Escape Velocity of a Body

[danny8522003]

Hey guys,

 

I was doing gravitation today in physics and we were looking at escape velocities. The velocity is given by:

 

[MATH]v = \sqrt\frac{2Gm}{x}[/MATH]

 

Now this equation only has the value of one of the masses in, so the velocity is only dependant upon one mass.

 

My question is why is this independant of the mass of the object trying to escape? If you were trying to jump off the Earth, you would attract the Earth and the Earth would attract you. Although if you had the mass of the moon your attractive force would be higher, so why does this not mean the velocity would need to be greater for the bodies to seperate compared to you having your mass or the mass of the moon?

 

Everything is relative after all.

[ecoli]

This escape velocity equation is derived by setting the KE equation equal to the minus gravitational potential energy equation.

 

 

when you solve for Ve, the mass of the object is cancelled out. Which makes sense. The escape velocity equation is assumed to have no outside forces acting on it, other then gravity.

[danny8522003]

But both the masses have gravitational fields that depend upon their mass.

 

A pencil and a planet would attract each other a lot less than say a star and a planet, so why doesnt the escape velocity represent this?

[swansont]

In your example, larger mass means larger force, but since F = ma, the acceleration remains the same.

 

Similarly, larger mass implies more energy, but since KE = 1/2 mv² the speed term is the same.

[danny8522003]

Ahh i see, makes sense now. Thank you.

[J.C.MacSwell]

Hey guys' date='

 

I was doing gravitation today in physics and we were looking at escape velocities. The velocity is given by:

 

[MATH']v = \sqrt\frac{2Gm}{x}[/MATH]

 

Now this equation only has the value of one of the masses in, so the velocity is only dependant upon one mass.

 

My question is why is this independant of the mass of the object trying to escape? If you were trying to jump off the Earth, you would attract the Earth and the Earth would attract you. Although if you had the mass of the moon your attractive force would be higher, so why does this not mean the velocity would need to be greater for the bodies to seperate compared to you having your mass or the mass of the moon?

 

Everything is relative after all.

 

I think this formula assumes the smaller mass is "negligible" compared to the larger mass, but I don't think it completely "cancels out". If it did the escape velocity of the larger mass from the smaller one would be much smaller than the escape velocity of the smaller from the larger... and obviously they must be equal.

 

So I think your point is valid.

[timo]

I think this formula assumes the smaller mass is "negligible" compared to the larger mass, but I don't think it completely "cancels out".

Thinking is great, knowing is sometimes better: They do cancel out: http://en.wikipedia.org/wiki/Escape_velocity under "Calculating an escape velocity".

 

If it did the escape velocity of the larger mass from the smaller one would be much smaller than the escape velocity of the smaller from the larger... and obviously they must be equal.

That´s neither obvious nor even true. It´s the energy required to seperate them that must be equal for both cases (moving away the heavy mass or moving the light mass). Since the velocity (squared) is the energy divided by the mass, the velocity must be uneven if the masses are.

 

EDIT: But I think I start to imagine where your problem lies: The calculation assumes that the mass not moved stays at the same place. This is obviously a good approximation for a small mass leaving a big mass. But if you strip away a big mass from a small one you´d need some ways of making sure that the small one stays in place and doesn´t follow the big one due to the gravitational attraction.

[J.C.MacSwell]

Thinking is great' date=' knowing is sometimes better: They do cancel out: http://en.wikipedia.org/wiki/Escape_velocity under "Calculating an escape velocity".

 

 

That´s neither obvious nor even true. It´s the energy required to seperate them that must be equal for [b']both cases [/b](moving away the heavy mass or moving the light mass). Since the velocity (squared) is the energy divided by the mass, the velocity must be uneven if the masses are.

 

I will check out the link, but how can it not be obvious? It is the same case calculated differently. If the second calculation yields a different answer then the formula cannot be valid for that calculation.

 

I therefore think that the formula only holds for a limited range of mass ratios as an approximation. I agree knowing would be better.

[timo]

Whether it´s an approx or not might to some point depend on your boundary conditions you enforce to your problem. If you assume that one of the masses stays in place, then the equations given in the wikipedia link simply hold true and you´ll have uneven escape velocities.

If you want the problem to be physically meaningfull (no artificial forces keeping one of the masses in place) and still want one of the masses to remain in the same position, then "yes": One of the masses must be much heavier than the other so that its movement can be neglected.

[swansont]

If you look at the situation in reverse, an object falling from infinite separation, you assume that the second body doesn't acquire a noticable speed. So you are assuming one mass is negligible compared to the other.

[J.C.MacSwell]

I checked the link and it says in the first paragraph that it is assuming the escape of a mass that is negligible compared to that of the source.

 

So is the escape velocity of a source mass independant of the mass of the object attempting to escape, but only negligibly small masses escape? (one and only one escape velocity for each source mass)

 

Or is the definition incomplete and they are they only defining escape velocity for the limited (most common) case?