escape velocity

[Primarygun]

Is it possible for an object to achieve the required escape velocity around a planet but not leaving the planet?

I still have some problems concerning the radial acceleration .

[swansont]

You could have the escape velocity such that you impacted with the planet, and you could have energy losses. Other than that, though, I think the object keeps receding from the planet.

[Locrian]

Is it possible for an object to achieve the required escape velocity around a planet but not leaving the planet?

 

Sure. A spacecraft could achieve escape velocity and then turn around and fly back.

 

Somehow that doesn't seem to be the answer you are looking for. Maybe you could elaborate on the question?

[insane_alien]

as long as ther was an additional force(a rocket engine or a magnetic field for example) that was acting towards the center of gravity then, yes yes it could.

[5614]

If you have a velocity greater than the escape velocity then you can "leave the planet" in your words.

 

However you don't technically have to.

 

Say you need speed x to escape from the Earths gravitational field:

 

1) I could travel at 5x towards the center of the Earth. 5x > x however I'm travelling towards the center, so I'm not going to escape.

 

2) I could orbit at velocity 5x, again 5x > x but if I have a rocket exerting a force on me towards the center of the Earth then it would be, in effect, the same as making it a stronger gravitational field. That is the rocket would supply more force towards the center of the planet and if it were great enough I would not escape the planet.

 

Now you could also get technical. Escape velocity. Velocity is a vector and so direction must be defined. If I was going at 5x but towards the center of the Earth I don't really have an escape velocity greater than x. I've got a speed greater than x, but not an escape velocity, because velocity, by definition, includes the direction (away from the planet) and my direction is towards the planet.

 

I just thought of that last para as I was replying, it seems a logical and correct technicality though.

[Martin]

I still have some problems concerning the radial acceleration .

 

Hi Primarygun' date='

swantsont and others have answered your question many times. I am still wondering, and some other people also may be wondering, what was your original confusion in the first place?

 

How does radial acceleration enter in and cause you problems?

 

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[b']this may be too simple for you, but I will say it anyway, in case there is some basic confusion[/b]

remember that as long as you don't hit the planet or its atmosphere, your escape velocity vector CAN BE POINTED IN ANY DIRECTION.

 

you don't have to analyse the velocity vector into its radial component and its horizontal component----just make sure that the speed is enough: so that you have the kinetic energy needed to escape.

 

Your escape velocity vector does not have to be pointed straight up, it can even be HORIZONTAL at the start, and you will still leave the planet.

 

It can even be slightly BELOW horizontal when you start, so that you begin by getting closer to planet before you swing out and leave. Only just make sure that it is not so much below horizontal that you crash into the planet.

 

basically all that matters is the comparison between your potential energy, which depends on initial distance from planet center----and your kinetic energy, which depends on speed.

[Primarygun]

thank you every body first.

I still have some problems concerning the radial acceleration .

If the radial acceleration is very large, does it affect the magnitude of the velocity of the object ?

Is it possible for an object to achieve the required escape velocity around a planet but not leaving the planet?

Actually, I had a similar question in a textbook.

Given the distance between the moon and the earth, the masses of them respectively, find the gravitational potential energy PE if the moon orbits the earth truly in a circular motion.

Then, find the kinetic energy of the moon.

so, do you expect PE > KE od KE > PE?

The answer said: PE > KE for magnitude since it is in a bound system.

I wonder if this is true every time.

That's a little bit different from the answer

2) I could orbit at velocity 5x, again 5x > x but if I have a rocket exerting a force on me towards the center of the Earth then it would be, in effect, the same as making it a stronger gravitational field. That is the rocket would supply more force towards the center of the planet and if it were great enough I would not escape the planet

Is an external force resulting the difference?

[5614]

remember that as long as you don't hit the planet or its atmosphere, your escape velocity vector CAN BE POINTED IN ANY DIRECTION.

I don't know if this was targeted at me, but I do know that. It was the "as long as you don't hit the planet" part which made my comment about vector vs. speed etc. but maybe I wasn't clear.

 

The closest you could get is a line where you are travelling at a tangent to a part of the planet. If you were travelling at a tangent and went even closer to the planet then you would collide with it.

 

That is; if you are starting at A, then decreasing the angle (marked as 'angle'!) means that you would collide with the planet. The angle shown there is the smallest angle, beyond which you would not escape the planet, as you would collide with it.

[J.C.MacSwell]

I don't know if this was targeted at me' date=' but I do know that. It was the [i']"as long as you don't hit the planet"[/i] part which made my comment about vector vs. speed etc. but maybe I wasn't clear.

 

The closest you could get is a line where you are travelling at a tangent to a part of the planet. If you were travelling at a tangent and went even closer to the planet then you would collide with it.

 

That is; if you are starting at A, then decreasing the angle (marked as 'angle'!) means that you would collide with the planet. The angle shown there is the smallest angle, beyond which you would not escape the planet, as you would collide with it.

 

You seem to be neglecting gravity when you choose your smallest angle

[Martin]

I don't know if this was targeted at me,...

hi Jonathan (5614). No, in fact, what I said was really addressed to Primarygun the OP. My post wasnt intended to question anything you said.

 

BTW MacSwell makes the point that it is not obvious how to give a formula for the allowed directions that escape velocity can be in. It would involve the radius R of the spherical planet, the mass M of the planet, and the satellite's initial distance D from the center of the planet.

I expect there is a simple formula, which someone here could write down for us, but offhand I do not know it.

Basically one would be looking at a equation for a parabola where the closest approach to the focus point is R

so the satellite just "grazes" the surface.

[5614]

You seem to be neglecting gravity when you choose your smallest angle

That is a good point!

 

Taking that into account the "line" would be curved, but otherwise the principle (that you touch the planet at a point) is still the same.

[Primarygun]

Last time, I stated that I was confused with the acceleration in a regular movement in reply 7.

Addition of two vectors in which the cos (theta) is not negative, then the resultant vector has a greater magnitude than each one, right?

The radial acceleration does not affect the magnitude of the velocity, why?

My suggestion is that it is because the time taken into account is very short, right?

[J.C.MacSwell]

Last time' date=' I stated that I was confused with the acceleration in a regular movement in reply 7.

Addition of two vectors in which the cos (theta) is not negative, then the resultant vector has a greater magnitude than each one, right?

The radial acceleration does not affect the magnitude of the velocity, why?

My suggestion is that it is because the time taken into account is very short, right?[/quote']

 

Acceleration at right angles to the velocity does not increase or decrease the speed, it just changes the direction of the velocity. If on changing the direction of the velocity the acceleration is no longer perpendicular to the velocity it will then increase or decrease the speed. So of short or long duration if it stays perpendicular it will not affect the speed (or magnitude of the velocity as you said)

[Primarygun]

Acceleration at right angles to the velocity

Why?

Isn't the resultant velocity is the vector sum of

v (the original velocity) and a x t ?

[J.C.MacSwell]

Why?

Isn't the resultant velocity is the vector sum of

v (the original velocity) and a x t ?

 

Yes but the addition from the a x t approaches (read is) zero at the margin and will only start to factor in as starts to accumulate non perpendicularly to the instantaneous velocity. Sorry if that is unclear, I'm not wording it very well. For , say, circular motion at constant speed there is never any acceleration other than perpendicular to the velocity so the acceleration doesn't change the speed, only the direction.

[Primarygun]

A restate of the true statement is very important.

Thank you very much!