How much mass with the centrifugal forces of earth to break gravity?

In summary: I missing here?You are missing the idea that centrifugal force is a rate of energy transfer. Centrifugal force is a rate of energy transfer because it is a force that opposes the acceleration of gravity on an object.
  • #1
d w
16
0
I'll start with this question, how much "mass" does something have to have before centrifugal force exceeds gravity on earth?

I used basic physics of centrifugal force and gravity, used force vectors, and the math doesn't jive. This is what I mean.

There are 2 forces on Earth that everything experiences if the Earth rotates about its axis and orbits around the sun.

Gravity and centrifugal force are quantified algebraically by acceleration (unit of distance)/(unit of time)^2 and thus they interact as force vectors.

When you look at the values for objects with a mass of 4250kg or more, you will see the centrifugal acceleration begins to exceed the acceleration of gravity.

I used these figures [when calculating centrifugal acceleration]:

Earth's orbit values:

Velocity 29,722ms−1

Radius 149,668,992,000m

Mass ≥22,679.6kg

Earth's rotation values:

Velocity 464.922ms−1 @ the equator (this value decreases down to zero as it approaches and reaches the "poles")

Radius 6,371,390m

Mass ≥4,250kg

Moon orbit values:

Velocity 1023.1ms−1

Radius 384,400,000m

Mass 7.34767309×10^22kg

With these values and their products in mind, notice that the force vectors begin to exceed gravity and counter any claim that the Earth is spinning or that the moon could orbit the earth. Is this right?

Value I used for constant of gravity is 143 N

Any value that exceeds 143 N from the centrifugal equation, is force away from Earth exceeding gravity towards earth.
 
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  • #2
d w said:
Is this right?

No.

The gravitational pull on the Moon is pretty much exactly what it needs to be to provide the correct centripetal acceleration.

It is also not clear why you think the mass has anything to do with things. Both the gravitational force and the required centripetal force are directly proportional to the mass and therefore cancels out in the comparison.
 
  • #3
Orodruin said:
No.

The gravitational pull on the Moon is pretty much exactly what it needs to be to provide the correct centripetal acceleration.

It is also not clear why you think the mass has anything to do with things. Both the gravitational force and the required centripetal force are directly proportional to the mass and therefore cancels out in the comparison.

Hi and thanks for responding, I was looking for someone who knew the math and physics to respond with an explanation though because I want them identify the error in the math, not the conclusion. The conclusion is stated in the math, so that is why I want to see the math error. Thank you :)
 
  • #4
Centripetal acceleration is independent of mass, and is found by [itex]\frac{v^2}{r}[/itex]

At the equator this works out to 0.0339 m/s^2 vs 9.8m/s^2 squared for the acceleration due to gravity.

For the Moon orbiting the Earth, the centripetal acceleration is ~0.0027 m/sec^2 and the acceleration due to gravity is ~0.0027, which is what you expect for an orbiting object.
 
  • #5
d w said:
Hi and thanks for responding, I was looking for someone who knew the math and physics to respond with an explanation though because I want them identify the error in the math, not the conclusion. The conclusion is stated in the math, so that is why I want to see the math error. Thank you :)
In order for us to show the math/physics error you made, you need to show the math you used to get your answers, not just the numbers you plugged in.
 
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  • #6
Janus said:
Centripetal acceleration is independent of mass, and is found by [itex]\frac{v^2}{r}[/itex]

At the equator this works out to 0.0339 m/s^2 vs 9.8m/s^2 squared for the acceleration due to gravity.

For the Moon orbiting the Earth, the centripetal acceleration is ~0.0027 m/sec^2 and the acceleration due to gravity is ~0.0027, which is what you expect for an orbiting object.

Hi, thank you also.

According to Newton's second law the centripetal force can be expressed as

Fc = m ac

= m v2 / r

Are you saying that the centrifugal force is not derived from the same equation and centripetal force? Or are you saying that the result of centrifugal force is not the actual acceleration multiplier on the mass? I get that F has the mass/acceleration mixed in, but even "mass" has mass/acceleration mixed in because the "mass" is a result of gravity on the mass. Gravity is a constant so they use that constant to form weights. I am saying the energy expressed in this equation is a rate of energy transfer. Newtons are really a rate of energy transfer, but it has to have time or distance for it to express the rate.

So far this is what I think, centrifugal force is on mass, so it isn't "independent" of mass. Centrifugal force expresses the acceleration on the weight in Newtons. Gravity force is expressed as weight in Newtons as 143 N. Since I am expressing gravity this way, I expressed centrifugal force the same way. Both my equations are expressing the effects of acceleration as force, no?
 
  • #7
You still have not specified how you have compared this with the gravitational force. You need to show what you have actually done and not assume that everybody is a mind reader, in particular as it is clear that you have misunderstood something along the way.
 
  • #8
Janus said:
In order for us to show the math/physics error you made, you need to show the math you used to get your answers, not just the numbers you plugged in.
Since it is all basic math and physics, I was hoping to discuss it with someone who already knew how to work the problem out. If I explain it, I have to explain centrifugal force, units, gravity, force vectors, and work that all out on a keyboard. I rather discuss it with someone who already knows that part of the problem. If you can't see how I arrived at my conclusion in the first place, you can't help.
 
  • #9
Orodruin said:
You still have not specified how you have compared this with the gravitational force. You need to show what you have actually done and not assume that everybody is a mind reader, in particular as it is clear that you have misunderstood something along the way.

Okay, I added some information. If you are looking for key numbers, sure, but if you want me to work it all out, it is really just 2 forces (expressed as acceleration in Newtons), 2 force vectors, 1 mass. Easy.
 
  • #10
d w said:
Since it is all basic math and physics, I was hoping to discuss it with someone who already knew how to work the problem out. If I explain it, I have to explain centrifugal force, units, gravity, force vectors, and work that all out on a keyboard. I rather discuss it with someone who already knows that part of the problem. If you can't see how I arrived at my conclusion in the first place, you can't help.
Since what you are doing is very basic and you are obviously doing it wrong, the task is with you to actually provide your computations. Most people here are familiar with basic gravity and mechanics, which you do not seem to be. You have already been provided with the correct arguments by Janus. It is up to you to either show your work or where you think he is wrong.
 
  • #11
Orodruin said:
Since what you are doing is very basic and you are obviously doing it wrong, the task is with you to actually provide your computations. Most people here are familiar with basic gravity and mechanics, which you do not seem to be. You have already been provided with the correct arguments by Janus. It is up to you to either show your work or where you think he is wrong.

Let me ask you a question. Have you worked out the math yet? IF you have, did you come to a different answer for the amount of weight that it would take for the centrifugal forces to equal the force of gravity?
 
  • #12
d w said:
Value I used for constant of gravity is 143 N
This is a value taken from nowhere and has nothing to do with how gravitation works on a general object. The gravitational force on a body depends on its mass, you cannot just assume it to be a particular value for all objects.
 
  • #13
Orodruin said:
This is a value taken from nowhere and has nothing to do with how gravitation works on a general object. The gravitational force on a body depends on its mass, you cannot just assume it to be a particular value for all objects.

From Wikipedia, "
One slug has a mass of 32.174049 lbm or 14.593903 kg based on standard gravity, the international foot, and the avoirdupois pound.[1] At the surface of the Earth, an object with a mass of 1 slug exerts a force of approximately 32.2 lbF or 143 N.[2][3]"

Notice that 143 N is not out of nowhere.
 
  • #14
d w said:
Notice that 143 N is not out of nowhere.
No, because it is specified which mass it belongs to. It is the force on 1 slug in a gravitational field equal to that at the Earth's surface. It is therefore completely arbitrary and will depend on what mass you are using as well as on the actual gravitational field, which changes with the distance from the gravitational source.

You cannot use it as the gravitational force on an arbitrary mass at an arbitrary distance.
 
  • #15
Orodruin said:
No, because it is specified which mass it belongs to. It is the force on 1 slug in a gravitational field equal to that at the Earth's surface. It is therefore completely arbitrary and will depend on what mass you are using as well as on the actual gravitational field, which changes with the distance from the gravitational source.

The constant of gravity is considered 9.8 m/s^2, no? IF that is a constant, the force of acceleration can be expressed as a force by converting to units using 1kg m/s^2 = 1 Newton. Since that is a measure of energy and acceleration is a demonstration of energy, they translate. Perhaps you have to see through the units a little, but the units represent ideas. The idea represented in acceleration and force is strength of energy force does it by saying "on mass" is all, and then over time it become rate of energy.
 
  • #16
d w said:
The constant of gravity is considered 9.8 m/s^2, no? IF that is a constant, the force of acceleration can be expressed as a force by converting to units using 1kg m/s^2 = 1 Newton.
No, you simply cannot do this. Force has physical dimension of mass*length/time^2 and the gravitational constant has physical dimensions of length/time^2 (just as acceleration). You will notice that m/s^2 is not the same as kg m/s^2... To find the gravitational force, you need to multiply by the mass of the object the gravitational field acts on, the result of which will have the correct physical dimension.

In addition, the gravitational field is 9.8 m/s^2 only at the Earth's surface. It varies with distance.

d w said:
The idea represented in acceleration and force is strength of energy and then over time it become rate of energy.
I am sorry, but these are just ramblings which have nothing to do with actual physics.
 
  • #17
Orodruin said:
No, you simply cannot do this. Force has physical dimension of mass*length/time^2 and the gravitational constant has physical dimensions of length/time^2 (just as acceleration). You will notice that m/s^2 is not the same as kg m/s^2... To find the gravitational force, you need to multiply by the mass of the object the gravitational field acts on, the result of which will have the correct physical dimension.I am sorry, but these are just ramblings which have nothing to do with actual physics.
I disagree with you, but, let us solve it this way. With the known Earth orbit and rotation values, how much mass is needed to equal the value of gravity under centrifugal force?
 
  • #18
@d w when you multiply the centripetal (or -fugal) acceleration by mass you end up with centripetal force. Similarly, when you multiply gravitational acceleration by mass you end up with gravitational force. Both are proportional to the mass you're multiplying by, so their ratio doesn't change in the process.
If one's greater than the other, it stays greater no matter the mass (unless it's really huge and messes up the orbit).

You can not compare acceleration with force and expect it to mean anything.
 
  • #19
d w said:
I disagree with you, but, let us solve it this way. With the known Earth orbit and rotation values, how much mass is needed to equal the value of gravity under centrifugal force?
There really is not anything to disagree with here, this has been well known since the time of Newton. If you have a different idea about how Newtonian gravity works, it has long since been disproved by observations.
 
  • #20
Bandersnatch said:
@d w when you multiply the centripetal (or -fugal) acceleration by mass you end up with centripetal force. Similarly, when you multiply gravitational acceleration by mass you end up with gravitational force. Both are proportional to the mass you're multiplying by, so their ratio doesn't change in the process.
If one's greater than the other, it stays greater no matter the mass (unless it's really huge and messes up the orbit).

You can not compare acceleration with force and expect it to mean anything.

Hi, the question I pose to you then is, what is the critical mass for when centrifugal force takes over gravity with the known values in our universe?

Earth's orbit values.

Velocity 29,722 m/s

Radius 149,668,992,000 m

Mass ? (I got 22,679.6 kg or greater)Earth's rotation values.

Velocity 464.922 m/s @ the equator (this value decreases down to zero as it approaches and reaches the "poles")

Radius 6,371,390 m

Mass ? (I got 4,250 kg or greater)
 
  • #21
Orodruin said:
There really is not anything to disagree with here, this has been well known since the time of Newton. If you have a different idea about how Newtonian gravity works, it has long since been disproved by observations.

Like I said, the question I pose to you then is, what is the critical mass for when centrifugal force takes over gravity with the known values in our universe?

Earth's orbit values.

Velocity 29,722 m/s

Radius 149,668,992,000 m

Mass ? (I got 22,679.6 kg or greater)Earth's rotation values.

Velocity 464.922 m/s @ the equator (this value decreases down to zero as it approaches and reaches the "poles")

Radius 6,371,390 m

Mass ? (I got 4,250 kg or greater)
 
  • #22
Orodruin said:
There really is not anything to disagree with here, this has been well known since the time of Newton. If you have a different idea about how Newtonian gravity works, it has long since been disproved by observations.

"There really is not anything to disagree with here, this has been well known since the time of Newton. If you have a different idea about how Newtonian gravity works, it has long since been disproved by observations."

I like how you assume your interpretation is absolutely the most correct it can be and that because I don't sound like you, I can't be right. Show me what numbers are needed for mass to cancel out gravity using centripetal force, on earth.
 
  • #23
d w said:
Hi, the question I pose to you then is, what is the critical mass for when centrifugal force takes over gravity with the known values in our universe?
There isn't one. This is because gravity grows with mass just as much as centripetal force does.
Look:
Centripetal force ##F_c=mV^2/R##
Centripetal acceleration ##a_c=F_c/m=V^2/R## - centripetal acceleration is then independent of mass of the object being accelerated.

Gravitational force ##F_g=GmM/R^2##
Gravitational acceleration ##a_g=F_g/m=GM/R^2## - gravitational acceleration is also independent of the mass of the object being accelerated (the other M is the mass of the planet/other object that attracts the first object).

edit: fixed the equations
 
  • #24
Bandersnatch said:
@d w when you multiply the centripetal (or -fugal) acceleration by mass you end up with centripetal force. Similarly, when you multiply gravitational acceleration by mass you end up with gravitational force. Both are proportional to the mass you're multiplying by, so their ratio doesn't change in the process.
If one's greater than the other, it stays greater no matter the mass (unless it's really huge and messes up the orbit).

You can not compare acceleration with force and expect it to mean anything.

Are you only going to say stuff based on assumptions? I thought this was a physics forum. Look at the actual physics and tell me the correct answer in terms of mass. If you can, thanks.
 
  • #25
Bandersnatch said:
There isn't one. This is because gravity grows with mass just as much as centripetal force does.
Look:
Centripetal force ##F_c=mV^2/2##
Centripetal acceleration ##a_c=F_c/m=V^2/2## - centripetal acceleration is then independent of mass of the object being accelerated.

Gravitational force ##F_g=GmM/R^2##
Gravitational acceleration ##a_g=F_g/m=GM/R^2## - gravitational acceleration is also independent of the mass of the object being accelerated (the other M is the mass of the planet/other object that attracts the first object).

"There isn't one. This is because gravity grows with mass just as much as centripetal force does" are you telling me that centripetal/centrifugal force CANNOT exceed the force of gravity?
 
  • #26
It can, but whether it does or not depends on the stuff that is in the equation - namely the tangential velocity and radius.

(by the way, there's an error in the equation - ##F_c## should be ##mV^2/R## not ##mV^2/2##; sorry 'bout that, I'll fix it right away.)
 
  • #27
Bandersnatch said:
It can, but whether it does or not depends on the stuff that is in the equation - namely the tangential velocity and radius.

(by the way, there's an error in the equation - ##F_c## should be ##mV^2/R## not ##mV^2/2##; sorry 'bout that, I'll fix it right away.)

So you agree that if the radius and velocity remain constant, that the size of the mass increasing will increase the force though, right?
 
  • #28
Yes. But so will the force of gravity, by the same factor. Look at the ##F_c## and ##F_g## equations - both have the mass ##m## in them, and both are directly proportional to it.
In other words, if you increase mass two times, the centrifugal force increases two times, and the gravitational force increases two times.
The ratio between the two forces remains unchanged in the process.
 
  • #29
Bandersnatch said:
Yes. But so will the force of gravity, by the same factor. Look at the ##F_c## and ##F_g## equations - both have the mass ##m## in them, and both are directly proportional to it.
In other words, if you increase mass two times, the centrifugal force increases two times, and the gravitational force increases two times.
The ratio between the two forces remains unchanged in the process.
Hold up, you just told me mass has nothing to do with centrifugal force, then you correct your equation, then you correct your statement saying, "yes, [increased mass does increase force]"

Then you proceed to brow beat me and offer no solution to a math problem that has to have a solution because MASS DOES MATTER. Please do not respond to this post anymore unless you have the answer. Thank you.
 
  • #30
Sorry, that was sloppy. The ratio of the forces doesn't depend on mass. The force itself does.

Your question is: for what values of mass ##m## the equation ##F_c/F_g>1##, and there is no ##m## in that equation, so the question doesn't make sense.
That's all there is to it.

If you can't see that the mass cancels out, just write it down and see for yourself. Equations for both forces are provided in the thread, or can be easily found elsewhere, if trust is an issue.
 
  • #31
d w said:
Hold up, you just told me mass has nothing to do with centrifugal force, then you correct your equation, then you correct your statement saying, "yes, [increased mass does increase force]"

Then you proceed to brow beat me and offer no solution to a math problem that has to have a solution because MASS DOES MATTER. Please do not respond to this post anymore unless you have the answer. Thank you.
Struggling to find an interpretation of the problem that results in something meaningful...

What mass would the Earth have to have for the centrifugal force on an object resting on its surface to match the gravitational force on that object?

Let M be the mass of the earth, m be the mass of the object on its surface, r be the Earth's radius and G be Newton's universal gravitational constant.
Let ##\omega## be the rotational speed of the Earth (in radians per unit of time).

Gravitational force = ##\frac{GmM}{r^2}##
Centrifugal force = ##m\omega^2r##

Solve for M.

##\frac{GmM}{r^2} = m\omega^2r##
##\frac{GM}{r^2} = \omega^2r##
##GM = \omega^2r^3##
##M = \frac{\omega^2r^3}{G}##
 
  • #32
d w said:
So you agree that if the radius and velocity remain constant, that the size of the mass increasing will increase the force though, right?
Yes, but so will the force of gravity, which is also proportional to the mass. What is constant is he gravitational acceleration and the centripetal acceleration, not the gravitational force or the centripetal force. You cannot just make an assumption and then stubbornly claim that it is correct when it is in contradiction to experimental observations. Physics is an empirical science, not philosophy.

Thread closed.
 
Last edited:
  • #33
d w said:
Hi, thank you also.

According to Newton's second law the centripetal force can be expressed as

Fc = m ac

= m v2 / r

Are you saying that the centrifugal force is not derived from the same equation and centripetal force? Or are you saying that the result of centrifugal force is not the actual acceleration multiplier on the mass? I get that F has the mass/acceleration mixed in, but even "mass" has mass/acceleration mixed in because the "mass" is a result of gravity on the mass. Gravity is a constant so they use that constant to form weights. I am saying the energy expressed in this equation is a rate of energy transfer. Newtons are really a rate of energy transfer, but it has to have time or distance for it to express the rate.

So far this is what I think, centrifugal force is on mass, so it isn't "independent" of mass. Centrifugal force expresses the acceleration on the weight in Newtons. Gravity force is expressed as weight in Newtons as 143 N. Since I am expressing gravity this way, I expressed centrifugal force the same way. Both my equations are expressing the effects of acceleration as force, no?

I am saying that when comparing the two, you have to stick to acceleration or force in both cases. I don't know where you got that 143 N figure from, as gravitational force would vary from object to object and does depend on mass. The 143 N force you give would be the gravitational force acting on a ~14.56 kg mass sitting on the surface of the Earth, but an object with a different mass would experience a different gravitational force. When we measure the weight of something we are actually measuring the gravitational force acting on it.

Gravitational force between two masses is
[tex]F_g = \frac{GMm}{d^2}[/tex]

Where G is the universal gravitational constant
M and m are the masses involved
d is the distance between the centers of the masses.

If M is the mass of the Earth, then GM has the value of 3.987x10^14 m3/s2
this is sometimes expressed as [itex]\mu[/itex]

Since F=ma and ergo a=F/m

The acceleration due to gravity is
Fg/m
or
[tex]A_g = \frac{\mu}{d^2}[/tex]

which depends of the distance from the center of the Earth where it is measured.

Using the figure for the radius of the Earth you gave, this gives an answer of 9.82 m/sec2 this holds for any mass near the surface of the Earth.

As you already noted, centripetal force can be found by

[tex]F_c = v^2m/r[/tex]

And as above centripetal acceleration is

[tex]A_c = v^2/r[/tex]

Which again holds for any mass.

So there is no critical mass for an object resting the surface of the Earth for which the centripetal force needed to hold object on the surface against the spin of the Earth exceeds the force of gravity acting on it. The centripetal force does increase with mass, but so does the gravitational force (weight).

And as I pointed out earlier, the gravitational acceleration far exceeds the required centripetal acceleration.
 

Related to How much mass with the centrifugal forces of earth to break gravity?

1. How does centrifugal force affect gravity on Earth?

The centrifugal force on Earth is a result of the Earth's rotation. It acts in the opposite direction of gravity and can slightly decrease the force of gravity at the equator, but the effect is very small and does not break gravity.

2. Can centrifugal force overcome gravity on Earth?

No, the centrifugal force on Earth is not strong enough to overcome the force of gravity. The Earth's gravity is much stronger and is what keeps objects on the surface of the Earth.

3. What is the mass needed to break gravity on Earth with centrifugal force?

The mass needed to break gravity on Earth with centrifugal force would be impossible to achieve. The Earth's gravity is a fundamental force that cannot be broken or overcome by any amount of mass.

4. Is there a limit to how much mass can be lifted off the Earth's surface with centrifugal force?

Yes, there is a limit to how much mass can be lifted off the Earth's surface with centrifugal force. This limit is determined by the Earth's escape velocity, which is the minimum speed an object needs to escape the Earth's gravitational pull.

5. Can centrifugal force be used to create artificial gravity on Earth?

No, centrifugal force cannot be used to create artificial gravity on Earth. While it can create a sensation of gravity, it is not a true gravitational force and would not be strong enough to mimic the effects of Earth's gravity.

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