Do orbiting planets have infinite energy?
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I know that planets can't have infinite energy, due to the law of conservation of energy.
However, I'm confused because I see a contradiction and it would be great if someone could explain it.
Energy is defined as the capacity to do work. Work is defined as Force x Distance. Force is defined as Mass x Acceleration. Thus, if we accelerate a mass for some distance by using some force, we are doing work, and we must have had energy in order to do that work.
In orbit, planets change direction, which is a change in velocity, which is an acceleration. Planets have mass, and they are moving over a particular distance. Thus, work is being done to move the planets.
In an ideal world, planets continue to orbit forever. Thus, infinite work will be done on the planets as they orbit.
How can infinite work be done (or finite work over an infinite time period, if you'd like to think of it that way) with a finite amount of energy?
Where is the flaw in this argument?
newtonian-mechanics newtonian-gravity energy-conservation orbital-motion planets
New contributor
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add a comment |
$begingroup$
I know that planets can't have infinite energy, due to the law of conservation of energy.
However, I'm confused because I see a contradiction and it would be great if someone could explain it.
Energy is defined as the capacity to do work. Work is defined as Force x Distance. Force is defined as Mass x Acceleration. Thus, if we accelerate a mass for some distance by using some force, we are doing work, and we must have had energy in order to do that work.
In orbit, planets change direction, which is a change in velocity, which is an acceleration. Planets have mass, and they are moving over a particular distance. Thus, work is being done to move the planets.
In an ideal world, planets continue to orbit forever. Thus, infinite work will be done on the planets as they orbit.
How can infinite work be done (or finite work over an infinite time period, if you'd like to think of it that way) with a finite amount of energy?
Where is the flaw in this argument?
newtonian-mechanics newtonian-gravity energy-conservation orbital-motion planets
New contributor
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1
$begingroup$
@PM2Ring That is not true unless the orbit is a circle.
$endgroup$
– alephzero
58 mins ago
add a comment |
$begingroup$
I know that planets can't have infinite energy, due to the law of conservation of energy.
However, I'm confused because I see a contradiction and it would be great if someone could explain it.
Energy is defined as the capacity to do work. Work is defined as Force x Distance. Force is defined as Mass x Acceleration. Thus, if we accelerate a mass for some distance by using some force, we are doing work, and we must have had energy in order to do that work.
In orbit, planets change direction, which is a change in velocity, which is an acceleration. Planets have mass, and they are moving over a particular distance. Thus, work is being done to move the planets.
In an ideal world, planets continue to orbit forever. Thus, infinite work will be done on the planets as they orbit.
How can infinite work be done (or finite work over an infinite time period, if you'd like to think of it that way) with a finite amount of energy?
Where is the flaw in this argument?
newtonian-mechanics newtonian-gravity energy-conservation orbital-motion planets
New contributor
$endgroup$
I know that planets can't have infinite energy, due to the law of conservation of energy.
However, I'm confused because I see a contradiction and it would be great if someone could explain it.
Energy is defined as the capacity to do work. Work is defined as Force x Distance. Force is defined as Mass x Acceleration. Thus, if we accelerate a mass for some distance by using some force, we are doing work, and we must have had energy in order to do that work.
In orbit, planets change direction, which is a change in velocity, which is an acceleration. Planets have mass, and they are moving over a particular distance. Thus, work is being done to move the planets.
In an ideal world, planets continue to orbit forever. Thus, infinite work will be done on the planets as they orbit.
How can infinite work be done (or finite work over an infinite time period, if you'd like to think of it that way) with a finite amount of energy?
Where is the flaw in this argument?
newtonian-mechanics newtonian-gravity energy-conservation orbital-motion planets
newtonian-mechanics newtonian-gravity energy-conservation orbital-motion planets
New contributor
New contributor
edited 2 hours ago
Qmechanic♦
105k121891202
105k121891202
New contributor
asked 2 hours ago
Pro QPro Q
1113
1113
New contributor
New contributor
1
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@PM2Ring That is not true unless the orbit is a circle.
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– alephzero
58 mins ago
add a comment |
1
$begingroup$
@PM2Ring That is not true unless the orbit is a circle.
$endgroup$
– alephzero
58 mins ago
1
1
$begingroup$
@PM2Ring That is not true unless the orbit is a circle.
$endgroup$
– alephzero
58 mins ago
$begingroup$
@PM2Ring That is not true unless the orbit is a circle.
$endgroup$
– alephzero
58 mins ago
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
Your definitions are incorrect. Force is rate of change of momentum and is a vector. More importantly, the work done by a force is not force x distance, it is the force resolved in the direction of the displacement x the magnitude of the displacement. This is more formally known as the scalar product of force and displacement.
In the case of a (circular orbit), the centripetal force supplied by gravity is at right angles to the displacement, so no work is done.
$endgroup$
$begingroup$
Perhaps you should add what happens in an elliptical or hyperbolic orbit
$endgroup$
– magma
2 hours ago
$begingroup$
To be clear, Newton's second lawF = m * a
is correct, but thinking of force as a change in momentum over time is useful. Here is a source with the correct definition of work, and another source with an explanation as to how you get fromF = m * a
toF = p / t
(where p is momentum).
$endgroup$
– Pro Q
1 hour ago
add a comment |
$begingroup$
Power expended when moving in orbit $vec {F}.vec {v}=-nabla phi .frac {dvec {r}}{dt}=-frac {dphi}{dt}$ , $phi$ is gravitational potential. Hence the work of gravitational forces is $W=int {vec {F}.vec {v} dt}=-int {frac {dphi}{dt}dt}$. For a periodic motion, the integral $W$ over the period is zero. For hyperbolic motion, the integral $W$ over the entire time of motion is zero.
$endgroup$
add a comment |
$begingroup$
In an ideal world, even then it is impossible to supply infinite energy.
I think the flaw here is that work done is only a measure of the change in energy. Obviously the satellite responds to the force of gravity acting on it (which in turn implies that the satellite does not have infinite energy).
What you have to understand is that work done on an object is the dot product of the force and displacement vector. This is where the $costheta$ term becomes vital. Remember that $theta$ is the angle between the force and the displacement vector. The object makes displacement, true. But that is a horizontal displacement. The force cannot change the magnitude as such, because you have to apply a force in the direction of motion to cause the change in speed. Since the force acts perpendicularly, the object would have to move towards the planet.
Which it does, but it does not fall to the planet either. Because the direction of force changes here. And hence, there is a circular motion.
Because the satellite has that constant speed $v$ (I am assuming simplest idea of a circular orbit), it has a kinetic energy. Because the satellite is attracted by gravity, there is a corresponding potential energy.
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$begingroup$
The speed of an orbiting body is only constant if the orbit is circular. See en.wikipedia.org/wiki/Elliptic_orbit
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– PM 2Ring
1 hour ago
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@PM 2Ring Yes. I have edited it.
$endgroup$
– KV18
1 hour ago
add a comment |
$begingroup$
Remember that work is force times displacement, not distance. Displacement is a vector, which means when a planet moves a full circle, its overall displacement is zero, resulting in a work of zero.
$endgroup$
1
$begingroup$
No, it isn't. It is the scalar product of force and displacement. The last paragraph is misleading and unhelpful.
$endgroup$
– Rob Jeffries
2 hours ago
1
$begingroup$
Yes, I just realized that now. For some reason I had a brain fart and thought that work was a vector.
$endgroup$
– Jason Chen
2 hours ago
$begingroup$
The combination of first and second sentence remains misleading. In the first sentence you are speaking of circular orbits. The second statement is correct but only for some non circular orbits.
$endgroup$
– GiorgioP
2 hours ago
$begingroup$
Answer is incorrect unfortunately. You can have an overall displacement of zero but a work that isn't zero, if the field isn't conservative. The gravitational field is conservative, but it need not be in general.
$endgroup$
– Allure
38 mins ago
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Your definitions are incorrect. Force is rate of change of momentum and is a vector. More importantly, the work done by a force is not force x distance, it is the force resolved in the direction of the displacement x the magnitude of the displacement. This is more formally known as the scalar product of force and displacement.
In the case of a (circular orbit), the centripetal force supplied by gravity is at right angles to the displacement, so no work is done.
$endgroup$
$begingroup$
Perhaps you should add what happens in an elliptical or hyperbolic orbit
$endgroup$
– magma
2 hours ago
$begingroup$
To be clear, Newton's second lawF = m * a
is correct, but thinking of force as a change in momentum over time is useful. Here is a source with the correct definition of work, and another source with an explanation as to how you get fromF = m * a
toF = p / t
(where p is momentum).
$endgroup$
– Pro Q
1 hour ago
add a comment |
$begingroup$
Your definitions are incorrect. Force is rate of change of momentum and is a vector. More importantly, the work done by a force is not force x distance, it is the force resolved in the direction of the displacement x the magnitude of the displacement. This is more formally known as the scalar product of force and displacement.
In the case of a (circular orbit), the centripetal force supplied by gravity is at right angles to the displacement, so no work is done.
$endgroup$
$begingroup$
Perhaps you should add what happens in an elliptical or hyperbolic orbit
$endgroup$
– magma
2 hours ago
$begingroup$
To be clear, Newton's second lawF = m * a
is correct, but thinking of force as a change in momentum over time is useful. Here is a source with the correct definition of work, and another source with an explanation as to how you get fromF = m * a
toF = p / t
(where p is momentum).
$endgroup$
– Pro Q
1 hour ago
add a comment |
$begingroup$
Your definitions are incorrect. Force is rate of change of momentum and is a vector. More importantly, the work done by a force is not force x distance, it is the force resolved in the direction of the displacement x the magnitude of the displacement. This is more formally known as the scalar product of force and displacement.
In the case of a (circular orbit), the centripetal force supplied by gravity is at right angles to the displacement, so no work is done.
$endgroup$
Your definitions are incorrect. Force is rate of change of momentum and is a vector. More importantly, the work done by a force is not force x distance, it is the force resolved in the direction of the displacement x the magnitude of the displacement. This is more formally known as the scalar product of force and displacement.
In the case of a (circular orbit), the centripetal force supplied by gravity is at right angles to the displacement, so no work is done.
answered 2 hours ago
Rob JeffriesRob Jeffries
68.8k7139235
68.8k7139235
$begingroup$
Perhaps you should add what happens in an elliptical or hyperbolic orbit
$endgroup$
– magma
2 hours ago
$begingroup$
To be clear, Newton's second lawF = m * a
is correct, but thinking of force as a change in momentum over time is useful. Here is a source with the correct definition of work, and another source with an explanation as to how you get fromF = m * a
toF = p / t
(where p is momentum).
$endgroup$
– Pro Q
1 hour ago
add a comment |
$begingroup$
Perhaps you should add what happens in an elliptical or hyperbolic orbit
$endgroup$
– magma
2 hours ago
$begingroup$
To be clear, Newton's second lawF = m * a
is correct, but thinking of force as a change in momentum over time is useful. Here is a source with the correct definition of work, and another source with an explanation as to how you get fromF = m * a
toF = p / t
(where p is momentum).
$endgroup$
– Pro Q
1 hour ago
$begingroup$
Perhaps you should add what happens in an elliptical or hyperbolic orbit
$endgroup$
– magma
2 hours ago
$begingroup$
Perhaps you should add what happens in an elliptical or hyperbolic orbit
$endgroup$
– magma
2 hours ago
$begingroup$
To be clear, Newton's second law
F = m * a
is correct, but thinking of force as a change in momentum over time is useful. Here is a source with the correct definition of work, and another source with an explanation as to how you get from F = m * a
to F = p / t
(where p is momentum).$endgroup$
– Pro Q
1 hour ago
$begingroup$
To be clear, Newton's second law
F = m * a
is correct, but thinking of force as a change in momentum over time is useful. Here is a source with the correct definition of work, and another source with an explanation as to how you get from F = m * a
to F = p / t
(where p is momentum).$endgroup$
– Pro Q
1 hour ago
add a comment |
$begingroup$
Power expended when moving in orbit $vec {F}.vec {v}=-nabla phi .frac {dvec {r}}{dt}=-frac {dphi}{dt}$ , $phi$ is gravitational potential. Hence the work of gravitational forces is $W=int {vec {F}.vec {v} dt}=-int {frac {dphi}{dt}dt}$. For a periodic motion, the integral $W$ over the period is zero. For hyperbolic motion, the integral $W$ over the entire time of motion is zero.
$endgroup$
add a comment |
$begingroup$
Power expended when moving in orbit $vec {F}.vec {v}=-nabla phi .frac {dvec {r}}{dt}=-frac {dphi}{dt}$ , $phi$ is gravitational potential. Hence the work of gravitational forces is $W=int {vec {F}.vec {v} dt}=-int {frac {dphi}{dt}dt}$. For a periodic motion, the integral $W$ over the period is zero. For hyperbolic motion, the integral $W$ over the entire time of motion is zero.
$endgroup$
add a comment |
$begingroup$
Power expended when moving in orbit $vec {F}.vec {v}=-nabla phi .frac {dvec {r}}{dt}=-frac {dphi}{dt}$ , $phi$ is gravitational potential. Hence the work of gravitational forces is $W=int {vec {F}.vec {v} dt}=-int {frac {dphi}{dt}dt}$. For a periodic motion, the integral $W$ over the period is zero. For hyperbolic motion, the integral $W$ over the entire time of motion is zero.
$endgroup$
Power expended when moving in orbit $vec {F}.vec {v}=-nabla phi .frac {dvec {r}}{dt}=-frac {dphi}{dt}$ , $phi$ is gravitational potential. Hence the work of gravitational forces is $W=int {vec {F}.vec {v} dt}=-int {frac {dphi}{dt}dt}$. For a periodic motion, the integral $W$ over the period is zero. For hyperbolic motion, the integral $W$ over the entire time of motion is zero.
answered 1 hour ago
Alex TrounevAlex Trounev
41216
41216
add a comment |
add a comment |
$begingroup$
In an ideal world, even then it is impossible to supply infinite energy.
I think the flaw here is that work done is only a measure of the change in energy. Obviously the satellite responds to the force of gravity acting on it (which in turn implies that the satellite does not have infinite energy).
What you have to understand is that work done on an object is the dot product of the force and displacement vector. This is where the $costheta$ term becomes vital. Remember that $theta$ is the angle between the force and the displacement vector. The object makes displacement, true. But that is a horizontal displacement. The force cannot change the magnitude as such, because you have to apply a force in the direction of motion to cause the change in speed. Since the force acts perpendicularly, the object would have to move towards the planet.
Which it does, but it does not fall to the planet either. Because the direction of force changes here. And hence, there is a circular motion.
Because the satellite has that constant speed $v$ (I am assuming simplest idea of a circular orbit), it has a kinetic energy. Because the satellite is attracted by gravity, there is a corresponding potential energy.
$endgroup$
$begingroup$
The speed of an orbiting body is only constant if the orbit is circular. See en.wikipedia.org/wiki/Elliptic_orbit
$endgroup$
– PM 2Ring
1 hour ago
$begingroup$
@PM 2Ring Yes. I have edited it.
$endgroup$
– KV18
1 hour ago
add a comment |
$begingroup$
In an ideal world, even then it is impossible to supply infinite energy.
I think the flaw here is that work done is only a measure of the change in energy. Obviously the satellite responds to the force of gravity acting on it (which in turn implies that the satellite does not have infinite energy).
What you have to understand is that work done on an object is the dot product of the force and displacement vector. This is where the $costheta$ term becomes vital. Remember that $theta$ is the angle between the force and the displacement vector. The object makes displacement, true. But that is a horizontal displacement. The force cannot change the magnitude as such, because you have to apply a force in the direction of motion to cause the change in speed. Since the force acts perpendicularly, the object would have to move towards the planet.
Which it does, but it does not fall to the planet either. Because the direction of force changes here. And hence, there is a circular motion.
Because the satellite has that constant speed $v$ (I am assuming simplest idea of a circular orbit), it has a kinetic energy. Because the satellite is attracted by gravity, there is a corresponding potential energy.
$endgroup$
$begingroup$
The speed of an orbiting body is only constant if the orbit is circular. See en.wikipedia.org/wiki/Elliptic_orbit
$endgroup$
– PM 2Ring
1 hour ago
$begingroup$
@PM 2Ring Yes. I have edited it.
$endgroup$
– KV18
1 hour ago
add a comment |
$begingroup$
In an ideal world, even then it is impossible to supply infinite energy.
I think the flaw here is that work done is only a measure of the change in energy. Obviously the satellite responds to the force of gravity acting on it (which in turn implies that the satellite does not have infinite energy).
What you have to understand is that work done on an object is the dot product of the force and displacement vector. This is where the $costheta$ term becomes vital. Remember that $theta$ is the angle between the force and the displacement vector. The object makes displacement, true. But that is a horizontal displacement. The force cannot change the magnitude as such, because you have to apply a force in the direction of motion to cause the change in speed. Since the force acts perpendicularly, the object would have to move towards the planet.
Which it does, but it does not fall to the planet either. Because the direction of force changes here. And hence, there is a circular motion.
Because the satellite has that constant speed $v$ (I am assuming simplest idea of a circular orbit), it has a kinetic energy. Because the satellite is attracted by gravity, there is a corresponding potential energy.
$endgroup$
In an ideal world, even then it is impossible to supply infinite energy.
I think the flaw here is that work done is only a measure of the change in energy. Obviously the satellite responds to the force of gravity acting on it (which in turn implies that the satellite does not have infinite energy).
What you have to understand is that work done on an object is the dot product of the force and displacement vector. This is where the $costheta$ term becomes vital. Remember that $theta$ is the angle between the force and the displacement vector. The object makes displacement, true. But that is a horizontal displacement. The force cannot change the magnitude as such, because you have to apply a force in the direction of motion to cause the change in speed. Since the force acts perpendicularly, the object would have to move towards the planet.
Which it does, but it does not fall to the planet either. Because the direction of force changes here. And hence, there is a circular motion.
Because the satellite has that constant speed $v$ (I am assuming simplest idea of a circular orbit), it has a kinetic energy. Because the satellite is attracted by gravity, there is a corresponding potential energy.
edited 1 hour ago
answered 2 hours ago
KV18KV18
547312
547312
$begingroup$
The speed of an orbiting body is only constant if the orbit is circular. See en.wikipedia.org/wiki/Elliptic_orbit
$endgroup$
– PM 2Ring
1 hour ago
$begingroup$
@PM 2Ring Yes. I have edited it.
$endgroup$
– KV18
1 hour ago
add a comment |
$begingroup$
The speed of an orbiting body is only constant if the orbit is circular. See en.wikipedia.org/wiki/Elliptic_orbit
$endgroup$
– PM 2Ring
1 hour ago
$begingroup$
@PM 2Ring Yes. I have edited it.
$endgroup$
– KV18
1 hour ago
$begingroup$
The speed of an orbiting body is only constant if the orbit is circular. See en.wikipedia.org/wiki/Elliptic_orbit
$endgroup$
– PM 2Ring
1 hour ago
$begingroup$
The speed of an orbiting body is only constant if the orbit is circular. See en.wikipedia.org/wiki/Elliptic_orbit
$endgroup$
– PM 2Ring
1 hour ago
$begingroup$
@PM 2Ring Yes. I have edited it.
$endgroup$
– KV18
1 hour ago
$begingroup$
@PM 2Ring Yes. I have edited it.
$endgroup$
– KV18
1 hour ago
add a comment |
$begingroup$
Remember that work is force times displacement, not distance. Displacement is a vector, which means when a planet moves a full circle, its overall displacement is zero, resulting in a work of zero.
$endgroup$
1
$begingroup$
No, it isn't. It is the scalar product of force and displacement. The last paragraph is misleading and unhelpful.
$endgroup$
– Rob Jeffries
2 hours ago
1
$begingroup$
Yes, I just realized that now. For some reason I had a brain fart and thought that work was a vector.
$endgroup$
– Jason Chen
2 hours ago
$begingroup$
The combination of first and second sentence remains misleading. In the first sentence you are speaking of circular orbits. The second statement is correct but only for some non circular orbits.
$endgroup$
– GiorgioP
2 hours ago
$begingroup$
Answer is incorrect unfortunately. You can have an overall displacement of zero but a work that isn't zero, if the field isn't conservative. The gravitational field is conservative, but it need not be in general.
$endgroup$
– Allure
38 mins ago
add a comment |
$begingroup$
Remember that work is force times displacement, not distance. Displacement is a vector, which means when a planet moves a full circle, its overall displacement is zero, resulting in a work of zero.
$endgroup$
1
$begingroup$
No, it isn't. It is the scalar product of force and displacement. The last paragraph is misleading and unhelpful.
$endgroup$
– Rob Jeffries
2 hours ago
1
$begingroup$
Yes, I just realized that now. For some reason I had a brain fart and thought that work was a vector.
$endgroup$
– Jason Chen
2 hours ago
$begingroup$
The combination of first and second sentence remains misleading. In the first sentence you are speaking of circular orbits. The second statement is correct but only for some non circular orbits.
$endgroup$
– GiorgioP
2 hours ago
$begingroup$
Answer is incorrect unfortunately. You can have an overall displacement of zero but a work that isn't zero, if the field isn't conservative. The gravitational field is conservative, but it need not be in general.
$endgroup$
– Allure
38 mins ago
add a comment |
$begingroup$
Remember that work is force times displacement, not distance. Displacement is a vector, which means when a planet moves a full circle, its overall displacement is zero, resulting in a work of zero.
$endgroup$
Remember that work is force times displacement, not distance. Displacement is a vector, which means when a planet moves a full circle, its overall displacement is zero, resulting in a work of zero.
edited 2 hours ago
answered 2 hours ago
Jason ChenJason Chen
255112
255112
1
$begingroup$
No, it isn't. It is the scalar product of force and displacement. The last paragraph is misleading and unhelpful.
$endgroup$
– Rob Jeffries
2 hours ago
1
$begingroup$
Yes, I just realized that now. For some reason I had a brain fart and thought that work was a vector.
$endgroup$
– Jason Chen
2 hours ago
$begingroup$
The combination of first and second sentence remains misleading. In the first sentence you are speaking of circular orbits. The second statement is correct but only for some non circular orbits.
$endgroup$
– GiorgioP
2 hours ago
$begingroup$
Answer is incorrect unfortunately. You can have an overall displacement of zero but a work that isn't zero, if the field isn't conservative. The gravitational field is conservative, but it need not be in general.
$endgroup$
– Allure
38 mins ago
add a comment |
1
$begingroup$
No, it isn't. It is the scalar product of force and displacement. The last paragraph is misleading and unhelpful.
$endgroup$
– Rob Jeffries
2 hours ago
1
$begingroup$
Yes, I just realized that now. For some reason I had a brain fart and thought that work was a vector.
$endgroup$
– Jason Chen
2 hours ago
$begingroup$
The combination of first and second sentence remains misleading. In the first sentence you are speaking of circular orbits. The second statement is correct but only for some non circular orbits.
$endgroup$
– GiorgioP
2 hours ago
$begingroup$
Answer is incorrect unfortunately. You can have an overall displacement of zero but a work that isn't zero, if the field isn't conservative. The gravitational field is conservative, but it need not be in general.
$endgroup$
– Allure
38 mins ago
1
1
$begingroup$
No, it isn't. It is the scalar product of force and displacement. The last paragraph is misleading and unhelpful.
$endgroup$
– Rob Jeffries
2 hours ago
$begingroup$
No, it isn't. It is the scalar product of force and displacement. The last paragraph is misleading and unhelpful.
$endgroup$
– Rob Jeffries
2 hours ago
1
1
$begingroup$
Yes, I just realized that now. For some reason I had a brain fart and thought that work was a vector.
$endgroup$
– Jason Chen
2 hours ago
$begingroup$
Yes, I just realized that now. For some reason I had a brain fart and thought that work was a vector.
$endgroup$
– Jason Chen
2 hours ago
$begingroup$
The combination of first and second sentence remains misleading. In the first sentence you are speaking of circular orbits. The second statement is correct but only for some non circular orbits.
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– GiorgioP
2 hours ago
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The combination of first and second sentence remains misleading. In the first sentence you are speaking of circular orbits. The second statement is correct but only for some non circular orbits.
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– GiorgioP
2 hours ago
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Answer is incorrect unfortunately. You can have an overall displacement of zero but a work that isn't zero, if the field isn't conservative. The gravitational field is conservative, but it need not be in general.
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– Allure
38 mins ago
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Answer is incorrect unfortunately. You can have an overall displacement of zero but a work that isn't zero, if the field isn't conservative. The gravitational field is conservative, but it need not be in general.
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– Allure
38 mins ago
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@PM2Ring That is not true unless the orbit is a circle.
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– alephzero
58 mins ago