Is Earth an inertial reference frame?
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Is earth considered as inertial frame? i was confused because we learned about coriolis effect. We know that earth spins therefore coriolis effect should take place . But does it have minimal effect for motion of balls etc when they move with respect to the ground?
newtonian-mechanics reference-frames inertial-frames earth coriolis-effect
edited Dec 2 '18 at 2:25
Qmechanic♦
102k121841171
asked Dec 1 '18 at 23:46
ado sarado sar
31310
$endgroup$
add a comment |
$begingroup$
Is earth considered as inertial frame? i was confused because we learned about coriolis effect. We know that earth spins therefore coriolis effect should take place . But does it have minimal effect for motion of balls etc when they move with respect to the ground?
newtonian-mechanics reference-frames inertial-frames earth coriolis-effect
edited Dec 2 '18 at 2:25
Qmechanic♦
102k121841171
asked Dec 1 '18 at 23:46
ado sarado sar
31310
$endgroup$
$begingroup$
you can calculate it's size, e.g, for a BMG 50 cal (853 m/s muzzle velocity) shot at its max effective range (1,800m) and tell us.
$endgroup$
– JEB
Dec 2 '18 at 0:05
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This question seems synonymous with "Does the earth rotate (with respect to the rest of the universe)?"
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– Geremia
Dec 2 '18 at 4:31
2
$begingroup$
That the Earth's surface is not an inertial frame is nicely demonstrated by the Foucault pendulum.
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– Ruslan
Dec 2 '18 at 15:51
1
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It depends on the physics problem you look at whether the Earth can be considered an inertial frame of reference or not.
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– jjack
Dec 2 '18 at 18:14
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To emphasize @jjack's point, it's notoriously hard work to get a Foucault pendulum to work properly because the non-inertial effects, while present, are very small.
$endgroup$
– dmckee♦
Dec 2 '18 at 19:28
add a comment |
$begingroup$
Is earth considered as inertial frame? i was confused because we learned about coriolis effect. We know that earth spins therefore coriolis effect should take place . But does it have minimal effect for motion of balls etc when they move with respect to the ground?
newtonian-mechanics reference-frames inertial-frames earth coriolis-effect
edited Dec 2 '18 at 2:25
Qmechanic♦
102k121841171
asked Dec 1 '18 at 23:46
ado sarado sar
31310
$endgroup$
Is earth considered as inertial frame? i was confused because we learned about coriolis effect. We know that earth spins therefore coriolis effect should take place . But does it have minimal effect for motion of balls etc when they move with respect to the ground?
newtonian-mechanics reference-frames inertial-frames earth coriolis-effect
newtonian-mechanics reference-frames inertial-frames earth coriolis-effect
edited Dec 2 '18 at 2:25
Qmechanic♦
102k121841171
asked Dec 1 '18 at 23:46
ado sarado sar
31310
edited Dec 2 '18 at 2:25
Qmechanic♦
102k121841171
asked Dec 1 '18 at 23:46
ado sarado sar
31310
edited Dec 2 '18 at 2:25
Qmechanic♦
102k121841171
edited Dec 2 '18 at 2:25
Qmechanic♦
102k121841171
edited Dec 2 '18 at 2:25
Qmechanic♦
102k121841171
102k121841171
asked Dec 1 '18 at 23:46
ado sarado sar
31310
asked Dec 1 '18 at 23:46
ado sarado sar
31310
asked Dec 1 '18 at 23:46
ado sarado sar
31310
31310
$begingroup$
you can calculate it's size, e.g, for a BMG 50 cal (853 m/s muzzle velocity) shot at its max effective range (1,800m) and tell us.
$endgroup$
– JEB
Dec 2 '18 at 0:05
$begingroup$
This question seems synonymous with "Does the earth rotate (with respect to the rest of the universe)?"
$endgroup$
– Geremia
Dec 2 '18 at 4:31
2
$begingroup$
That the Earth's surface is not an inertial frame is nicely demonstrated by the Foucault pendulum.
$endgroup$
– Ruslan
Dec 2 '18 at 15:51
1
$begingroup$
It depends on the physics problem you look at whether the Earth can be considered an inertial frame of reference or not.
$endgroup$
– jjack
Dec 2 '18 at 18:14
$begingroup$
To emphasize @jjack's point, it's notoriously hard work to get a Foucault pendulum to work properly because the non-inertial effects, while present, are very small.
$endgroup$
– dmckee♦
Dec 2 '18 at 19:28
add a comment |
$begingroup$
you can calculate it's size, e.g, for a BMG 50 cal (853 m/s muzzle velocity) shot at its max effective range (1,800m) and tell us.
$endgroup$
– JEB
Dec 2 '18 at 0:05
$begingroup$
This question seems synonymous with "Does the earth rotate (with respect to the rest of the universe)?"
$endgroup$
– Geremia
Dec 2 '18 at 4:31
2
$begingroup$
That the Earth's surface is not an inertial frame is nicely demonstrated by the Foucault pendulum.
$endgroup$
– Ruslan
Dec 2 '18 at 15:51
1
$begingroup$
It depends on the physics problem you look at whether the Earth can be considered an inertial frame of reference or not.
$endgroup$
– jjack
Dec 2 '18 at 18:14
$begingroup$
To emphasize @jjack's point, it's notoriously hard work to get a Foucault pendulum to work properly because the non-inertial effects, while present, are very small.
$endgroup$
– dmckee♦
Dec 2 '18 at 19:28
$begingroup$
you can calculate it's size, e.g, for a BMG 50 cal (853 m/s muzzle velocity) shot at its max effective range (1,800m) and tell us.
$endgroup$
– JEB
Dec 2 '18 at 0:05
$begingroup$
you can calculate it's size, e.g, for a BMG 50 cal (853 m/s muzzle velocity) shot at its max effective range (1,800m) and tell us.
$endgroup$
– JEB
Dec 2 '18 at 0:05
$begingroup$
This question seems synonymous with "Does the earth rotate (with respect to the rest of the universe)?"
$endgroup$
– Geremia
Dec 2 '18 at 4:31
$begingroup$
This question seems synonymous with "Does the earth rotate (with respect to the rest of the universe)?"
$endgroup$
– Geremia
Dec 2 '18 at 4:31
2
2
$begingroup$
That the Earth's surface is not an inertial frame is nicely demonstrated by the Foucault pendulum.
$endgroup$
– Ruslan
Dec 2 '18 at 15:51
$begingroup$
That the Earth's surface is not an inertial frame is nicely demonstrated by the Foucault pendulum.
$endgroup$
– Ruslan
Dec 2 '18 at 15:51
1
1
$begingroup$
It depends on the physics problem you look at whether the Earth can be considered an inertial frame of reference or not.
$endgroup$
– jjack
Dec 2 '18 at 18:14
$begingroup$
It depends on the physics problem you look at whether the Earth can be considered an inertial frame of reference or not.
$endgroup$
– jjack
Dec 2 '18 at 18:14
$begingroup$
To emphasize @jjack's point, it's notoriously hard work to get a Foucault pendulum to work properly because the non-inertial effects, while present, are very small.
$endgroup$
– dmckee♦
Dec 2 '18 at 19:28
$begingroup$
To emphasize @jjack's point, it's notoriously hard work to get a Foucault pendulum to work properly because the non-inertial effects, while present, are very small.
$endgroup$
– dmckee♦
Dec 2 '18 at 19:28
add a comment |
4 Answers
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The surface of the Earth is not, rigorously speaking, an inertial frame of reference. Objects at rest relative to Earth's surface are actually subject to a series of inertial effects, like the ficticious forces (Coriolis, centrifugal etc.) because of Earth's rotation, precession and other kinds of acceleration.
When solving physics problems, however, we usually take the Earth frame as being inertial. This is because the inertial effects are minuscule for most of our day-to-day experiences and experiments. For example, objects in the Equator are the ones subject to the strongest centrifugal force and it is only about $3 times10^{-3}$ or $0.3%$ of their weight.
So for the most part, if an experiment is short enough and happens in a small enough region, the surface of Earth can indeed be approximated to an inertial frame of reference since the effects on the experiment's results are very, very tiny.
This of course has exceptions, as cited in njspeer's answer.
If however by "Earth" you mean the reference frame in Earth's center, it is an inertial frame according to General Relativity (GR), since observers in free fall are inertial in GR. The Earth actually does have some proper acceleration due to external forces such as radiation pressure, but these are also minuscule effects.
answered Dec 2 '18 at 0:10
João Vítor G. LimaJoão Vítor G. Lima
857319
$endgroup$
2
$begingroup$
The earth is in free-fall, so its acceleration due to the sun's gravity (or the Milky Way's) does not cause the earth-centered frame to be non-inertial. (Except for even more miniscule effects, e.g. solar radiation pressure.)
$endgroup$
– David
Dec 2 '18 at 8:23
$begingroup$
I wasn't really sure about this. I mean, it would be considered inertial in GR, but I was using a more classical idea. I'm editing the post to address this issue. Thank you.
$endgroup$
– João Vítor G. Lima
Dec 2 '18 at 12:52
add a comment |
$begingroup$
Because the earth is rotating, it is never strictly an inertial reference frame. However, because the effects are small in many situations, it can often be approximated as one. When to use Coriolis forces will have to be determined on a case-by-case basis. E.g. ballistic problems that cover large distances will most certainly require Coriolis-force corrections, and pendulums that swings for a long time would also require Coriolis-force corrections. For a block sliding down an inclined plane, or a spring on a mass, or a vibrating string, you should not need to take it into consideration.
answered Dec 2 '18 at 0:01
njspeernjspeer
4374
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add a comment |
$begingroup$
Mach would say that non-inertial effects are due to the relative motion between the earth and the rest of the universe.
See:
• Assis, André K. T. Relational Mechanics and Implementation of Mach’s Principle with Weber’s Gravitational Force. Montréal: Apeiron, 2014.
answered Dec 2 '18 at 4:35
GeremiaGeremia
1,1752927
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add a comment |
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I know it's being a little pedantic, but I would say "earth" is a thing, not a reference frame. You could define an inertial reference frame that contains the earth.
But suppose you mean a reference frame defined with reference to the earth: Z is perpendicular to the ground where you are standing, X and Y are parallel to the ground and perpendicular to each other.
If you stand at rest in this reference frame, you may notice you feel acceleration (gravity). It is therefore not an inertial reference frame.
answered Dec 2 '18 at 15:51
OwenOwen
27729
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add a comment |
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4 Answers
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4 Answers
4
active
oldest
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active
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active
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$begingroup$
The surface of the Earth is not, rigorously speaking, an inertial frame of reference. Objects at rest relative to Earth's surface are actually subject to a series of inertial effects, like the ficticious forces (Coriolis, centrifugal etc.) because of Earth's rotation, precession and other kinds of acceleration.
When solving physics problems, however, we usually take the Earth frame as being inertial. This is because the inertial effects are minuscule for most of our day-to-day experiences and experiments. For example, objects in the Equator are the ones subject to the strongest centrifugal force and it is only about $3 times10^{-3}$ or $0.3%$ of their weight.
So for the most part, if an experiment is short enough and happens in a small enough region, the surface of Earth can indeed be approximated to an inertial frame of reference since the effects on the experiment's results are very, very tiny.
This of course has exceptions, as cited in njspeer's answer.
If however by "Earth" you mean the reference frame in Earth's center, it is an inertial frame according to General Relativity (GR), since observers in free fall are inertial in GR. The Earth actually does have some proper acceleration due to external forces such as radiation pressure, but these are also minuscule effects.
answered Dec 2 '18 at 0:10
João Vítor G. LimaJoão Vítor G. Lima
857319
$endgroup$
2
$begingroup$
The earth is in free-fall, so its acceleration due to the sun's gravity (or the Milky Way's) does not cause the earth-centered frame to be non-inertial. (Except for even more miniscule effects, e.g. solar radiation pressure.)
$endgroup$
– David
Dec 2 '18 at 8:23
$begingroup$
I wasn't really sure about this. I mean, it would be considered inertial in GR, but I was using a more classical idea. I'm editing the post to address this issue. Thank you.
$endgroup$
– João Vítor G. Lima
Dec 2 '18 at 12:52
add a comment |
$begingroup$
The surface of the Earth is not, rigorously speaking, an inertial frame of reference. Objects at rest relative to Earth's surface are actually subject to a series of inertial effects, like the ficticious forces (Coriolis, centrifugal etc.) because of Earth's rotation, precession and other kinds of acceleration.
When solving physics problems, however, we usually take the Earth frame as being inertial. This is because the inertial effects are minuscule for most of our day-to-day experiences and experiments. For example, objects in the Equator are the ones subject to the strongest centrifugal force and it is only about $3 times10^{-3}$ or $0.3%$ of their weight.
So for the most part, if an experiment is short enough and happens in a small enough region, the surface of Earth can indeed be approximated to an inertial frame of reference since the effects on the experiment's results are very, very tiny.
This of course has exceptions, as cited in njspeer's answer.
If however by "Earth" you mean the reference frame in Earth's center, it is an inertial frame according to General Relativity (GR), since observers in free fall are inertial in GR. The Earth actually does have some proper acceleration due to external forces such as radiation pressure, but these are also minuscule effects.
answered Dec 2 '18 at 0:10
João Vítor G. LimaJoão Vítor G. Lima
857319
$endgroup$
2
$begingroup$
The earth is in free-fall, so its acceleration due to the sun's gravity (or the Milky Way's) does not cause the earth-centered frame to be non-inertial. (Except for even more miniscule effects, e.g. solar radiation pressure.)
$endgroup$
– David
Dec 2 '18 at 8:23
$begingroup$
I wasn't really sure about this. I mean, it would be considered inertial in GR, but I was using a more classical idea. I'm editing the post to address this issue. Thank you.
$endgroup$
– João Vítor G. Lima
Dec 2 '18 at 12:52
add a comment |
$begingroup$
The surface of the Earth is not, rigorously speaking, an inertial frame of reference. Objects at rest relative to Earth's surface are actually subject to a series of inertial effects, like the ficticious forces (Coriolis, centrifugal etc.) because of Earth's rotation, precession and other kinds of acceleration.
When solving physics problems, however, we usually take the Earth frame as being inertial. This is because the inertial effects are minuscule for most of our day-to-day experiences and experiments. For example, objects in the Equator are the ones subject to the strongest centrifugal force and it is only about $3 times10^{-3}$ or $0.3%$ of their weight.
So for the most part, if an experiment is short enough and happens in a small enough region, the surface of Earth can indeed be approximated to an inertial frame of reference since the effects on the experiment's results are very, very tiny.
This of course has exceptions, as cited in njspeer's answer.
If however by "Earth" you mean the reference frame in Earth's center, it is an inertial frame according to General Relativity (GR), since observers in free fall are inertial in GR. The Earth actually does have some proper acceleration due to external forces such as radiation pressure, but these are also minuscule effects.
answered Dec 2 '18 at 0:10
João Vítor G. LimaJoão Vítor G. Lima
857319
$endgroup$
The surface of the Earth is not, rigorously speaking, an inertial frame of reference. Objects at rest relative to Earth's surface are actually subject to a series of inertial effects, like the ficticious forces (Coriolis, centrifugal etc.) because of Earth's rotation, precession and other kinds of acceleration.
When solving physics problems, however, we usually take the Earth frame as being inertial. This is because the inertial effects are minuscule for most of our day-to-day experiences and experiments. For example, objects in the Equator are the ones subject to the strongest centrifugal force and it is only about $3 times10^{-3}$ or $0.3%$ of their weight.
So for the most part, if an experiment is short enough and happens in a small enough region, the surface of Earth can indeed be approximated to an inertial frame of reference since the effects on the experiment's results are very, very tiny.
This of course has exceptions, as cited in njspeer's answer.
If however by "Earth" you mean the reference frame in Earth's center, it is an inertial frame according to General Relativity (GR), since observers in free fall are inertial in GR. The Earth actually does have some proper acceleration due to external forces such as radiation pressure, but these are also minuscule effects.
answered Dec 2 '18 at 0:10
João Vítor G. LimaJoão Vítor G. Lima
857319
edited Dec 2 '18 at 12:54
answered Dec 2 '18 at 0:10
João Vítor G. LimaJoão Vítor G. Lima
857319
answered Dec 2 '18 at 0:10
João Vítor G. LimaJoão Vítor G. Lima
857319
answered Dec 2 '18 at 0:10
João Vítor G. LimaJoão Vítor G. Lima
857319
857319
2
$begingroup$
The earth is in free-fall, so its acceleration due to the sun's gravity (or the Milky Way's) does not cause the earth-centered frame to be non-inertial. (Except for even more miniscule effects, e.g. solar radiation pressure.)
$endgroup$
– David
Dec 2 '18 at 8:23
$begingroup$
I wasn't really sure about this. I mean, it would be considered inertial in GR, but I was using a more classical idea. I'm editing the post to address this issue. Thank you.
$endgroup$
– João Vítor G. Lima
Dec 2 '18 at 12:52
add a comment |
2
$begingroup$
The earth is in free-fall, so its acceleration due to the sun's gravity (or the Milky Way's) does not cause the earth-centered frame to be non-inertial. (Except for even more miniscule effects, e.g. solar radiation pressure.)
$endgroup$
– David
Dec 2 '18 at 8:23
$begingroup$
I wasn't really sure about this. I mean, it would be considered inertial in GR, but I was using a more classical idea. I'm editing the post to address this issue. Thank you.
$endgroup$
– João Vítor G. Lima
Dec 2 '18 at 12:52
2
2
$begingroup$
The earth is in free-fall, so its acceleration due to the sun's gravity (or the Milky Way's) does not cause the earth-centered frame to be non-inertial. (Except for even more miniscule effects, e.g. solar radiation pressure.)
$endgroup$
– David
Dec 2 '18 at 8:23
$begingroup$
The earth is in free-fall, so its acceleration due to the sun's gravity (or the Milky Way's) does not cause the earth-centered frame to be non-inertial. (Except for even more miniscule effects, e.g. solar radiation pressure.)
$endgroup$
– David
Dec 2 '18 at 8:23
$begingroup$
I wasn't really sure about this. I mean, it would be considered inertial in GR, but I was using a more classical idea. I'm editing the post to address this issue. Thank you.
$endgroup$
– João Vítor G. Lima
Dec 2 '18 at 12:52
$begingroup$
I wasn't really sure about this. I mean, it would be considered inertial in GR, but I was using a more classical idea. I'm editing the post to address this issue. Thank you.
$endgroup$
– João Vítor G. Lima
Dec 2 '18 at 12:52
add a comment |
$begingroup$
Because the earth is rotating, it is never strictly an inertial reference frame. However, because the effects are small in many situations, it can often be approximated as one. When to use Coriolis forces will have to be determined on a case-by-case basis. E.g. ballistic problems that cover large distances will most certainly require Coriolis-force corrections, and pendulums that swings for a long time would also require Coriolis-force corrections. For a block sliding down an inclined plane, or a spring on a mass, or a vibrating string, you should not need to take it into consideration.
answered Dec 2 '18 at 0:01
njspeernjspeer
4374
$endgroup$
add a comment |
$begingroup$
Because the earth is rotating, it is never strictly an inertial reference frame. However, because the effects are small in many situations, it can often be approximated as one. When to use Coriolis forces will have to be determined on a case-by-case basis. E.g. ballistic problems that cover large distances will most certainly require Coriolis-force corrections, and pendulums that swings for a long time would also require Coriolis-force corrections. For a block sliding down an inclined plane, or a spring on a mass, or a vibrating string, you should not need to take it into consideration.
answered Dec 2 '18 at 0:01
njspeernjspeer
4374
$endgroup$
add a comment |
$begingroup$
Because the earth is rotating, it is never strictly an inertial reference frame. However, because the effects are small in many situations, it can often be approximated as one. When to use Coriolis forces will have to be determined on a case-by-case basis. E.g. ballistic problems that cover large distances will most certainly require Coriolis-force corrections, and pendulums that swings for a long time would also require Coriolis-force corrections. For a block sliding down an inclined plane, or a spring on a mass, or a vibrating string, you should not need to take it into consideration.
answered Dec 2 '18 at 0:01
njspeernjspeer
4374
$endgroup$
Because the earth is rotating, it is never strictly an inertial reference frame. However, because the effects are small in many situations, it can often be approximated as one. When to use Coriolis forces will have to be determined on a case-by-case basis. E.g. ballistic problems that cover large distances will most certainly require Coriolis-force corrections, and pendulums that swings for a long time would also require Coriolis-force corrections. For a block sliding down an inclined plane, or a spring on a mass, or a vibrating string, you should not need to take it into consideration.
answered Dec 2 '18 at 0:01
njspeernjspeer
4374
edited Dec 2 '18 at 3:20
answered Dec 2 '18 at 0:01
njspeernjspeer
4374
answered Dec 2 '18 at 0:01
njspeernjspeer
4374
answered Dec 2 '18 at 0:01
njspeernjspeer
4374
4374
add a comment |
add a comment |
$begingroup$
Mach would say that non-inertial effects are due to the relative motion between the earth and the rest of the universe.
See:
• Assis, André K. T. Relational Mechanics and Implementation of Mach’s Principle with Weber’s Gravitational Force. Montréal: Apeiron, 2014.
answered Dec 2 '18 at 4:35
GeremiaGeremia
1,1752927
$endgroup$
add a comment |
$begingroup$
Mach would say that non-inertial effects are due to the relative motion between the earth and the rest of the universe.
See:
• Assis, André K. T. Relational Mechanics and Implementation of Mach’s Principle with Weber’s Gravitational Force. Montréal: Apeiron, 2014.
answered Dec 2 '18 at 4:35
GeremiaGeremia
1,1752927
$endgroup$
add a comment |
$begingroup$
Mach would say that non-inertial effects are due to the relative motion between the earth and the rest of the universe.
See:
• Assis, André K. T. Relational Mechanics and Implementation of Mach’s Principle with Weber’s Gravitational Force. Montréal: Apeiron, 2014.
answered Dec 2 '18 at 4:35
GeremiaGeremia
1,1752927
$endgroup$
Mach would say that non-inertial effects are due to the relative motion between the earth and the rest of the universe.
See:
• Assis, André K. T. Relational Mechanics and Implementation of Mach’s Principle with Weber’s Gravitational Force. Montréal: Apeiron, 2014.
answered Dec 2 '18 at 4:35
GeremiaGeremia
1,1752927
edited Dec 4 '18 at 16:46
answered Dec 2 '18 at 4:35
GeremiaGeremia
1,1752927
answered Dec 2 '18 at 4:35
GeremiaGeremia
1,1752927
answered Dec 2 '18 at 4:35
GeremiaGeremia
1,1752927
1,1752927
add a comment |
add a comment |
$begingroup$
I know it's being a little pedantic, but I would say "earth" is a thing, not a reference frame. You could define an inertial reference frame that contains the earth.
But suppose you mean a reference frame defined with reference to the earth: Z is perpendicular to the ground where you are standing, X and Y are parallel to the ground and perpendicular to each other.
If you stand at rest in this reference frame, you may notice you feel acceleration (gravity). It is therefore not an inertial reference frame.
answered Dec 2 '18 at 15:51
OwenOwen
27729
$endgroup$
add a comment |
$begingroup$
I know it's being a little pedantic, but I would say "earth" is a thing, not a reference frame. You could define an inertial reference frame that contains the earth.
But suppose you mean a reference frame defined with reference to the earth: Z is perpendicular to the ground where you are standing, X and Y are parallel to the ground and perpendicular to each other.
If you stand at rest in this reference frame, you may notice you feel acceleration (gravity). It is therefore not an inertial reference frame.
answered Dec 2 '18 at 15:51
OwenOwen
27729
$endgroup$
add a comment |
$begingroup$
I know it's being a little pedantic, but I would say "earth" is a thing, not a reference frame. You could define an inertial reference frame that contains the earth.
But suppose you mean a reference frame defined with reference to the earth: Z is perpendicular to the ground where you are standing, X and Y are parallel to the ground and perpendicular to each other.
If you stand at rest in this reference frame, you may notice you feel acceleration (gravity). It is therefore not an inertial reference frame.
answered Dec 2 '18 at 15:51
OwenOwen
27729
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I know it's being a little pedantic, but I would say "earth" is a thing, not a reference frame. You could define an inertial reference frame that contains the earth.
But suppose you mean a reference frame defined with reference to the earth: Z is perpendicular to the ground where you are standing, X and Y are parallel to the ground and perpendicular to each other.
If you stand at rest in this reference frame, you may notice you feel acceleration (gravity). It is therefore not an inertial reference frame.
answered Dec 2 '18 at 15:51
OwenOwen
27729
answered Dec 2 '18 at 15:51
OwenOwen
27729
answered Dec 2 '18 at 15:51
OwenOwen
27729
answered Dec 2 '18 at 15:51
OwenOwen
27729
27729
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add a comment |
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you can calculate it's size, e.g, for a BMG 50 cal (853 m/s muzzle velocity) shot at its max effective range (1,800m) and tell us.
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– JEB
Dec 2 '18 at 0:05
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This question seems synonymous with "Does the earth rotate (with respect to the rest of the universe)?"
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– Geremia
Dec 2 '18 at 4:31
2
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That the Earth's surface is not an inertial frame is nicely demonstrated by the Foucault pendulum.
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– Ruslan
Dec 2 '18 at 15:51
1
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It depends on the physics problem you look at whether the Earth can be considered an inertial frame of reference or not.
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– jjack
Dec 2 '18 at 18:14
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To emphasize @jjack's point, it's notoriously hard work to get a Foucault pendulum to work properly because the non-inertial effects, while present, are very small.
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– dmckee♦
Dec 2 '18 at 19:28