The Problem With Centripetal Force

[Ilyas Elbadaoui]

If the Earth is in fact orbiting the Sun and hurtling around a super-massive black hole, then wouldn't centripetal force be at work thus creating a slightly more intense pull towards one side of the Earth and vice versa for the other side. I came up with an equation for this and it seems to make complete sense due to how the weight (not to be confused with mass) of an object seems to change depending on the proximity to its planet's equator. Centripetal Force divided by the rotational velocity of the object determines the difference in weight of objects on either side of the object if the object is moving. Could this be possible?

[Strange]

30 minutes ago, Ilyas Elbadaoui said:

If the Earth is in fact orbiting the Sun and hurtling around a super-massive black hole

There is no doubt that the Earth is orbiting the Sun.

But there is no black hole near enough to have any effect on the Earth.

But you are right that centripetal force changes the weight of objects at the equator (because of the rotation of the Earth).

The centripetal force on the Earth from the Sun is the Sun's gravity. This is pretty small at this distance, but it does cause a tidal effect. However, this is much smaller than the Moon's. The difference in gravitational force (ie weight) from one side of the Earth to another is, I think, about a factor of 1.00017.

What is your equation?

[Ilyas Elbadaoui]

Difference = Centripetal Force Divided by Rotational Velocity.

Also, there is a super-massive black hole at the centre of the Milky Way Galaxy in which the stars orbit.

Sagittarius A*.

I wound up with the answer of 0.00013 lb difference between the side of the Earth that is currently experiencing dawn to the side experiencing dusk.

Edited October 26, 2018 by Ilyas Elbadaoui

[Janus]

The Earth is orbiting the Sun, so the centripetal force needed to maintain a circular orbit is equal to the force of gravity provided by the Sun at our distance from it.  An object sitting at the center of the Earth would feel no net effect from either.

If you move to the side of the Earth near the Sun, the gravity from the Sun is stronger than at the center and if you move to opposite side it becomes less.  What you would experience is the net difference between the sun's gravity at these points and that at the center of the Earth. (thus you would be lighter on both the near and far side than you would standing at points 90 degrees way from those points.  This difference is called tidal force, and it is this, plus the same effect caused by the Moon's gravity that produces tides.  

Tidal forces are proportional to the mass creating them and inversely proportional to the distance from the mass.  So even though the Moon is much less massive than the Sun, because it is 1/400th the distance of the Sun, its tidal force on the Earth is ~2 times that of the Sun's.

The only supermassive BH in our galaxy that we are aware of is the one at the center of the Galaxy some 26,490 light years away, it is some 4.3 million times more massive than the Sun, but is some 1.67e11 times further away than the Sun is.  Thus its tidal force across the Earth would only be 4.3e6/1.67e11^3 = 9.2e-27 that of the Sun tidal influence; Too small to even measure.

[Janus]

49 minutes ago, Ilyas Elbadaoui said:

Difference = Centripetal Force Divided by Rotational Velocity.

 

Your equation as written makes no sense:

First off, by Rotational velocity do you mean tangential velocity (ft/s) or angular velocity (rad/sec)?

In either case you don't get an answer in lbs. (since you specified weight, I assume you mean lb_force )

using ft/sec, you are taking a force  (lb_f) divided by ft/sec  which gives an answer in lb_f-sec/ft, which is neither a dimensionless ratio or lb_f

Using rad/sec, you get  lb_f-sec/rad, which isn't any better.

 

[Ilyas Elbadaoui]

I’m not talking about tidal force, I’m talking about how if the sun is orbiting Sagitarius A* then the Earth is moving in a motion that implies that objects on Earth should be being forced onto the planet further, much like how you might be pulled back if when you’re inside a car if it abruptly starts going fast. But it should be different depending on where the object is positioned on the planet, in the direction the Earth is moving (Where dusk is being experienced) and in the direction the Earth is moving away from (Where dawn is currently being experienced). Also, by rotational velocity I mean the speed at which the Earth spins.

And by weight I am talking about the difference in the intensity in which an object is being pulled towards the Earth when it is in the direction that the Earth is moving to when it is on the opposite side. It’s divided by rotational velocity because the spinning keeps the object on the Earth due to inertia, like how a motorcycle can drive across the a ceiling, defying gravity, if it’s moving fast enough.

[Strange]

2 hours ago, Ilyas Elbadaoui said:

I’m not talking about tidal force, I’m talking about how if the sun is orbiting Sagitarius A*

It isn't. If that black hole were not there it would make no difference at all to the Sun.

2 hours ago, Ilyas Elbadaoui said:

much like how you might be pulled back if when you’re inside a car if it abruptly starts going fast.

That is because it is accelerating. When the car is moving at a constant speed, there is no such force.

And, importantly for this thread, when the car goes round the corner you are pushed/pulled to the side of the car.

3 hours ago, Ilyas Elbadaoui said:

And by weight I am talking about the difference in the intensity in which an object is being pulled towards the Earth when it is in the direction that the Earth is moving to when it is on the opposite side.

But, as with your example of the car, it doesn't work like that. Any force from the Earth going in circles round the Sun acts "sideways" ie. in the direction away from the Sun.

3 hours ago, Ilyas Elbadaoui said:

spinning keeps the object on the Earth due to inertia, like how a motorcycle can drive across the a ceiling, defying gravity, if it’s moving fast enough.

This is the wrong way round. A motorcycle can ride around the inside of a cylinder, for example because the centrifugal force (*) pushes it against the surface of the cylinder.

However, on the surface of the Earth the centrifugal force (*) is pushing things away from the surface. This is why things weigh less at the equator (ignoring the flattened shape of the Earth).

(*) Centrifugal force? Yes, I know it doesn't exist but it is a useful shorthand in these examples.

[Janus]

7 hours ago, Ilyas Elbadaoui said:

I’m not talking about tidal force, I’m talking about how if the sun is orbiting Sagitarius A* then the Earth is moving in a motion that implies that objects on Earth should be being forced onto the planet further, much like how you might be pulled back if when you’re inside a car if it abruptly starts going fast. But it should be different depending on where the object is positioned on the planet, in the direction the Earth is moving (Where dusk is being experienced) and in the direction the Earth is moving away from (Where dawn is currently being experienced). Also, by rotational velocity I mean the speed at which the Earth spins.

 

In the car example, you feel the backward "force" because the car is accelerating forward.   The wheels push against the ground accelerating the car forward and this force propagates through the car to you which you feels as the eat pushing forward on you.  With the Earth orbiting the Sun, the acceleration vector isn't in the direction of the Earth orbital motion, but towards the Sun.   In addition, this acceleration is caused by the Sun's gravity, which acts on you just as much as it does the Earth.  So we don't have the case like the car where the Earth is accelerated and that acceleration is transferred to you through the body of the Earth.

To illustrate imagine a mass attached to a spring.   In the car example, it would be like pushing on one end of the spring, the spring compresses and then exerts of push on the mass which accelerates it.  You can measure how much the spring compresses to get the force across it.

With the Earth-Sun gravity example, it is like you are pushing all points of the spring and the mass at the same time.   Since the force is accelerating all parts of the spring and mass the same amount, the spring doesn't compress and you measure zero force across it.

The spin of the Earth will make you a bit lighter at the equator than at the poles, but this difference remains constants over the entire equator and won't change due to the position of the Sun ( except for that amount due to the aforementioned tidal force.)

[swansont]

12 hours ago, Ilyas Elbadaoui said:

I’m not talking about tidal force, I’m talking about how if the sun is orbiting Sagitarius A* then the Earth is moving in a motion that implies that objects on Earth should be being forced onto the planet further, much like how you might be pulled back if when you’re inside a car if it abruptly starts going fast. But it should be different depending on where the object is positioned on the planet, in the direction the Earth is moving (Where dusk is being experienced) and in the direction the Earth is moving away from (Where dawn is currently being experienced). Also, by rotational velocity I mean the speed at which the Earth spins.

And by weight I am talking about the difference in the intensity in which an object is being pulled towards the Earth when it is in the direction that the Earth is moving to when it is on the opposite side. It’s divided by rotational velocity because the spinning keeps the object on the Earth due to inertia, like how a motorcycle can drive across the a ceiling, defying gravity, if it’s moving fast enough.

Show a calculation. You will see that any effect from bodies outside the solar system are negligible. 

[Ilyas Elbadaoui]

The entire galaxy is held together by the gravity of Sagittarius A*. The sun is orbiting it.

[Strange]

1 hour ago, Ilyas Elbadaoui said:

The entire galaxy is held together by the gravity of Sagittarius A*. The sun is orbiting it.

Don't be silly. The mass of the black hole is about 4 million solar masses (1) while the mass of the galaxy is about 10¹² solar masses (2). In other words, the black hole contributes about one millionth of the mass. We are not orbiting the black hole, we are orbiting the mass of the galaxy. If the black hole disappeared it would make no difference at all.

(1) https://en.wikipedia.org/wiki/Sagittarius_A*

(2) https://en.wikipedia.org/wiki/Milky_Way

[StringJunky]

1 hour ago, Strange said:

Don't be silly. The mass of the black hole is about 4 million solar masses (1) while the mass of the galaxy is about 10¹² solar masses (2). In other words, the black hole contributes about one millionth of the mass. We are not orbiting the black hole, we are orbiting the mass of the galaxy. If the black hole disappeared it would make no difference at all.

(1) https://en.wikipedia.org/wiki/Sagittarius_A*

(2) https://en.wikipedia.org/wiki/Milky_Way

The SMBH was probably influential in the the early formation of the galaxy, so I understand. 

[Janus]

4 hours ago, Ilyas Elbadaoui said:

The entire galaxy is held together by the gravity of Sagittarius A*. The sun is orbiting it.

Even if this were true,  (which it isn't due to the vast difference between Sagittarius A and the galaxy as a whole. Remove Sag A, and the galaxy would hardly notice it), your claim of a difference in weight for objects on the leading and trailing sides of the Earth is wrong. 

In this statement:

On 10/27/2018 at 1:35 AM, Ilyas Elbadaoui said:

the Earth is moving in a motion that implies that objects on Earth should be being forced onto the planet further, much like how you might be pulled back if when you’re inside a car if it abruptly starts going fast.

The key phrase is "abruptly starts going fast".   You feel pulled back due to the fact that the car is changing velocity in that direction.  Once you reach speed, you no longer feel that pull.

The Earth, is not abruptly changing is velocity in the direction it is traveling around the galaxy, it maintains a pretty constant pace. The only acceleration is has with respect to the galaxy center is towards the center. ( if you want to use the car example, it would like when you feel pushed to the side when you go around a sharp corner at speed.)

But even then, the magnitude of that acceleration depends not only on the velocity, but the degree of "bend" in the curve.  Even with the high speed that the solar system travels with respect to the center of the galaxy, the curve is so gentle, that the acceleration only works out to 0.000000000018 g (For a 175 lb person, this would result in a difference of 0.000000005 ounces).

And even then, the other key word is "orbiting" 

Objects in orbit are in a free fall path. they may be accelerating but all parts are accelerating equally in response to gravity, so they feel no net difference between their individual parts ( other than that due to the early mentioned tidal force).  So even that small acceleration in the above paragraph isn't felt.  ( a falling elevator is constantly accelerating downward, but someone in it would not be pined against the ceiling, but would just float around like they were weightless)

Edited November 10, 2018 by Janus

[Strange]

3 hours ago, StringJunky said:

The SMBH was probably influential in the the early formation of the galaxy, so I understand. 

That's possible. But I think it is still undecided if the black holes spurred the formation of galaxies or the formed because of the matter in the galaxy. Or maybe some combination of both.

[beecee]

On 10/27/2018 at 7:35 PM, Ilyas Elbadaoui said:

 I’m talking about how if the sun is orbiting Sagitarius A*

While you have been given some really good answers, I would like to add something with regards to the MW galaxy and most other galaxies. The SMBH would certainly keep the stars in the very inner part of the galaxy in orbit, and the gravity of those stars will keep other stars further out in orbit, and they in turn keep those stars in our region of the MW, including the Sun, in orbit. In other words it is the total gravitational effects of all the stars, on each other that keep them in orbit....Then of course we have DM! Remove the SMBH at the core, and the orbital parameters of the stars would generally remain as is....remove the stars that orbit closer to the core then the Sun, and we would probably fly off into inter-galactic space...remove all the DM, and likewise, we would again probably go flying off into oblivion.

The galactic orbital  motions of all the stars in the galaxy is complicated, each affected by another, and far different then the orbital reason for the orbital parameters of our solar system, where the Sun is indeed the dominant player whose mass exceeds all the planets and asteroids together.

[tinkerer]

Most interesting discussion! Seems to contain "posturing and conjecturing" in areas mostly beyond my comprehension, however. Being rather pragmatic, I would view a superficial difference of unbalanced force acting upon the Earth as quite insignificant when compared to precessionary forces, for one.