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AbhiFromXtraZ
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Can you please explain why the paths of objects are hyperbolic, parabolic and elliptic for energies positive, zero and negative respectively?
What will be the trajectory of space objects?
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Discussion Overview
The discussion revolves around the trajectories of space objects and the relationship between their paths (hyperbolic, parabolic, and elliptical) and their total mechanical energy (positive, zero, and negative). Participants explore the implications of energy levels in gravitational fields and the conditions under which different types of orbits occur.
Discussion Character
- Technical explanation
- Conceptual clarification
- Debate/contested
- Mathematical reasoning
Main Points Raised
- Some participants explain that total mechanical energy, defined as kinetic energy plus gravitational potential energy, determines the shape of the trajectory: positive energy leads to hyperbolic paths, zero energy to parabolic paths, and negative energy to elliptical paths.
- There is a discussion about the definition of potential energy being relative to an arbitrary zero-level, often chosen as "at infinity."
- One participant questions why a parabolic path is associated with zero total energy, suggesting that it could also be hyperbolic since both are unbounded paths that approach infinity.
- Another participant emphasizes that negative total energy indicates an elliptical trajectory but does not specify the eccentricity of the ellipse, which can vary significantly.
- Concerns are raised about the conditions under which an object with positive total energy would interact with a gravitational center, such as Earth, and where the turning point of the trajectory would be.
- Clarifications are made regarding the terms used in the energy equation, particularly the role of centrifugal potential energy and its effects on motion.
Areas of Agreement / Disagreement
Participants express differing views on the nature of trajectories associated with zero total energy and the implications of positive total energy. The discussion remains unresolved regarding the specifics of how these energy levels influence the trajectories and the conditions under which they apply.
Contextual Notes
Participants note that the total energy in an attractive field can be complex and that the definitions and assumptions regarding potential energy can affect the interpretation of trajectories. There is also mention of the need to consider eccentricity in elliptical orbits, which adds to the complexity of the discussion.
- #2
tiny-tim
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Hi AbhiFromXtraZ! Welcome to PF! 
By "energy", you mean total mechanical energy, = kinetic energy plus gravitational potential energy.
Potential energy has to be measured relative to an arbitrary zero-level.
In this case, we choose "at infinity" to be the zero-level of potential energy.
Consider a parabolic orbit.
As it "approaches infinity", its speed becomes smaller and smaller, and tends to zero … ie, its speed (and KE) is zero "at infinity".
So, anywhere along its trajectory, its KE is minus its PE, ie its total energy is 0 (and its speed is always equal to the local escape velocity).
If its speed "at infinity" is positive, then it's hyperbolic.
And if its speed "at infinity" is negative, then obviously it can't reach infinity! So it's ellipitc.
AbhiFromXtraZ said:
By "energy", you mean total mechanical energy, = kinetic energy plus gravitational potential energy.
Potential energy has to be measured relative to an arbitrary zero-level.
In this case, we choose "at infinity" to be the zero-level of potential energy.
Consider a parabolic orbit.
As it "approaches infinity", its speed becomes smaller and smaller, and tends to zero … ie, its speed (and KE) is zero "at infinity".
So, anywhere along its trajectory, its KE is minus its PE, ie its total energy is 0 (and its speed is always equal to the local escape velocity).
If its speed "at infinity" is positive, then it's hyperbolic.
And if its speed "at infinity" is negative, then obviously it can't reach infinity! So it's ellipitc.
- #3
AbhiFromXtraZ
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Thank you a lot for responding to my thread.
According to your answer, if the total energy is zero, then the path will be parabolic...but why? It could be hyperbolic...as it also an unbounded path and ends at infinity just like a parabola.
The total energy in an attractive field is given by,
E = 1/2mv^2 + 1/2Iw^2 - k/r
now if the total energy is positive such that 1/2mv^2 term is greater than the sum of other two terms (negative)...Then??...if the centre of force is earth, it will attack earth...then where will be the turning point??
tiny-tim said:By "energy", you mean total mechanical energy, = kinetic energy plus gravitational potential energy.
Potential energy has to be measured relative to an arbitrary zero-level.
In this case, we choose "at infinity" to be the zero-level of potential energy.
Consider a parabolic orbit.
As it "approaches infinity", its speed becomes smaller and smaller, and tends to zero ¡ ie, its speed (and KE) is zero "at infinity".
So, anywhere along its trajectory, its KE is minus its PE, ie its total energy is 0 (and its speed is always equal to the local escape velocity).
According to your answer, if the total energy is zero, then the path will be parabolic...but why? It could be hyperbolic...as it also an unbounded path and ends at infinity just like a parabola.
The total energy in an attractive field is given by,
E = 1/2mv^2 + 1/2Iw^2 - k/r
now if the total energy is positive such that 1/2mv^2 term is greater than the sum of other two terms (negative)...Then??...if the centre of force is earth, it will attack earth...then where will be the turning point??
- #4
tiny-tim
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Hi AbhiFromXtraZ!
We can ignore the 1/2Iω2, it makes no difference since it is constant over the whole trajectory
negative energy is obviously an ellipse
positive energy is obviously a hyperbola
zero energy is the limiting case, and therefore has to be the limit between an ellipse and a hyperbola, which is a parabola
negative total energy tells you that the trajectory is an ellipse, but it does not tell you the eccentricity of the ellipse
if two ellipses have the same energy but different eccentricities, one may have a "turning point" outside the earth, and the other inside the earth: so the first is an orbit while the second is a crash
AbhiFromXtraZ said:
We can ignore the 1/2Iω2, it makes no difference since it is constant over the whole trajectory
negative energy is obviously an ellipse
positive energy is obviously a hyperbola
zero energy is the limiting case, and therefore has to be the limit between an ellipse and a hyperbola, which is a parabola
negative total energy tells you that the trajectory is an ellipse, but it does not tell you the eccentricity of the ellipse
if two ellipses have the same energy but different eccentricities, one may have a "turning point" outside the earth, and the other inside the earth: so the first is an orbit while the second is a crash
- #5
AbhiFromXtraZ
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tiny-tim said:
Oh...you fell in a misunderstanding...sorry...it was my fault..
Actually I asked for positive mechanical energy...and ''(negative)'' means the sum of 1/2Iw2 and k/r is negative...and the magnitude is less than 1/2mv2 such that E becomes positive...
And 1/2Iw2 is actually 1/2m(wr)^2...my book says this the centrifugal potential energy (sorry for my fault)...I think this term prevents the object from moving along straight line...
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