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Why does an orbit become hyperbolic when total orbital energy is positive?
Missing something basic about simple orbital mechanicsHow badly could someone be injured by concentrated sunlight?What could cause an asymmetric orbit in a symmetric potential?How does The work done in going around any path in a gravitational field is zero implies conservation of energy?Homework question: calculating radius and mass of planet from velocity and timeMathematical relationship between escape speed and orbital speedEnergy to lift and then MOVE an objectWork done to change circular orbit and orbital speed
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$begingroup$
I stumbled across this page describing the energy of a given object in orbit.
It describes 'total energy' as:
$$E_k + E_p = E_mathrmtotal $$
where
$$E_k = frac12mv^2$$
and
$$E_p = frac-GMmr$$
Towards the bottom of the page it states that if total orbital energy exceeds zero, the orbit becomes hyperbolic. Is there any empirical reasoning as to why this is true?
energy orbital-motion
edited 6 hours ago
SuperCiocia
7,0807 gold badges41 silver badges97 bronze badges
asked 8 hours ago
Alex ApplebyAlex Appleby
61 bronze badge
New contributor
Alex Appleby is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment
|
$begingroup$
I stumbled across this page describing the energy of a given object in orbit.
It describes 'total energy' as:
$$E_k + E_p = E_mathrmtotal $$
where
$$E_k = frac12mv^2$$
and
$$E_p = frac-GMmr$$
Towards the bottom of the page it states that if total orbital energy exceeds zero, the orbit becomes hyperbolic. Is there any empirical reasoning as to why this is true?
energy orbital-motion
edited 6 hours ago
SuperCiocia
7,0807 gold badges41 silver badges97 bronze badges
asked 8 hours ago
Alex ApplebyAlex Appleby
61 bronze badge
New contributor
Alex Appleby is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
Why are you looking for empirical reasoning rather than the mathematical derivation of the trajectory from the equations of motion?
$endgroup$
– G. Smith
8 hours ago
1
$begingroup$
Do you understand that if the energy is positive the trajectory must go out to infinity rather than being a bound orbit? (But this argument doesn’t tell you that the trajectory is a hyperbola rather than some other curve that goes to infinity.)
$endgroup$
– G. Smith
8 hours ago
add a comment
|
$begingroup$
I stumbled across this page describing the energy of a given object in orbit.
It describes 'total energy' as:
$$E_k + E_p = E_mathrmtotal $$
where
$$E_k = frac12mv^2$$
and
$$E_p = frac-GMmr$$
Towards the bottom of the page it states that if total orbital energy exceeds zero, the orbit becomes hyperbolic. Is there any empirical reasoning as to why this is true?
energy orbital-motion
edited 6 hours ago
SuperCiocia
7,0807 gold badges41 silver badges97 bronze badges
asked 8 hours ago
Alex ApplebyAlex Appleby
61 bronze badge
New contributor
Alex Appleby is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
I stumbled across this page describing the energy of a given object in orbit.
It describes 'total energy' as:
$$E_k + E_p = E_mathrmtotal $$
where
$$E_k = frac12mv^2$$
and
$$E_p = frac-GMmr$$
Towards the bottom of the page it states that if total orbital energy exceeds zero, the orbit becomes hyperbolic. Is there any empirical reasoning as to why this is true?
energy orbital-motion
energy orbital-motion
edited 6 hours ago
SuperCiocia
7,0807 gold badges41 silver badges97 bronze badges
asked 8 hours ago
Alex ApplebyAlex Appleby
61 bronze badge
New contributor
Alex Appleby is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 6 hours ago
SuperCiocia
7,0807 gold badges41 silver badges97 bronze badges
asked 8 hours ago
Alex ApplebyAlex Appleby
61 bronze badge
New contributor
Alex Appleby is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 6 hours ago
SuperCiocia
7,0807 gold badges41 silver badges97 bronze badges
edited 6 hours ago
SuperCiocia
7,0807 gold badges41 silver badges97 bronze badges
edited 6 hours ago
SuperCiocia
7,0807 gold badges41 silver badges97 bronze badges
7,0807 gold badges41 silver badges97 bronze badges
asked 8 hours ago
Alex ApplebyAlex Appleby
61 bronze badge
New contributor
Alex Appleby is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 8 hours ago
Alex ApplebyAlex Appleby
61 bronze badge
asked 8 hours ago
Alex ApplebyAlex Appleby
61 bronze badge
61 bronze badge
New contributor
Alex Appleby is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Alex Appleby is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
Why are you looking for empirical reasoning rather than the mathematical derivation of the trajectory from the equations of motion?
$endgroup$
– G. Smith
8 hours ago
1
$begingroup$
Do you understand that if the energy is positive the trajectory must go out to infinity rather than being a bound orbit? (But this argument doesn’t tell you that the trajectory is a hyperbola rather than some other curve that goes to infinity.)
$endgroup$
– G. Smith
8 hours ago
add a comment
|
$begingroup$
Why are you looking for empirical reasoning rather than the mathematical derivation of the trajectory from the equations of motion?
$endgroup$
– G. Smith
8 hours ago
1
$begingroup$
Do you understand that if the energy is positive the trajectory must go out to infinity rather than being a bound orbit? (But this argument doesn’t tell you that the trajectory is a hyperbola rather than some other curve that goes to infinity.)
$endgroup$
– G. Smith
8 hours ago
$begingroup$
Why are you looking for empirical reasoning rather than the mathematical derivation of the trajectory from the equations of motion?
$endgroup$
– G. Smith
8 hours ago
$begingroup$
Why are you looking for empirical reasoning rather than the mathematical derivation of the trajectory from the equations of motion?
$endgroup$
– G. Smith
8 hours ago
1
1
$begingroup$
Do you understand that if the energy is positive the trajectory must go out to infinity rather than being a bound orbit? (But this argument doesn’t tell you that the trajectory is a hyperbola rather than some other curve that goes to infinity.)
$endgroup$
– G. Smith
8 hours ago
$begingroup$
Do you understand that if the energy is positive the trajectory must go out to infinity rather than being a bound orbit? (But this argument doesn’t tell you that the trajectory is a hyperbola rather than some other curve that goes to infinity.)
$endgroup$
– G. Smith
8 hours ago
add a comment
|
2 Answers
2
active
oldest
votes
$begingroup$
Intuitively, you expect that if your total energy $E$ is positive, then you have more kinetic energy than potential energy. Hence, the potential is not strong enough to ever bind the object into a closed orbit. Instead, the object will feel the effect of the pulling gravitational body, merely changing its trajectory but still remaining open, i.e. going to infinity.
Why a hyperbola, specifically?
I am afraid the particular shape of the function is a result of the algebra.
This is treated here for instance.
The orbit equation (in polar coordinates) is:
$$ r = fracr_01-epsiloncostheta ,, $$
which you can re-write in Cartesian coordinates as:
$$ x^2(1-epsilon^2)-2epsilon x r_0+y^2 = r_0^2. $$
The $epsilon$ parameter is the key of the solution, and it known as eccentricity, related to the energy $E$ as:
$$ E = fracG m_1 m_2 (epsilon^2-1)2r_0.$$
If the energy is positive, $E>0$, then the eccentricity $epsilon >1$ which results in an orbit equation of the form:
$$ y^2 - ax^2 - bx = k,$$
with $a, b$ and $k$ all positive. This is the equation for a hyperbola.
answered 7 hours ago
SuperCiociaSuperCiocia
7,0807 gold badges41 silver badges97 bronze badges
$endgroup$
add a comment
|
$begingroup$
Is there any empirical reasoning as to why this is true?
I'm not sure what you have in mind, but I'll give it a try. See from the equations that:
$$E_kge 0$$
and
$$E_p lt 0$$
So, if the total energy is negative, the kinetic energy can be (momentarily) zero, i.e., the particle can be momentarily stationary in the gravitational field for finite radial coordinate $r$.
However, if the total energy is positive, the particle cannot have zero speed, i.e., the particle can attain an arbitrarily large radial coordinate $r$ with non-zero speed away from the central force source.
I believe it is intuitive that, as the particle gets arbitrarily far away, the particle becomes arbitrarily close to being free of the field, i.e., the particle's trajectory asymptotically approaches the trajectory of a free particle with kinetic energy $E_total$.
Now, consider the hyperbola and its asymptotes.
answered 1 hour ago
Alfred CentauriAlfred Centauri
51.4k3 gold badges54 silver badges166 bronze badges
$endgroup$
add a comment
|
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Intuitively, you expect that if your total energy $E$ is positive, then you have more kinetic energy than potential energy. Hence, the potential is not strong enough to ever bind the object into a closed orbit. Instead, the object will feel the effect of the pulling gravitational body, merely changing its trajectory but still remaining open, i.e. going to infinity.
Why a hyperbola, specifically?
I am afraid the particular shape of the function is a result of the algebra.
This is treated here for instance.
The orbit equation (in polar coordinates) is:
$$ r = fracr_01-epsiloncostheta ,, $$
which you can re-write in Cartesian coordinates as:
$$ x^2(1-epsilon^2)-2epsilon x r_0+y^2 = r_0^2. $$
The $epsilon$ parameter is the key of the solution, and it known as eccentricity, related to the energy $E$ as:
$$ E = fracG m_1 m_2 (epsilon^2-1)2r_0.$$
If the energy is positive, $E>0$, then the eccentricity $epsilon >1$ which results in an orbit equation of the form:
$$ y^2 - ax^2 - bx = k,$$
with $a, b$ and $k$ all positive. This is the equation for a hyperbola.
answered 7 hours ago
SuperCiociaSuperCiocia
7,0807 gold badges41 silver badges97 bronze badges
$endgroup$
add a comment
|
$begingroup$
Intuitively, you expect that if your total energy $E$ is positive, then you have more kinetic energy than potential energy. Hence, the potential is not strong enough to ever bind the object into a closed orbit. Instead, the object will feel the effect of the pulling gravitational body, merely changing its trajectory but still remaining open, i.e. going to infinity.
Why a hyperbola, specifically?
I am afraid the particular shape of the function is a result of the algebra.
This is treated here for instance.
The orbit equation (in polar coordinates) is:
$$ r = fracr_01-epsiloncostheta ,, $$
which you can re-write in Cartesian coordinates as:
$$ x^2(1-epsilon^2)-2epsilon x r_0+y^2 = r_0^2. $$
The $epsilon$ parameter is the key of the solution, and it known as eccentricity, related to the energy $E$ as:
$$ E = fracG m_1 m_2 (epsilon^2-1)2r_0.$$
If the energy is positive, $E>0$, then the eccentricity $epsilon >1$ which results in an orbit equation of the form:
$$ y^2 - ax^2 - bx = k,$$
with $a, b$ and $k$ all positive. This is the equation for a hyperbola.
answered 7 hours ago
SuperCiociaSuperCiocia
7,0807 gold badges41 silver badges97 bronze badges
$endgroup$
add a comment
|
$begingroup$
Intuitively, you expect that if your total energy $E$ is positive, then you have more kinetic energy than potential energy. Hence, the potential is not strong enough to ever bind the object into a closed orbit. Instead, the object will feel the effect of the pulling gravitational body, merely changing its trajectory but still remaining open, i.e. going to infinity.
Why a hyperbola, specifically?
I am afraid the particular shape of the function is a result of the algebra.
This is treated here for instance.
The orbit equation (in polar coordinates) is:
$$ r = fracr_01-epsiloncostheta ,, $$
which you can re-write in Cartesian coordinates as:
$$ x^2(1-epsilon^2)-2epsilon x r_0+y^2 = r_0^2. $$
The $epsilon$ parameter is the key of the solution, and it known as eccentricity, related to the energy $E$ as:
$$ E = fracG m_1 m_2 (epsilon^2-1)2r_0.$$
If the energy is positive, $E>0$, then the eccentricity $epsilon >1$ which results in an orbit equation of the form:
$$ y^2 - ax^2 - bx = k,$$
with $a, b$ and $k$ all positive. This is the equation for a hyperbola.
answered 7 hours ago
SuperCiociaSuperCiocia
7,0807 gold badges41 silver badges97 bronze badges
$endgroup$
Intuitively, you expect that if your total energy $E$ is positive, then you have more kinetic energy than potential energy. Hence, the potential is not strong enough to ever bind the object into a closed orbit. Instead, the object will feel the effect of the pulling gravitational body, merely changing its trajectory but still remaining open, i.e. going to infinity.
Why a hyperbola, specifically?
I am afraid the particular shape of the function is a result of the algebra.
This is treated here for instance.
The orbit equation (in polar coordinates) is:
$$ r = fracr_01-epsiloncostheta ,, $$
which you can re-write in Cartesian coordinates as:
$$ x^2(1-epsilon^2)-2epsilon x r_0+y^2 = r_0^2. $$
The $epsilon$ parameter is the key of the solution, and it known as eccentricity, related to the energy $E$ as:
$$ E = fracG m_1 m_2 (epsilon^2-1)2r_0.$$
If the energy is positive, $E>0$, then the eccentricity $epsilon >1$ which results in an orbit equation of the form:
$$ y^2 - ax^2 - bx = k,$$
with $a, b$ and $k$ all positive. This is the equation for a hyperbola.
answered 7 hours ago
SuperCiociaSuperCiocia
7,0807 gold badges41 silver badges97 bronze badges
answered 7 hours ago
SuperCiociaSuperCiocia
7,0807 gold badges41 silver badges97 bronze badges
answered 7 hours ago
SuperCiociaSuperCiocia
7,0807 gold badges41 silver badges97 bronze badges
answered 7 hours ago
SuperCiociaSuperCiocia
7,0807 gold badges41 silver badges97 bronze badges
7,0807 gold badges41 silver badges97 bronze badges
add a comment
|
add a comment
|
$begingroup$
Is there any empirical reasoning as to why this is true?
I'm not sure what you have in mind, but I'll give it a try. See from the equations that:
$$E_kge 0$$
and
$$E_p lt 0$$
So, if the total energy is negative, the kinetic energy can be (momentarily) zero, i.e., the particle can be momentarily stationary in the gravitational field for finite radial coordinate $r$.
However, if the total energy is positive, the particle cannot have zero speed, i.e., the particle can attain an arbitrarily large radial coordinate $r$ with non-zero speed away from the central force source.
I believe it is intuitive that, as the particle gets arbitrarily far away, the particle becomes arbitrarily close to being free of the field, i.e., the particle's trajectory asymptotically approaches the trajectory of a free particle with kinetic energy $E_total$.
Now, consider the hyperbola and its asymptotes.
answered 1 hour ago
Alfred CentauriAlfred Centauri
51.4k3 gold badges54 silver badges166 bronze badges
$endgroup$
add a comment
|
$begingroup$
Is there any empirical reasoning as to why this is true?
I'm not sure what you have in mind, but I'll give it a try. See from the equations that:
$$E_kge 0$$
and
$$E_p lt 0$$
So, if the total energy is negative, the kinetic energy can be (momentarily) zero, i.e., the particle can be momentarily stationary in the gravitational field for finite radial coordinate $r$.
However, if the total energy is positive, the particle cannot have zero speed, i.e., the particle can attain an arbitrarily large radial coordinate $r$ with non-zero speed away from the central force source.
I believe it is intuitive that, as the particle gets arbitrarily far away, the particle becomes arbitrarily close to being free of the field, i.e., the particle's trajectory asymptotically approaches the trajectory of a free particle with kinetic energy $E_total$.
Now, consider the hyperbola and its asymptotes.
answered 1 hour ago
Alfred CentauriAlfred Centauri
51.4k3 gold badges54 silver badges166 bronze badges
$endgroup$
add a comment
|
$begingroup$
Is there any empirical reasoning as to why this is true?
I'm not sure what you have in mind, but I'll give it a try. See from the equations that:
$$E_kge 0$$
and
$$E_p lt 0$$
So, if the total energy is negative, the kinetic energy can be (momentarily) zero, i.e., the particle can be momentarily stationary in the gravitational field for finite radial coordinate $r$.
However, if the total energy is positive, the particle cannot have zero speed, i.e., the particle can attain an arbitrarily large radial coordinate $r$ with non-zero speed away from the central force source.
I believe it is intuitive that, as the particle gets arbitrarily far away, the particle becomes arbitrarily close to being free of the field, i.e., the particle's trajectory asymptotically approaches the trajectory of a free particle with kinetic energy $E_total$.
Now, consider the hyperbola and its asymptotes.
answered 1 hour ago
Alfred CentauriAlfred Centauri
51.4k3 gold badges54 silver badges166 bronze badges
$endgroup$
Is there any empirical reasoning as to why this is true?
I'm not sure what you have in mind, but I'll give it a try. See from the equations that:
$$E_kge 0$$
and
$$E_p lt 0$$
So, if the total energy is negative, the kinetic energy can be (momentarily) zero, i.e., the particle can be momentarily stationary in the gravitational field for finite radial coordinate $r$.
However, if the total energy is positive, the particle cannot have zero speed, i.e., the particle can attain an arbitrarily large radial coordinate $r$ with non-zero speed away from the central force source.
I believe it is intuitive that, as the particle gets arbitrarily far away, the particle becomes arbitrarily close to being free of the field, i.e., the particle's trajectory asymptotically approaches the trajectory of a free particle with kinetic energy $E_total$.
Now, consider the hyperbola and its asymptotes.
answered 1 hour ago
Alfred CentauriAlfred Centauri
51.4k3 gold badges54 silver badges166 bronze badges
answered 1 hour ago
Alfred CentauriAlfred Centauri
51.4k3 gold badges54 silver badges166 bronze badges
answered 1 hour ago
Alfred CentauriAlfred Centauri
51.4k3 gold badges54 silver badges166 bronze badges
answered 1 hour ago
Alfred CentauriAlfred Centauri
51.4k3 gold badges54 silver badges166 bronze badges
51.4k3 gold badges54 silver badges166 bronze badges
add a comment
|
add a comment
|
Alex Appleby is a new contributor. Be nice, and check out our Code of Conduct.
Alex Appleby is a new contributor. Be nice, and check out our Code of Conduct.
Alex Appleby is a new contributor. Be nice, and check out our Code of Conduct.
Alex Appleby is a new contributor. Be nice, and check out our Code of Conduct.
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$begingroup$
Why are you looking for empirical reasoning rather than the mathematical derivation of the trajectory from the equations of motion?
$endgroup$
– G. Smith
8 hours ago
1
$begingroup$
Do you understand that if the energy is positive the trajectory must go out to infinity rather than being a bound orbit? (But this argument doesn’t tell you that the trajectory is a hyperbola rather than some other curve that goes to infinity.)
$endgroup$
– G. Smith
8 hours ago