Intuition for why escape velocity doesn't depend on angle

[RubinLicht]

Homework Statement

solve for escape velocity from Earth's surface

Homework Equations

just either use the line integral, and the tangential term disappears, or just use the energy equations

The Attempt at a Solution

I've solved it, but I'm having some trouble just coming to grips with this.

in terms of motion, isn't it possible to have circular motion around the Earth at any radius? so, if you launch at some arbitrary angle from the earth, could there be an angle that you launch at where the mass ends up more in an orbit rather than escaping Earth's gravitational field? (or maybe the total energy associated with circular motion is less than escape velocity, in which case I guess this is not possible, I'll work this out in a bit)

now that I think about it, if you solve the conservation of energy equation, isn't there an implicit assumption that the mass will take the path such that it ends up at an infinite distance from earth? is this a valid assumption? does energy guarantee that a particle will take a path that arrives at "infinite" radius? I thought it only guaranteed that there's enough energy to reach that state, but it might not necessarily ever get there?

please ask for clarification if my reasoning is hazy

 

[haruspex]

isn't it possible to have circular motion around the Earth at any radius?

Yes, but how does the KE depend on that? Write the equations.

 

[RubinLicht]

Yes, but how does the KE depend on that? Write the equations.

KE= (m*Re^2*g) /(2*r)

I'm not sure what this implies, though.

 

[haruspex]

KE= (m*Re^2*g) /(2*r)

I'm not sure what this implies, though.

So as r increases without limit, what happens to the KE?

 

[RubinLicht]

v

So as r increases without limit, what happens to the KE?

vanishes, but isn't that the same end behavior as a mass launched at escape velocity? it vanishes as r - > infinite?

 

[haruspex]

isn't that the same end behavior as a mass launched at escape velocity?

At escape velocity, yes, but what about more?
But sorry, that's not answering your question...

 

[haruspex]

maybe the total energy associated with circular motion is less than escape velocity

Again, write the equation. What is the total energy, and what is the limit? How does that compare with the total energy at escape velocity? What does that imply for any finite orbit?

 

[RubinLicht]

Oh I see, it's double that of a mass in circular orbit. makes sense, thanks.

 

[haruspex]

Oh I see, it's double that of a mass in circular orbit. makes sense, thanks.

Not quite sure what you mean there.
By convention, we take GPE at infinity as zero, making it negative for all lesser distances.
In a circular orbit at infinity, its KE is also zero, so zero total energy.
In a finite circular orbit, the KE has half the magnitude of the GPE, so net negative energy.
Since it is just enough to reach infinity, an object at escape velocity has total zero energy.
An object faster than escape velocity has positive energy, so will not enter a circular orbit even at infinity.

 

[RubinLicht]

Not quite sure what you mean there.
By convention, we take GPE at infinity as zero, making it negative for all lesser distances.
In a circular orbit at infinity, its KE is also zero, so zero total energy.
In a finite circular orbit, the KE has half the magnitude of the GPE, so net negative energy.
Since it is just enough to reach infinity, an object at escape velocity has total zero energy.
An object faster than escape velocity has positive energy, so will not enter a circular orbit even at infinity.

I see. that helps me interpret my math more clearly, thanks.