which you locate the movement.

For example in a frame of reference linked to the Earth, the Moon revolves around the Earth. But in a frame of reference linked to the Sun, well it turns .. around the Sun! : Wink:
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Re: Release speed




by Exnihiloest » 27/05/21, 21:39

ABC2019 wrote:...
both points of view are valid, but they should not be mixed. This is why in Mechanics the first thing to do is to define the frame of reference against which you locate the movement.
...


To clarify what you said and what I completely agree with, the easiest way is almost always to use an inertial frame of reference, therefore without acceleration or rotation, like the center of the earth to follow the movements around.
Centrifugal force is indeed a pseudo force which only exists in non-inertial frames of reference. On the other hand we can speak of centrifugal acceleration, it is also seen from an inertial frame of reference since we know the speed v (vector) so a = dv / dt also vector, constant in modulus but not in direction therefore acceleration, radial outwards when spinning in orbit. We then see the two vectors g and a canceling each other out, I find it more elegant than talking about a pseudo-force which would cancel out the weight. We do not need the forces for analysis.
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Re: Release speed




by dede2002 » 28/05/21, 07:16

Totally agree.

Moreover the formula of the centrifugal force f = m * v2 / r can be transformed into g = v2 / r because f = m * g, and f and m disappear.
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dede2002
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Re: Release speed




by dede2002 » 28/05/21, 14:03

In fact the formula is very simple:
To calculate the takeoff speed it is v = root of r * g
And to calculate the release rate it is v = root of 2 * r * g

Between these two speeds, we are in orbit around the frame of reference, above we move away and we can get back into orbit by slowing down, below we crash.

But I don't know how to explain the "2" ...
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