Why doesn't the Earth accelerate towards us?
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According to Newton's third law of motion that states that every action has an equal and opposite reaction.
So, if the Earth exerts a gravitational pull on us(people) then even we should exert a force equal and opposite (in terms of direction) on the Earth.
It is intuitive to think that this force is really small to get the Earth to move.But, if we take a look at the second law of motion that states that F = ma we see that however small the force, there will be some amount of acceleration. Therefore, even though we exert a very small gravitational force on the Earth it should be enough to get the Earth to move even though the acceleration is a very small amount.
But what I said clearly does not happen.So there must be some flaw with my reasoning.Can someone point out that flaw.
newtonian-mechanics forces newtonian-gravity reference-frames acceleration
New contributor
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add a comment |
$begingroup$
According to Newton's third law of motion that states that every action has an equal and opposite reaction.
So, if the Earth exerts a gravitational pull on us(people) then even we should exert a force equal and opposite (in terms of direction) on the Earth.
It is intuitive to think that this force is really small to get the Earth to move.But, if we take a look at the second law of motion that states that F = ma we see that however small the force, there will be some amount of acceleration. Therefore, even though we exert a very small gravitational force on the Earth it should be enough to get the Earth to move even though the acceleration is a very small amount.
But what I said clearly does not happen.So there must be some flaw with my reasoning.Can someone point out that flaw.
newtonian-mechanics forces newtonian-gravity reference-frames acceleration
New contributor
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4
$begingroup$
What do you mean that it clearly does not happen? If the acceleration is exceptionally tiny (because the mass is exceptionally large), then maybe you just don't notice?
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– Steeven
2 hours ago
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@Steeven Acceleration is indeed exceptionally tiny because of the mass being exceptionally large.In this case the gravitational force that we exert on the Earth is constant.Therefore, there will be a constant acceleration.The acceleration is and will remain tiny but its velocity will slowly build up due to a constant acceleration. That's what I think
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– Aditya Bharadwaj
2 hours ago
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@Aditya Bharadwaj I agree with Steeven completely. It's tiny so you can't notice it in regular life
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– Theoretical
2 hours ago
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@Theoretical Acceleration is tiny, but that does not mean that its velocity will be tiny. Its velocity will slowly build up due to the constant acceleration.
$endgroup$
– Aditya Bharadwaj
2 hours ago
add a comment |
$begingroup$
According to Newton's third law of motion that states that every action has an equal and opposite reaction.
So, if the Earth exerts a gravitational pull on us(people) then even we should exert a force equal and opposite (in terms of direction) on the Earth.
It is intuitive to think that this force is really small to get the Earth to move.But, if we take a look at the second law of motion that states that F = ma we see that however small the force, there will be some amount of acceleration. Therefore, even though we exert a very small gravitational force on the Earth it should be enough to get the Earth to move even though the acceleration is a very small amount.
But what I said clearly does not happen.So there must be some flaw with my reasoning.Can someone point out that flaw.
newtonian-mechanics forces newtonian-gravity reference-frames acceleration
New contributor
$endgroup$
According to Newton's third law of motion that states that every action has an equal and opposite reaction.
So, if the Earth exerts a gravitational pull on us(people) then even we should exert a force equal and opposite (in terms of direction) on the Earth.
It is intuitive to think that this force is really small to get the Earth to move.But, if we take a look at the second law of motion that states that F = ma we see that however small the force, there will be some amount of acceleration. Therefore, even though we exert a very small gravitational force on the Earth it should be enough to get the Earth to move even though the acceleration is a very small amount.
But what I said clearly does not happen.So there must be some flaw with my reasoning.Can someone point out that flaw.
newtonian-mechanics forces newtonian-gravity reference-frames acceleration
newtonian-mechanics forces newtonian-gravity reference-frames acceleration
New contributor
New contributor
edited 1 hour ago
Qmechanic♦
102k121831166
102k121831166
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asked 2 hours ago
Aditya BharadwajAditya Bharadwaj
182
182
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New contributor
4
$begingroup$
What do you mean that it clearly does not happen? If the acceleration is exceptionally tiny (because the mass is exceptionally large), then maybe you just don't notice?
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– Steeven
2 hours ago
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@Steeven Acceleration is indeed exceptionally tiny because of the mass being exceptionally large.In this case the gravitational force that we exert on the Earth is constant.Therefore, there will be a constant acceleration.The acceleration is and will remain tiny but its velocity will slowly build up due to a constant acceleration. That's what I think
$endgroup$
– Aditya Bharadwaj
2 hours ago
$begingroup$
@Aditya Bharadwaj I agree with Steeven completely. It's tiny so you can't notice it in regular life
$endgroup$
– Theoretical
2 hours ago
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@Theoretical Acceleration is tiny, but that does not mean that its velocity will be tiny. Its velocity will slowly build up due to the constant acceleration.
$endgroup$
– Aditya Bharadwaj
2 hours ago
add a comment |
4
$begingroup$
What do you mean that it clearly does not happen? If the acceleration is exceptionally tiny (because the mass is exceptionally large), then maybe you just don't notice?
$endgroup$
– Steeven
2 hours ago
$begingroup$
@Steeven Acceleration is indeed exceptionally tiny because of the mass being exceptionally large.In this case the gravitational force that we exert on the Earth is constant.Therefore, there will be a constant acceleration.The acceleration is and will remain tiny but its velocity will slowly build up due to a constant acceleration. That's what I think
$endgroup$
– Aditya Bharadwaj
2 hours ago
$begingroup$
@Aditya Bharadwaj I agree with Steeven completely. It's tiny so you can't notice it in regular life
$endgroup$
– Theoretical
2 hours ago
$begingroup$
@Theoretical Acceleration is tiny, but that does not mean that its velocity will be tiny. Its velocity will slowly build up due to the constant acceleration.
$endgroup$
– Aditya Bharadwaj
2 hours ago
4
4
$begingroup$
What do you mean that it clearly does not happen? If the acceleration is exceptionally tiny (because the mass is exceptionally large), then maybe you just don't notice?
$endgroup$
– Steeven
2 hours ago
$begingroup$
What do you mean that it clearly does not happen? If the acceleration is exceptionally tiny (because the mass is exceptionally large), then maybe you just don't notice?
$endgroup$
– Steeven
2 hours ago
$begingroup$
@Steeven Acceleration is indeed exceptionally tiny because of the mass being exceptionally large.In this case the gravitational force that we exert on the Earth is constant.Therefore, there will be a constant acceleration.The acceleration is and will remain tiny but its velocity will slowly build up due to a constant acceleration. That's what I think
$endgroup$
– Aditya Bharadwaj
2 hours ago
$begingroup$
@Steeven Acceleration is indeed exceptionally tiny because of the mass being exceptionally large.In this case the gravitational force that we exert on the Earth is constant.Therefore, there will be a constant acceleration.The acceleration is and will remain tiny but its velocity will slowly build up due to a constant acceleration. That's what I think
$endgroup$
– Aditya Bharadwaj
2 hours ago
$begingroup$
@Aditya Bharadwaj I agree with Steeven completely. It's tiny so you can't notice it in regular life
$endgroup$
– Theoretical
2 hours ago
$begingroup$
@Aditya Bharadwaj I agree with Steeven completely. It's tiny so you can't notice it in regular life
$endgroup$
– Theoretical
2 hours ago
$begingroup$
@Theoretical Acceleration is tiny, but that does not mean that its velocity will be tiny. Its velocity will slowly build up due to the constant acceleration.
$endgroup$
– Aditya Bharadwaj
2 hours ago
$begingroup$
@Theoretical Acceleration is tiny, but that does not mean that its velocity will be tiny. Its velocity will slowly build up due to the constant acceleration.
$endgroup$
– Aditya Bharadwaj
2 hours ago
add a comment |
1 Answer
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The acceleration that your gravitational pull causes in the Earth is tiny, tiny, tiny because the Earth's mass is enormous. If your mass is, say, $70;mathrm{kg}$, then you cause an acceleration of $aapprox 1.1times 10^{-22};mathrm{m/s^2}$.
A tiny, tiny, tiny acceleration does not necessarily mean a tiny, tiny, tiny speed, since, as you mention in comments, the velocity accumulates. True. It doesn't necessarily mean that - but in this case it does. The speed gained after 1 year at this acceleration is only $vapprox 3.6×10^{-15};mathrm{m/s}$. And after a lifetime of 100 years it is still only around $vapprox 3.6×10^{-13};mathrm{m/s}$.
If all 7.6 billion people on the planet suddenly lifted off of Earth and stayed hanging in the air on the same side of the planet for 100 years, the planet would reach no more than $vapprox 2.8times 10^{^-3};mathrm{m/s}$; that is around 3 millimeters-per-second in this obscure scenario of 100 years and billions of people's masses.
Now, with all that being said, note that I have to assumed that all those people are not just standing on the ground - they must be levitating above the ground.
Because, while levitating (i.e. during free-fall), they only exert the gravitational force $F_g$:
$$sum F=maquadLeftrightarrowquad F_g=ma$$
and there is a net acceleration. But when standing on the ground, they also exert a downwards pushing force which is their weight $w$:
$$sum F=maquadLeftrightarrowquad F_g-w=ma$$
Now there are two forces on the Earth, pushing in opposite directions. And in fact, that force called weight exactly equals the gravitational force (because those two are the action-reaction pair from Newton's 3rd law). So the above formula gives zero acceleration. The forces cancel out and nothing accelerates any further.
In general, any system can never accelerate purely by it's own internal forces. If we consider the Earth-and-people as one system, then their gravitational forces on each other are internal. Each part of the system may move individually - the Earth can move towards the people and the free-falling people can move towards the Earth. But the system as a whole - defined by the centre-of-mass of the system - will not move anywhere.
So, the Earth can move a tiny, tiny, tiny bit towards you while you move a much larger distance towards the Earth during your free-fall so the combined centre-of-mass is still stationary. But when standing on the ground, nothing can move because that would require you to break through the ground and move inwards into the Earth. If the Earth was moving but you weren't, then the centre of mass would be moving (accelerating) and that is simply impossible. The system would be gaining kinetic energy without any external energy input; creating free energy out of thin air is just not possible. So this never happens.
$endgroup$
1
$begingroup$
Alright so if I am standing on the Earth's surface then I get squeezed to the Earth's surface and exert a pushing force on the Earth's surface that cancels out the pulling gravitational force that I exert on the Earth. If I somehow manage to levitate then the velocity will take a really really really long time to build up for us to notice.
$endgroup$
– Aditya Bharadwaj
1 hour ago
1
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@AdityaBharadwaj Exactly.
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– Steeven
1 hour ago
1
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Thanks a lot man.
$endgroup$
– Aditya Bharadwaj
1 hour ago
1
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@Aditya note also that, unless you're at the North or South poles, if you were levitating over a constant spot of the ground, then the acceleration you'd cause on the Earth would be rotating once a day, and the bulk of the contributions would cancel, as the velocity accumulated during the day would point opposite to that accumulated at night. Only the North-South component, proportional to the sine of your latitude, would survive.
$endgroup$
– Emilio Pisanty
40 mins ago
$begingroup$
Reminds me on what-if.xkcd.com/8
$endgroup$
– Holger
14 mins ago
add a comment |
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$begingroup$
The acceleration that your gravitational pull causes in the Earth is tiny, tiny, tiny because the Earth's mass is enormous. If your mass is, say, $70;mathrm{kg}$, then you cause an acceleration of $aapprox 1.1times 10^{-22};mathrm{m/s^2}$.
A tiny, tiny, tiny acceleration does not necessarily mean a tiny, tiny, tiny speed, since, as you mention in comments, the velocity accumulates. True. It doesn't necessarily mean that - but in this case it does. The speed gained after 1 year at this acceleration is only $vapprox 3.6×10^{-15};mathrm{m/s}$. And after a lifetime of 100 years it is still only around $vapprox 3.6×10^{-13};mathrm{m/s}$.
If all 7.6 billion people on the planet suddenly lifted off of Earth and stayed hanging in the air on the same side of the planet for 100 years, the planet would reach no more than $vapprox 2.8times 10^{^-3};mathrm{m/s}$; that is around 3 millimeters-per-second in this obscure scenario of 100 years and billions of people's masses.
Now, with all that being said, note that I have to assumed that all those people are not just standing on the ground - they must be levitating above the ground.
Because, while levitating (i.e. during free-fall), they only exert the gravitational force $F_g$:
$$sum F=maquadLeftrightarrowquad F_g=ma$$
and there is a net acceleration. But when standing on the ground, they also exert a downwards pushing force which is their weight $w$:
$$sum F=maquadLeftrightarrowquad F_g-w=ma$$
Now there are two forces on the Earth, pushing in opposite directions. And in fact, that force called weight exactly equals the gravitational force (because those two are the action-reaction pair from Newton's 3rd law). So the above formula gives zero acceleration. The forces cancel out and nothing accelerates any further.
In general, any system can never accelerate purely by it's own internal forces. If we consider the Earth-and-people as one system, then their gravitational forces on each other are internal. Each part of the system may move individually - the Earth can move towards the people and the free-falling people can move towards the Earth. But the system as a whole - defined by the centre-of-mass of the system - will not move anywhere.
So, the Earth can move a tiny, tiny, tiny bit towards you while you move a much larger distance towards the Earth during your free-fall so the combined centre-of-mass is still stationary. But when standing on the ground, nothing can move because that would require you to break through the ground and move inwards into the Earth. If the Earth was moving but you weren't, then the centre of mass would be moving (accelerating) and that is simply impossible. The system would be gaining kinetic energy without any external energy input; creating free energy out of thin air is just not possible. So this never happens.
$endgroup$
1
$begingroup$
Alright so if I am standing on the Earth's surface then I get squeezed to the Earth's surface and exert a pushing force on the Earth's surface that cancels out the pulling gravitational force that I exert on the Earth. If I somehow manage to levitate then the velocity will take a really really really long time to build up for us to notice.
$endgroup$
– Aditya Bharadwaj
1 hour ago
1
$begingroup$
@AdityaBharadwaj Exactly.
$endgroup$
– Steeven
1 hour ago
1
$begingroup$
Thanks a lot man.
$endgroup$
– Aditya Bharadwaj
1 hour ago
1
$begingroup$
@Aditya note also that, unless you're at the North or South poles, if you were levitating over a constant spot of the ground, then the acceleration you'd cause on the Earth would be rotating once a day, and the bulk of the contributions would cancel, as the velocity accumulated during the day would point opposite to that accumulated at night. Only the North-South component, proportional to the sine of your latitude, would survive.
$endgroup$
– Emilio Pisanty
40 mins ago
$begingroup$
Reminds me on what-if.xkcd.com/8
$endgroup$
– Holger
14 mins ago
add a comment |
$begingroup$
The acceleration that your gravitational pull causes in the Earth is tiny, tiny, tiny because the Earth's mass is enormous. If your mass is, say, $70;mathrm{kg}$, then you cause an acceleration of $aapprox 1.1times 10^{-22};mathrm{m/s^2}$.
A tiny, tiny, tiny acceleration does not necessarily mean a tiny, tiny, tiny speed, since, as you mention in comments, the velocity accumulates. True. It doesn't necessarily mean that - but in this case it does. The speed gained after 1 year at this acceleration is only $vapprox 3.6×10^{-15};mathrm{m/s}$. And after a lifetime of 100 years it is still only around $vapprox 3.6×10^{-13};mathrm{m/s}$.
If all 7.6 billion people on the planet suddenly lifted off of Earth and stayed hanging in the air on the same side of the planet for 100 years, the planet would reach no more than $vapprox 2.8times 10^{^-3};mathrm{m/s}$; that is around 3 millimeters-per-second in this obscure scenario of 100 years and billions of people's masses.
Now, with all that being said, note that I have to assumed that all those people are not just standing on the ground - they must be levitating above the ground.
Because, while levitating (i.e. during free-fall), they only exert the gravitational force $F_g$:
$$sum F=maquadLeftrightarrowquad F_g=ma$$
and there is a net acceleration. But when standing on the ground, they also exert a downwards pushing force which is their weight $w$:
$$sum F=maquadLeftrightarrowquad F_g-w=ma$$
Now there are two forces on the Earth, pushing in opposite directions. And in fact, that force called weight exactly equals the gravitational force (because those two are the action-reaction pair from Newton's 3rd law). So the above formula gives zero acceleration. The forces cancel out and nothing accelerates any further.
In general, any system can never accelerate purely by it's own internal forces. If we consider the Earth-and-people as one system, then their gravitational forces on each other are internal. Each part of the system may move individually - the Earth can move towards the people and the free-falling people can move towards the Earth. But the system as a whole - defined by the centre-of-mass of the system - will not move anywhere.
So, the Earth can move a tiny, tiny, tiny bit towards you while you move a much larger distance towards the Earth during your free-fall so the combined centre-of-mass is still stationary. But when standing on the ground, nothing can move because that would require you to break through the ground and move inwards into the Earth. If the Earth was moving but you weren't, then the centre of mass would be moving (accelerating) and that is simply impossible. The system would be gaining kinetic energy without any external energy input; creating free energy out of thin air is just not possible. So this never happens.
$endgroup$
1
$begingroup$
Alright so if I am standing on the Earth's surface then I get squeezed to the Earth's surface and exert a pushing force on the Earth's surface that cancels out the pulling gravitational force that I exert on the Earth. If I somehow manage to levitate then the velocity will take a really really really long time to build up for us to notice.
$endgroup$
– Aditya Bharadwaj
1 hour ago
1
$begingroup$
@AdityaBharadwaj Exactly.
$endgroup$
– Steeven
1 hour ago
1
$begingroup$
Thanks a lot man.
$endgroup$
– Aditya Bharadwaj
1 hour ago
1
$begingroup$
@Aditya note also that, unless you're at the North or South poles, if you were levitating over a constant spot of the ground, then the acceleration you'd cause on the Earth would be rotating once a day, and the bulk of the contributions would cancel, as the velocity accumulated during the day would point opposite to that accumulated at night. Only the North-South component, proportional to the sine of your latitude, would survive.
$endgroup$
– Emilio Pisanty
40 mins ago
$begingroup$
Reminds me on what-if.xkcd.com/8
$endgroup$
– Holger
14 mins ago
add a comment |
$begingroup$
The acceleration that your gravitational pull causes in the Earth is tiny, tiny, tiny because the Earth's mass is enormous. If your mass is, say, $70;mathrm{kg}$, then you cause an acceleration of $aapprox 1.1times 10^{-22};mathrm{m/s^2}$.
A tiny, tiny, tiny acceleration does not necessarily mean a tiny, tiny, tiny speed, since, as you mention in comments, the velocity accumulates. True. It doesn't necessarily mean that - but in this case it does. The speed gained after 1 year at this acceleration is only $vapprox 3.6×10^{-15};mathrm{m/s}$. And after a lifetime of 100 years it is still only around $vapprox 3.6×10^{-13};mathrm{m/s}$.
If all 7.6 billion people on the planet suddenly lifted off of Earth and stayed hanging in the air on the same side of the planet for 100 years, the planet would reach no more than $vapprox 2.8times 10^{^-3};mathrm{m/s}$; that is around 3 millimeters-per-second in this obscure scenario of 100 years and billions of people's masses.
Now, with all that being said, note that I have to assumed that all those people are not just standing on the ground - they must be levitating above the ground.
Because, while levitating (i.e. during free-fall), they only exert the gravitational force $F_g$:
$$sum F=maquadLeftrightarrowquad F_g=ma$$
and there is a net acceleration. But when standing on the ground, they also exert a downwards pushing force which is their weight $w$:
$$sum F=maquadLeftrightarrowquad F_g-w=ma$$
Now there are two forces on the Earth, pushing in opposite directions. And in fact, that force called weight exactly equals the gravitational force (because those two are the action-reaction pair from Newton's 3rd law). So the above formula gives zero acceleration. The forces cancel out and nothing accelerates any further.
In general, any system can never accelerate purely by it's own internal forces. If we consider the Earth-and-people as one system, then their gravitational forces on each other are internal. Each part of the system may move individually - the Earth can move towards the people and the free-falling people can move towards the Earth. But the system as a whole - defined by the centre-of-mass of the system - will not move anywhere.
So, the Earth can move a tiny, tiny, tiny bit towards you while you move a much larger distance towards the Earth during your free-fall so the combined centre-of-mass is still stationary. But when standing on the ground, nothing can move because that would require you to break through the ground and move inwards into the Earth. If the Earth was moving but you weren't, then the centre of mass would be moving (accelerating) and that is simply impossible. The system would be gaining kinetic energy without any external energy input; creating free energy out of thin air is just not possible. So this never happens.
$endgroup$
The acceleration that your gravitational pull causes in the Earth is tiny, tiny, tiny because the Earth's mass is enormous. If your mass is, say, $70;mathrm{kg}$, then you cause an acceleration of $aapprox 1.1times 10^{-22};mathrm{m/s^2}$.
A tiny, tiny, tiny acceleration does not necessarily mean a tiny, tiny, tiny speed, since, as you mention in comments, the velocity accumulates. True. It doesn't necessarily mean that - but in this case it does. The speed gained after 1 year at this acceleration is only $vapprox 3.6×10^{-15};mathrm{m/s}$. And after a lifetime of 100 years it is still only around $vapprox 3.6×10^{-13};mathrm{m/s}$.
If all 7.6 billion people on the planet suddenly lifted off of Earth and stayed hanging in the air on the same side of the planet for 100 years, the planet would reach no more than $vapprox 2.8times 10^{^-3};mathrm{m/s}$; that is around 3 millimeters-per-second in this obscure scenario of 100 years and billions of people's masses.
Now, with all that being said, note that I have to assumed that all those people are not just standing on the ground - they must be levitating above the ground.
Because, while levitating (i.e. during free-fall), they only exert the gravitational force $F_g$:
$$sum F=maquadLeftrightarrowquad F_g=ma$$
and there is a net acceleration. But when standing on the ground, they also exert a downwards pushing force which is their weight $w$:
$$sum F=maquadLeftrightarrowquad F_g-w=ma$$
Now there are two forces on the Earth, pushing in opposite directions. And in fact, that force called weight exactly equals the gravitational force (because those two are the action-reaction pair from Newton's 3rd law). So the above formula gives zero acceleration. The forces cancel out and nothing accelerates any further.
In general, any system can never accelerate purely by it's own internal forces. If we consider the Earth-and-people as one system, then their gravitational forces on each other are internal. Each part of the system may move individually - the Earth can move towards the people and the free-falling people can move towards the Earth. But the system as a whole - defined by the centre-of-mass of the system - will not move anywhere.
So, the Earth can move a tiny, tiny, tiny bit towards you while you move a much larger distance towards the Earth during your free-fall so the combined centre-of-mass is still stationary. But when standing on the ground, nothing can move because that would require you to break through the ground and move inwards into the Earth. If the Earth was moving but you weren't, then the centre of mass would be moving (accelerating) and that is simply impossible. The system would be gaining kinetic energy without any external energy input; creating free energy out of thin air is just not possible. So this never happens.
edited 59 mins ago
answered 1 hour ago
SteevenSteeven
26.3k560107
26.3k560107
1
$begingroup$
Alright so if I am standing on the Earth's surface then I get squeezed to the Earth's surface and exert a pushing force on the Earth's surface that cancels out the pulling gravitational force that I exert on the Earth. If I somehow manage to levitate then the velocity will take a really really really long time to build up for us to notice.
$endgroup$
– Aditya Bharadwaj
1 hour ago
1
$begingroup$
@AdityaBharadwaj Exactly.
$endgroup$
– Steeven
1 hour ago
1
$begingroup$
Thanks a lot man.
$endgroup$
– Aditya Bharadwaj
1 hour ago
1
$begingroup$
@Aditya note also that, unless you're at the North or South poles, if you were levitating over a constant spot of the ground, then the acceleration you'd cause on the Earth would be rotating once a day, and the bulk of the contributions would cancel, as the velocity accumulated during the day would point opposite to that accumulated at night. Only the North-South component, proportional to the sine of your latitude, would survive.
$endgroup$
– Emilio Pisanty
40 mins ago
$begingroup$
Reminds me on what-if.xkcd.com/8
$endgroup$
– Holger
14 mins ago
add a comment |
1
$begingroup$
Alright so if I am standing on the Earth's surface then I get squeezed to the Earth's surface and exert a pushing force on the Earth's surface that cancels out the pulling gravitational force that I exert on the Earth. If I somehow manage to levitate then the velocity will take a really really really long time to build up for us to notice.
$endgroup$
– Aditya Bharadwaj
1 hour ago
1
$begingroup$
@AdityaBharadwaj Exactly.
$endgroup$
– Steeven
1 hour ago
1
$begingroup$
Thanks a lot man.
$endgroup$
– Aditya Bharadwaj
1 hour ago
1
$begingroup$
@Aditya note also that, unless you're at the North or South poles, if you were levitating over a constant spot of the ground, then the acceleration you'd cause on the Earth would be rotating once a day, and the bulk of the contributions would cancel, as the velocity accumulated during the day would point opposite to that accumulated at night. Only the North-South component, proportional to the sine of your latitude, would survive.
$endgroup$
– Emilio Pisanty
40 mins ago
$begingroup$
Reminds me on what-if.xkcd.com/8
$endgroup$
– Holger
14 mins ago
1
1
$begingroup$
Alright so if I am standing on the Earth's surface then I get squeezed to the Earth's surface and exert a pushing force on the Earth's surface that cancels out the pulling gravitational force that I exert on the Earth. If I somehow manage to levitate then the velocity will take a really really really long time to build up for us to notice.
$endgroup$
– Aditya Bharadwaj
1 hour ago
$begingroup$
Alright so if I am standing on the Earth's surface then I get squeezed to the Earth's surface and exert a pushing force on the Earth's surface that cancels out the pulling gravitational force that I exert on the Earth. If I somehow manage to levitate then the velocity will take a really really really long time to build up for us to notice.
$endgroup$
– Aditya Bharadwaj
1 hour ago
1
1
$begingroup$
@AdityaBharadwaj Exactly.
$endgroup$
– Steeven
1 hour ago
$begingroup$
@AdityaBharadwaj Exactly.
$endgroup$
– Steeven
1 hour ago
1
1
$begingroup$
Thanks a lot man.
$endgroup$
– Aditya Bharadwaj
1 hour ago
$begingroup$
Thanks a lot man.
$endgroup$
– Aditya Bharadwaj
1 hour ago
1
1
$begingroup$
@Aditya note also that, unless you're at the North or South poles, if you were levitating over a constant spot of the ground, then the acceleration you'd cause on the Earth would be rotating once a day, and the bulk of the contributions would cancel, as the velocity accumulated during the day would point opposite to that accumulated at night. Only the North-South component, proportional to the sine of your latitude, would survive.
$endgroup$
– Emilio Pisanty
40 mins ago
$begingroup$
@Aditya note also that, unless you're at the North or South poles, if you were levitating over a constant spot of the ground, then the acceleration you'd cause on the Earth would be rotating once a day, and the bulk of the contributions would cancel, as the velocity accumulated during the day would point opposite to that accumulated at night. Only the North-South component, proportional to the sine of your latitude, would survive.
$endgroup$
– Emilio Pisanty
40 mins ago
$begingroup$
Reminds me on what-if.xkcd.com/8
$endgroup$
– Holger
14 mins ago
$begingroup$
Reminds me on what-if.xkcd.com/8
$endgroup$
– Holger
14 mins ago
add a comment |
Aditya Bharadwaj is a new contributor. Be nice, and check out our Code of Conduct.
Aditya Bharadwaj is a new contributor. Be nice, and check out our Code of Conduct.
Aditya Bharadwaj is a new contributor. Be nice, and check out our Code of Conduct.
Aditya Bharadwaj is a new contributor. Be nice, and check out our Code of Conduct.
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$begingroup$
What do you mean that it clearly does not happen? If the acceleration is exceptionally tiny (because the mass is exceptionally large), then maybe you just don't notice?
$endgroup$
– Steeven
2 hours ago
$begingroup$
@Steeven Acceleration is indeed exceptionally tiny because of the mass being exceptionally large.In this case the gravitational force that we exert on the Earth is constant.Therefore, there will be a constant acceleration.The acceleration is and will remain tiny but its velocity will slowly build up due to a constant acceleration. That's what I think
$endgroup$
– Aditya Bharadwaj
2 hours ago
$begingroup$
@Aditya Bharadwaj I agree with Steeven completely. It's tiny so you can't notice it in regular life
$endgroup$
– Theoretical
2 hours ago
$begingroup$
@Theoretical Acceleration is tiny, but that does not mean that its velocity will be tiny. Its velocity will slowly build up due to the constant acceleration.
$endgroup$
– Aditya Bharadwaj
2 hours ago