Is Earth an inertial reference frame?
up vote
2
down vote
favorite
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
add a comment |
up vote
2
down vote
favorite
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
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.
– JEB
4 hours ago
This question seems synonymous with "Does the earth rotate (with respect to the rest of the universe)?"
– Geremia
31 mins ago
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
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
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 2 hours ago
Qmechanic♦
100k121791123
100k121791123
asked 5 hours ago
ado sar
1027
1027
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.
– JEB
4 hours ago
This question seems synonymous with "Does the earth rotate (with respect to the rest of the universe)?"
– Geremia
31 mins ago
add a comment |
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.
– JEB
4 hours ago
This question seems synonymous with "Does the earth rotate (with respect to the rest of the universe)?"
– Geremia
31 mins ago
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.
– JEB
4 hours ago
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.
– JEB
4 hours ago
This question seems synonymous with "Does the earth rotate (with respect to the rest of the universe)?"
– Geremia
31 mins ago
This question seems synonymous with "Does the earth rotate (with respect to the rest of the universe)?"
– Geremia
31 mins ago
add a comment |
3 Answers
3
active
oldest
votes
up vote
3
down vote
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 also not inertial due to the fact that the planet accelerates while orbiting the Sun and the Milky Way.
add a comment |
up vote
1
down vote
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.
add a comment |
up vote
0
down vote
Mach would say that the earth could be considered at rest and non-inertial effects are due to 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.
add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
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 also not inertial due to the fact that the planet accelerates while orbiting the Sun and the Milky Way.
add a comment |
up vote
3
down vote
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 also not inertial due to the fact that the planet accelerates while orbiting the Sun and the Milky Way.
add a comment |
up vote
3
down vote
up vote
3
down vote
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 also not inertial due to the fact that the planet accelerates while orbiting the Sun and the Milky Way.
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 also not inertial due to the fact that the planet accelerates while orbiting the Sun and the Milky Way.
answered 4 hours ago
João Vítor G. Lima
618118
618118
add a comment |
add a comment |
up vote
1
down vote
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.
add a comment |
up vote
1
down vote
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.
add a comment |
up vote
1
down vote
up vote
1
down vote
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.
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.
edited 1 hour ago
answered 5 hours ago
njspeer
3594
3594
add a comment |
add a comment |
up vote
0
down vote
Mach would say that the earth could be considered at rest and non-inertial effects are due to 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.
add a comment |
up vote
0
down vote
Mach would say that the earth could be considered at rest and non-inertial effects are due to 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.
add a comment |
up vote
0
down vote
up vote
0
down vote
Mach would say that the earth could be considered at rest and non-inertial effects are due to 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.
Mach would say that the earth could be considered at rest and non-inertial effects are due to 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 27 mins ago
Geremia
1,1322927
1,1322927
add a comment |
add a comment |
Thanks for contributing an answer to Physics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fphysics.stackexchange.com%2fquestions%2f444580%2fis-earth-an-inertial-reference-frame%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
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.
– JEB
4 hours ago
This question seems synonymous with "Does the earth rotate (with respect to the rest of the universe)?"
– Geremia
31 mins ago