Mixing
Shouldn't ball has to move in the circle even after cutting the rope by which it was attached? [closed]
$begingroup$
Since the tension force is a central acting force, the torque on an orbiting ball about that center is zero. But if the rope is cut down during motion the torque would still remain zero. This would mean that angular momentum of the orbiting ball should be conserved, but I find everywhere that ball will move in a straight line tangentially. What is wrong in my reasoning?
newtonian-mechanics newtonian-gravity rotational-dynamics reference-frames orbital-motion
asked Jan 5 at 12:09
Piyush GalavPiyush Galav
324
$endgroup$
closed as unclear what you're asking by Norbert Schuch, Chair, Ben Crowell, John Rennie, ZeroTheHero Jan 5 at 22:44
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
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show 13 more comments
$begingroup$
Since the tension force is a central acting force, the torque on an orbiting ball about that center is zero. But if the rope is cut down during motion the torque would still remain zero. This would mean that angular momentum of the orbiting ball should be conserved, but I find everywhere that ball will move in a straight line tangentially. What is wrong in my reasoning?
newtonian-mechanics newtonian-gravity rotational-dynamics reference-frames orbital-motion
asked Jan 5 at 12:09
Piyush GalavPiyush Galav
324
$endgroup$
closed as unclear what you're asking by Norbert Schuch, Chair, Ben Crowell, John Rennie, ZeroTheHero Jan 5 at 22:44
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
13
$begingroup$
"what's wrong in my reasoning?" - wouldn't the Sun disappearing violate one or more conservation laws?
$endgroup$
– Hal Hollis
Jan 5 at 12:52
3
$begingroup$
Have you tried calculating the angular momentum of an object moving in a straight line?
$endgroup$
– By Symmetry
Jan 5 at 13:22
4
$begingroup$
@AaronStevens, but my point is that the OP begins by stipulating something that violates one or more conservation laws and then draws a conclusion based on conservation laws holding. That, it seems to me, is a significant flaw in reasoning.
$endgroup$
– Hal Hollis
Jan 5 at 13:34
2
$begingroup$
@Piyush Might I suggest changing your scenario to the one I mentioned earlier about the ball on a rope so this question isn't closed by those who can't understand your intentions of this question?
$endgroup$
– Aaron Stevens
Jan 5 at 13:43
4
$begingroup$
A theory can't answer questions about what happens when the theory is violated. In both Newtonian gravity and general relativity, it's forbidden for the sun to suddenly disappear. Therefore neither theory can answer this question.
$endgroup$
– Ben Crowell
Jan 5 at 15:58
|
show 13 more comments
$begingroup$
Since the tension force is a central acting force, the torque on an orbiting ball about that center is zero. But if the rope is cut down during motion the torque would still remain zero. This would mean that angular momentum of the orbiting ball should be conserved, but I find everywhere that ball will move in a straight line tangentially. What is wrong in my reasoning?
newtonian-mechanics newtonian-gravity rotational-dynamics reference-frames orbital-motion
asked Jan 5 at 12:09
Piyush GalavPiyush Galav
324
$endgroup$
Since the tension force is a central acting force, the torque on an orbiting ball about that center is zero. But if the rope is cut down during motion the torque would still remain zero. This would mean that angular momentum of the orbiting ball should be conserved, but I find everywhere that ball will move in a straight line tangentially. What is wrong in my reasoning?
newtonian-mechanics newtonian-gravity rotational-dynamics reference-frames orbital-motion
newtonian-mechanics newtonian-gravity rotational-dynamics reference-frames orbital-motion
asked Jan 5 at 12:09
Piyush GalavPiyush Galav
324
asked Jan 5 at 12:09
Piyush GalavPiyush Galav
324
edited Jan 6 at 5:38
Piyush Galav
asked Jan 5 at 12:09
Piyush GalavPiyush Galav
324
asked Jan 5 at 12:09
Piyush GalavPiyush Galav
324
asked Jan 5 at 12:09
Piyush GalavPiyush Galav
324
324
closed as unclear what you're asking by Norbert Schuch, Chair, Ben Crowell, John Rennie, ZeroTheHero Jan 5 at 22:44
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
closed as unclear what you're asking by Norbert Schuch, Chair, Ben Crowell, John Rennie, ZeroTheHero Jan 5 at 22:44
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
13
$begingroup$
"what's wrong in my reasoning?" - wouldn't the Sun disappearing violate one or more conservation laws?
$endgroup$
– Hal Hollis
Jan 5 at 12:52
3
$begingroup$
Have you tried calculating the angular momentum of an object moving in a straight line?
$endgroup$
– By Symmetry
Jan 5 at 13:22
4
$begingroup$
@AaronStevens, but my point is that the OP begins by stipulating something that violates one or more conservation laws and then draws a conclusion based on conservation laws holding. That, it seems to me, is a significant flaw in reasoning.
$endgroup$
– Hal Hollis
Jan 5 at 13:34
2
$begingroup$
@Piyush Might I suggest changing your scenario to the one I mentioned earlier about the ball on a rope so this question isn't closed by those who can't understand your intentions of this question?
$endgroup$
– Aaron Stevens
Jan 5 at 13:43
4
$begingroup$
A theory can't answer questions about what happens when the theory is violated. In both Newtonian gravity and general relativity, it's forbidden for the sun to suddenly disappear. Therefore neither theory can answer this question.
$endgroup$
– Ben Crowell
Jan 5 at 15:58
|
show 13 more comments
13
$begingroup$
"what's wrong in my reasoning?" - wouldn't the Sun disappearing violate one or more conservation laws?
$endgroup$
– Hal Hollis
Jan 5 at 12:52
3
$begingroup$
Have you tried calculating the angular momentum of an object moving in a straight line?
$endgroup$
– By Symmetry
Jan 5 at 13:22
4
$begingroup$
@AaronStevens, but my point is that the OP begins by stipulating something that violates one or more conservation laws and then draws a conclusion based on conservation laws holding. That, it seems to me, is a significant flaw in reasoning.
$endgroup$
– Hal Hollis
Jan 5 at 13:34
2
$begingroup$
@Piyush Might I suggest changing your scenario to the one I mentioned earlier about the ball on a rope so this question isn't closed by those who can't understand your intentions of this question?
$endgroup$
– Aaron Stevens
Jan 5 at 13:43
4
$begingroup$
A theory can't answer questions about what happens when the theory is violated. In both Newtonian gravity and general relativity, it's forbidden for the sun to suddenly disappear. Therefore neither theory can answer this question.
$endgroup$
– Ben Crowell
Jan 5 at 15:58
13
13
$begingroup$
"what's wrong in my reasoning?" - wouldn't the Sun disappearing violate one or more conservation laws?
$endgroup$
– Hal Hollis
Jan 5 at 12:52
$begingroup$
"what's wrong in my reasoning?" - wouldn't the Sun disappearing violate one or more conservation laws?
$endgroup$
– Hal Hollis
Jan 5 at 12:52
3
3
$begingroup$
Have you tried calculating the angular momentum of an object moving in a straight line?
$endgroup$
– By Symmetry
Jan 5 at 13:22
$begingroup$
Have you tried calculating the angular momentum of an object moving in a straight line?
$endgroup$
– By Symmetry
Jan 5 at 13:22
4
4
$begingroup$
@AaronStevens, but my point is that the OP begins by stipulating something that violates one or more conservation laws and then draws a conclusion based on conservation laws holding. That, it seems to me, is a significant flaw in reasoning.
$endgroup$
– Hal Hollis
Jan 5 at 13:34
$begingroup$
@AaronStevens, but my point is that the OP begins by stipulating something that violates one or more conservation laws and then draws a conclusion based on conservation laws holding. That, it seems to me, is a significant flaw in reasoning.
$endgroup$
– Hal Hollis
Jan 5 at 13:34
2
2
$begingroup$
@Piyush Might I suggest changing your scenario to the one I mentioned earlier about the ball on a rope so this question isn't closed by those who can't understand your intentions of this question?
$endgroup$
– Aaron Stevens
Jan 5 at 13:43
$begingroup$
@Piyush Might I suggest changing your scenario to the one I mentioned earlier about the ball on a rope so this question isn't closed by those who can't understand your intentions of this question?
$endgroup$
– Aaron Stevens
Jan 5 at 13:43
4
4
$begingroup$
A theory can't answer questions about what happens when the theory is violated. In both Newtonian gravity and general relativity, it's forbidden for the sun to suddenly disappear. Therefore neither theory can answer this question.
$endgroup$
– Ben Crowell
Jan 5 at 15:58
$begingroup$
A theory can't answer questions about what happens when the theory is violated. In both Newtonian gravity and general relativity, it's forbidden for the sun to suddenly disappear. Therefore neither theory can answer this question.
$endgroup$
– Ben Crowell
Jan 5 at 15:58
|
show 13 more comments
2 Answers
2
active
oldest
votes
$begingroup$
The flaw in your reasoning is thinking that straight line motion at constant velocity does not constitute constant angular momentum about some point, but it actually does.
Angular momentum is given by$^*$
$$mathbf L=mathbf rtimesmathbf p$$
Without loss of generality, let's assume after the rope is cut our object is moving along the line $y=1$ in the x-y plane, and we are looking at the angular momentum about the origin. Then our angular momentum must always be perpendicular to the x-y plane, so it will be sufficient to just look at the magnitude of the angular momentum
$$L=rpsintheta$$
where $theta$ is the angle between the position vector and the momentum vector (which is the angle between the position vector and the x-axis based on the set up above).
Now, since there are no forces acting on our object, $p$ is constant. Also, $rsintheta$ is just the constant $y=1$ value given by the line the object is moving along. Therefore, it must be that $L$ is constant.
This shows that absence of a net torque (conserved angular momentum) is not enough to uniquely determine the motion. While in circular motion, there is still a net force acting on our object. Without the rope, there is no net force. The motions are different.
$^*$Note that this applies to any type of motion, not just circular motion.
answered Jan 5 at 13:19
Aaron StevensAaron Stevens
9,71931741
$endgroup$
$begingroup$
So, angular momentum can be applied for all motions and not just circular type?
$endgroup$
– Piyush Galav
Jan 5 at 19:55
$begingroup$
@PiyushGalav Yes. Angular momentum about some point is $mathbf L=mathbf rtimesmathbf p$. It can be applied to any motion.
$endgroup$
– Aaron Stevens
Jan 5 at 20:39
2
$begingroup$
Straight lines are just circles about the point at infinity...
$endgroup$
– Derek Elkins
Jan 5 at 20:49
add a comment |
$begingroup$
The planets will continue moving with the momentum they have at the moment the Sun disappears, apart from mutual attraction. These orbitals conserve planetary angular momentum is conserved. This is clear from the formula $L = vec r times vec p$. $r sin theta$ and $mv$ remain constant after such an event.
The presence of gravity means that the orbit has to be a conic section to conserve angular momentum, the ellipticity depending on a combination of L and energy. In the absence of gravity straight inertial movement does the job.
answered Jan 5 at 13:48
my2ctsmy2cts
4,8792618
$endgroup$
$begingroup$
What do you mean by conical?
$endgroup$
– Karthik V
Jan 5 at 16:32
1
$begingroup$
@KarthikV the orbits must trace a conic section (circle, elipse, or parabola) in order to conserve angular momentum in the presence of a single large point source of gravity.
$endgroup$
– Vaelus
Jan 5 at 16:41
1
$begingroup$
Note: the question has been edited to refer to a less dramatic angular momentum scenario.
$endgroup$
– rob♦
Jan 6 at 5:52
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The flaw in your reasoning is thinking that straight line motion at constant velocity does not constitute constant angular momentum about some point, but it actually does.
Angular momentum is given by$^*$
$$mathbf L=mathbf rtimesmathbf p$$
Without loss of generality, let's assume after the rope is cut our object is moving along the line $y=1$ in the x-y plane, and we are looking at the angular momentum about the origin. Then our angular momentum must always be perpendicular to the x-y plane, so it will be sufficient to just look at the magnitude of the angular momentum
$$L=rpsintheta$$
where $theta$ is the angle between the position vector and the momentum vector (which is the angle between the position vector and the x-axis based on the set up above).
Now, since there are no forces acting on our object, $p$ is constant. Also, $rsintheta$ is just the constant $y=1$ value given by the line the object is moving along. Therefore, it must be that $L$ is constant.
This shows that absence of a net torque (conserved angular momentum) is not enough to uniquely determine the motion. While in circular motion, there is still a net force acting on our object. Without the rope, there is no net force. The motions are different.
$^*$Note that this applies to any type of motion, not just circular motion.
answered Jan 5 at 13:19
Aaron StevensAaron Stevens
9,71931741
$endgroup$
$begingroup$
So, angular momentum can be applied for all motions and not just circular type?
$endgroup$
– Piyush Galav
Jan 5 at 19:55
$begingroup$
@PiyushGalav Yes. Angular momentum about some point is $mathbf L=mathbf rtimesmathbf p$. It can be applied to any motion.
$endgroup$
– Aaron Stevens
Jan 5 at 20:39
2
$begingroup$
Straight lines are just circles about the point at infinity...
$endgroup$
– Derek Elkins
Jan 5 at 20:49
add a comment |
$begingroup$
The flaw in your reasoning is thinking that straight line motion at constant velocity does not constitute constant angular momentum about some point, but it actually does.
Angular momentum is given by$^*$
$$mathbf L=mathbf rtimesmathbf p$$
Without loss of generality, let's assume after the rope is cut our object is moving along the line $y=1$ in the x-y plane, and we are looking at the angular momentum about the origin. Then our angular momentum must always be perpendicular to the x-y plane, so it will be sufficient to just look at the magnitude of the angular momentum
$$L=rpsintheta$$
where $theta$ is the angle between the position vector and the momentum vector (which is the angle between the position vector and the x-axis based on the set up above).
Now, since there are no forces acting on our object, $p$ is constant. Also, $rsintheta$ is just the constant $y=1$ value given by the line the object is moving along. Therefore, it must be that $L$ is constant.
This shows that absence of a net torque (conserved angular momentum) is not enough to uniquely determine the motion. While in circular motion, there is still a net force acting on our object. Without the rope, there is no net force. The motions are different.
$^*$Note that this applies to any type of motion, not just circular motion.
answered Jan 5 at 13:19
Aaron StevensAaron Stevens
9,71931741
$endgroup$
$begingroup$
So, angular momentum can be applied for all motions and not just circular type?
$endgroup$
– Piyush Galav
Jan 5 at 19:55
$begingroup$
@PiyushGalav Yes. Angular momentum about some point is $mathbf L=mathbf rtimesmathbf p$. It can be applied to any motion.
$endgroup$
– Aaron Stevens
Jan 5 at 20:39
2
$begingroup$
Straight lines are just circles about the point at infinity...
$endgroup$
– Derek Elkins
Jan 5 at 20:49
add a comment |
$begingroup$
The flaw in your reasoning is thinking that straight line motion at constant velocity does not constitute constant angular momentum about some point, but it actually does.
Angular momentum is given by$^*$
$$mathbf L=mathbf rtimesmathbf p$$
Without loss of generality, let's assume after the rope is cut our object is moving along the line $y=1$ in the x-y plane, and we are looking at the angular momentum about the origin. Then our angular momentum must always be perpendicular to the x-y plane, so it will be sufficient to just look at the magnitude of the angular momentum
$$L=rpsintheta$$
where $theta$ is the angle between the position vector and the momentum vector (which is the angle between the position vector and the x-axis based on the set up above).
Now, since there are no forces acting on our object, $p$ is constant. Also, $rsintheta$ is just the constant $y=1$ value given by the line the object is moving along. Therefore, it must be that $L$ is constant.
This shows that absence of a net torque (conserved angular momentum) is not enough to uniquely determine the motion. While in circular motion, there is still a net force acting on our object. Without the rope, there is no net force. The motions are different.
$^*$Note that this applies to any type of motion, not just circular motion.
answered Jan 5 at 13:19
Aaron StevensAaron Stevens
9,71931741
$endgroup$
The flaw in your reasoning is thinking that straight line motion at constant velocity does not constitute constant angular momentum about some point, but it actually does.
Angular momentum is given by$^*$
$$mathbf L=mathbf rtimesmathbf p$$
Without loss of generality, let's assume after the rope is cut our object is moving along the line $y=1$ in the x-y plane, and we are looking at the angular momentum about the origin. Then our angular momentum must always be perpendicular to the x-y plane, so it will be sufficient to just look at the magnitude of the angular momentum
$$L=rpsintheta$$
where $theta$ is the angle between the position vector and the momentum vector (which is the angle between the position vector and the x-axis based on the set up above).
Now, since there are no forces acting on our object, $p$ is constant. Also, $rsintheta$ is just the constant $y=1$ value given by the line the object is moving along. Therefore, it must be that $L$ is constant.
This shows that absence of a net torque (conserved angular momentum) is not enough to uniquely determine the motion. While in circular motion, there is still a net force acting on our object. Without the rope, there is no net force. The motions are different.
$^*$Note that this applies to any type of motion, not just circular motion.
answered Jan 5 at 13:19
Aaron StevensAaron Stevens
9,71931741
edited Jan 6 at 21:21
answered Jan 5 at 13:19
Aaron StevensAaron Stevens
9,71931741
answered Jan 5 at 13:19
Aaron StevensAaron Stevens
9,71931741
answered Jan 5 at 13:19
Aaron StevensAaron Stevens
9,71931741
9,71931741
$begingroup$
So, angular momentum can be applied for all motions and not just circular type?
$endgroup$
– Piyush Galav
Jan 5 at 19:55
$begingroup$
@PiyushGalav Yes. Angular momentum about some point is $mathbf L=mathbf rtimesmathbf p$. It can be applied to any motion.
$endgroup$
– Aaron Stevens
Jan 5 at 20:39
2
$begingroup$
Straight lines are just circles about the point at infinity...
$endgroup$
– Derek Elkins
Jan 5 at 20:49
add a comment |
$begingroup$
So, angular momentum can be applied for all motions and not just circular type?
$endgroup$
– Piyush Galav
Jan 5 at 19:55
$begingroup$
@PiyushGalav Yes. Angular momentum about some point is $mathbf L=mathbf rtimesmathbf p$. It can be applied to any motion.
$endgroup$
– Aaron Stevens
Jan 5 at 20:39
2
$begingroup$
Straight lines are just circles about the point at infinity...
$endgroup$
– Derek Elkins
Jan 5 at 20:49
$begingroup$
So, angular momentum can be applied for all motions and not just circular type?
$endgroup$
– Piyush Galav
Jan 5 at 19:55
$begingroup$
So, angular momentum can be applied for all motions and not just circular type?
$endgroup$
– Piyush Galav
Jan 5 at 19:55
$begingroup$
@PiyushGalav Yes. Angular momentum about some point is $mathbf L=mathbf rtimesmathbf p$. It can be applied to any motion.
$endgroup$
– Aaron Stevens
Jan 5 at 20:39
$begingroup$
@PiyushGalav Yes. Angular momentum about some point is $mathbf L=mathbf rtimesmathbf p$. It can be applied to any motion.
$endgroup$
– Aaron Stevens
Jan 5 at 20:39
2
2
$begingroup$
Straight lines are just circles about the point at infinity...
$endgroup$
– Derek Elkins
Jan 5 at 20:49
$begingroup$
Straight lines are just circles about the point at infinity...
$endgroup$
– Derek Elkins
Jan 5 at 20:49
add a comment |
$begingroup$
The planets will continue moving with the momentum they have at the moment the Sun disappears, apart from mutual attraction. These orbitals conserve planetary angular momentum is conserved. This is clear from the formula $L = vec r times vec p$. $r sin theta$ and $mv$ remain constant after such an event.
The presence of gravity means that the orbit has to be a conic section to conserve angular momentum, the ellipticity depending on a combination of L and energy. In the absence of gravity straight inertial movement does the job.
answered Jan 5 at 13:48
my2ctsmy2cts
4,8792618
$endgroup$
$begingroup$
What do you mean by conical?
$endgroup$
– Karthik V
Jan 5 at 16:32
1
$begingroup$
@KarthikV the orbits must trace a conic section (circle, elipse, or parabola) in order to conserve angular momentum in the presence of a single large point source of gravity.
$endgroup$
– Vaelus
Jan 5 at 16:41
1
$begingroup$
Note: the question has been edited to refer to a less dramatic angular momentum scenario.
$endgroup$
– rob♦
Jan 6 at 5:52
add a comment |
$begingroup$
The planets will continue moving with the momentum they have at the moment the Sun disappears, apart from mutual attraction. These orbitals conserve planetary angular momentum is conserved. This is clear from the formula $L = vec r times vec p$. $r sin theta$ and $mv$ remain constant after such an event.
The presence of gravity means that the orbit has to be a conic section to conserve angular momentum, the ellipticity depending on a combination of L and energy. In the absence of gravity straight inertial movement does the job.
answered Jan 5 at 13:48
my2ctsmy2cts
4,8792618
$endgroup$
$begingroup$
What do you mean by conical?
$endgroup$
– Karthik V
Jan 5 at 16:32
1
$begingroup$
@KarthikV the orbits must trace a conic section (circle, elipse, or parabola) in order to conserve angular momentum in the presence of a single large point source of gravity.
$endgroup$
– Vaelus
Jan 5 at 16:41
1
$begingroup$
Note: the question has been edited to refer to a less dramatic angular momentum scenario.
$endgroup$
– rob♦
Jan 6 at 5:52
add a comment |
$begingroup$
The planets will continue moving with the momentum they have at the moment the Sun disappears, apart from mutual attraction. These orbitals conserve planetary angular momentum is conserved. This is clear from the formula $L = vec r times vec p$. $r sin theta$ and $mv$ remain constant after such an event.
The presence of gravity means that the orbit has to be a conic section to conserve angular momentum, the ellipticity depending on a combination of L and energy. In the absence of gravity straight inertial movement does the job.
answered Jan 5 at 13:48
my2ctsmy2cts
4,8792618
$endgroup$
The planets will continue moving with the momentum they have at the moment the Sun disappears, apart from mutual attraction. These orbitals conserve planetary angular momentum is conserved. This is clear from the formula $L = vec r times vec p$. $r sin theta$ and $mv$ remain constant after such an event.
The presence of gravity means that the orbit has to be a conic section to conserve angular momentum, the ellipticity depending on a combination of L and energy. In the absence of gravity straight inertial movement does the job.
answered Jan 5 at 13:48
my2ctsmy2cts
4,8792618
edited Jan 5 at 17:00
answered Jan 5 at 13:48
my2ctsmy2cts
4,8792618
answered Jan 5 at 13:48
my2ctsmy2cts
4,8792618
answered Jan 5 at 13:48
my2ctsmy2cts
4,8792618
4,8792618
$begingroup$
What do you mean by conical?
$endgroup$
– Karthik V
Jan 5 at 16:32
1
$begingroup$
@KarthikV the orbits must trace a conic section (circle, elipse, or parabola) in order to conserve angular momentum in the presence of a single large point source of gravity.
$endgroup$
– Vaelus
Jan 5 at 16:41
1
$begingroup$
Note: the question has been edited to refer to a less dramatic angular momentum scenario.
$endgroup$
– rob♦
Jan 6 at 5:52
add a comment |
$begingroup$
What do you mean by conical?
$endgroup$
– Karthik V
Jan 5 at 16:32
1
$begingroup$
@KarthikV the orbits must trace a conic section (circle, elipse, or parabola) in order to conserve angular momentum in the presence of a single large point source of gravity.
$endgroup$
– Vaelus
Jan 5 at 16:41
1
$begingroup$
Note: the question has been edited to refer to a less dramatic angular momentum scenario.
$endgroup$
– rob♦
Jan 6 at 5:52
$begingroup$
What do you mean by conical?
$endgroup$
– Karthik V
Jan 5 at 16:32
$begingroup$
What do you mean by conical?
$endgroup$
– Karthik V
Jan 5 at 16:32
1
1
$begingroup$
@KarthikV the orbits must trace a conic section (circle, elipse, or parabola) in order to conserve angular momentum in the presence of a single large point source of gravity.
$endgroup$
– Vaelus
Jan 5 at 16:41
$begingroup$
@KarthikV the orbits must trace a conic section (circle, elipse, or parabola) in order to conserve angular momentum in the presence of a single large point source of gravity.
$endgroup$
– Vaelus
Jan 5 at 16:41
1
1
$begingroup$
Note: the question has been edited to refer to a less dramatic angular momentum scenario.
$endgroup$
– rob♦
Jan 6 at 5:52
$begingroup$
Note: the question has been edited to refer to a less dramatic angular momentum scenario.
$endgroup$
– rob♦
Jan 6 at 5:52
add a comment |
13
$begingroup$
"what's wrong in my reasoning?" - wouldn't the Sun disappearing violate one or more conservation laws?
$endgroup$
– Hal Hollis
Jan 5 at 12:52
3
$begingroup$
Have you tried calculating the angular momentum of an object moving in a straight line?
$endgroup$
– By Symmetry
Jan 5 at 13:22
4
$begingroup$
@AaronStevens, but my point is that the OP begins by stipulating something that violates one or more conservation laws and then draws a conclusion based on conservation laws holding. That, it seems to me, is a significant flaw in reasoning.
$endgroup$
– Hal Hollis
Jan 5 at 13:34
2
$begingroup$
@Piyush Might I suggest changing your scenario to the one I mentioned earlier about the ball on a rope so this question isn't closed by those who can't understand your intentions of this question?
$endgroup$
– Aaron Stevens
Jan 5 at 13:43
4
$begingroup$
A theory can't answer questions about what happens when the theory is violated. In both Newtonian gravity and general relativity, it's forbidden for the sun to suddenly disappear. Therefore neither theory can answer this question.
$endgroup$
– Ben Crowell
Jan 5 at 15:58