Release speed

[Obamot]

A priori it does not matter the mass of the object to be sent:

The rate of release is a scalar, not a vector quantity: it just specifies an amplitude, not a direction. An object which moves at the speed of release can escape the gravitational field whatever its initial direction (insofar as the trajectory does not meet the surface of the star). It does not depend either on the mass of the object, only on that of the star.

https://fr.wikipedia.org/wiki/Vitesse_d ... 9ristiques

Yes, but that's a bit lame as a remark, you can also theoretically leave the orbit of a planet with an ionic engine!

And why didn't you say then that you wanted to trap us by knowing the answer to the question you are asking (and not to try to understand?) And that you bounce back each time by finding loopholes in the answers that are data and which are not “false”, while preserving suspense.

I was just saying that AS LONG AS YOU DON'T GIVE THE MASS of what to get out of Earth's orbit, it is not possible to calculate this: DO YOU DEMAND that you have to deduct the 9.81 m / s necessary (or whatever the number) to escape the earth's gravity of the total thrust necessary to get a machine of a given mass out of the gravity field? That's what this phrase from wikipedia IMHO means.

If I throw you a sheet of paper folded in the shape of an airplane wing, I make it fly, for a moment my “plane” escapes the earth's attraction. But I don't think I can do the same by throwing a Soyuz ship “by hand”.

[ABC2019]

Impossible according to Wikipedia, The rocket with the probe will have to reach a speed of 40 km / h to leave Earth ...

I would be surprised if wikipedia said that, since it is false, do you have a link?

(...)

For an object launched from the surface of the Earth, the speed of release allowing it to escape the Earth's attraction is 11,2 km / s (or 40 km / h).

https://fr.wikipedia.org/wiki/Vitesse_d ... 3%A9ration

That's what I'm trying to understand ...

this is what I explained to you from the beginning, it only concerns thrown objects without own propulsion. It's not about airplanes or rockets.

[Obamot]

Even better said, thank you!

[gildas]

This also concerns and objects with propulsion ...

[Obamot]

You did not answer my question.

[ABC2019]

This also concerns and objects with propulsion ...
Screenshot_2021-05-10-21-06-43-500_com.google.android.googlequicksearchbox.jpg

mash ... the link on wikipedia is right, but it does say an object launched from the face of the earth , therefore without propulsion. The second is false, a rocket does not need to reach 40 km / h to escape the attraction of the earth.

the fundamental formula is the mechanical energy theorem which says that the change in mechanical energy is equal to the work of non-conservative forces, the mechanical energy being the sum of kinetic energy and potential energy.

Kinetic energy is 1/2 mv ^ 2

The potential energy is - GMm / r, at infinity it is zero.

The mechanical energy is therefore worth 1/2 mv ^ 2- GMm / r and at infinity it is just worth the kinetic energy 1/2 m v_infini ^ 2

The kinetic energy theorem therefore says that 1/2 m v_infinity ^ 2 - 1/2 mv ^ 2 + GMm / r = W (forces)

If there is no thrust, W (forces) = 0 and therefore v ^ 2 = v_infinity ^ 2 + 2GM / r which is only possible if v> root (2GM / r) is the speed release.

On the other hand, if there is a push, you cannot say anything about v, the only condition is that v ^ 2 + 2W / m> 2GM / r but that does not give a condition on v, v can be as small as wants and it is the work W which ensures the extraction of the field of gravity.

[gildas]

This also concerns and objects with propulsion ...
Screenshot_2021-05-10-21-06-43-500_com.google.android.googlequicksearchbox.jpg

mash ... the link on wikipedia is right, but it does say an object launched from the face of the earth , therefore without propulsion. The second is false, a rocket does not need to reach 40 km / h to escape the attraction of the earth.
(...)

To overcome the gravitational force and not fall back to Earth, Saturn V must reach a speed of about 28 km / h. To achieve this, the launcher essentially consists of three powerful reactors as well as a suitable fuel tank. These stages are successively ignited and then dropped once consumed. The actual Apollo spacecraft sits at the front at the tip of the rocket. 000 minutes after takeoff, Apollo 12 reaches orbit and circles the Earth one and a half times. In this “parking orbit”, the crew controls all the systems again, then the control tower gives the instruction: “Go!”. The third-stage reactor then ignites and sends Apollo out of Earth's orbit at a speed of 40 km / h towards the Moon. Now, Apollo 11 flies on a "free return path": even if the reactors failed, the spacecraft, after having made a loop around the Moon, would again be attracted by Earth's gravity and would find its way back to Earth. our planet.

https://m.simplyscience.ch/archives-jeu ... terri.html

[Obamot]

There's no doubt, I'm gonna have to throw my paper plane a lot harder and farther

[GuyGadeboisTheBack]

Why an airplane (while heading immediately vertically) .....

Yes, it is heading vertically (sic) but it is not going in a straight line!

[gildas]

yes, seen from space.

But as said previously if it also embeds its oxidizer and it stop at 40, 50 or 60 km altitude, it must be funny, engine at full power ...