Tag: Newton

Some thoughts on Gravity and Newton’s Laws

Over the past few weeks I’ve been thinking about gravity and the various laws that Newton proposed.  One of the main questions I have pertains to the equation for F (force) which is F=ma and the so-called law of universal gravitation which is:

F1 = F2 = G⋅m1 x m2 / r(squared)

And this is put into normal language as:

The force is proportional to the product of the two masses and inversely proportional to the square of the distance between them

meaning, if the mass increases then the product increases.  For example, as a mass accelerates the value of F will increase.  This can be seen in the extreme example of a mass that approaches the speed of light;  as it approaches, it’s mass approaches an infinite value.  The difficulty with this concept (as it is only a concept without any empirical evidence) is Newton’s 3nd law which states that “for every force there is an equal and opposite force”.

If you take the 2nd law and apply it to any accelerating mass, you increase the total amount of force which also an increase in energy (e=mc2).

Now if you take the law of universal gravitation (as shown above), and apply the 2nd law to m2, the total force between the two objects will increase.  you can ignore the increase of r(radius) since the increase in radius is insignificant relative to the size of m1.  In other words, an object like a rocket could not possibly launch if gravity acts according to the prevailing theory since the acceleration adds to the objects total energy (as shown above).  By increasing energy you increase the force between m1 and m2 with a “…a force that is proportional to the product of the two masses” and with equal force (as per the 2nd law).  Therefore for every pound of thrust an equal amount of force is brought to bear between the rocket and the earth.

This would also apply to anything accelerating away from the center of the earth (ie. rapidly raising my hand above my head).  In other words, gravity should be acting like a brake against the accelerating body.  To “break free” (think of an inverted pendulum flywheel) of the gravitational force, an object would have to accelerate with a force greater than m1 * acceleration.   Therefore, if we take two equal masses (m3 = m4) and accelerate one of them (m4) it would require the second mass (m4) to accelerate at a greater value than m3 * acceleration.  Essentially, it is an application of e=mc2.  This a far better explanation as to why the mass of an object increases as it approaches the speed of light – the mass itself is not increasing but the affect of gravity increases proportionally to the accelerated mass  therefore the effective mass increases.

The great irony here is that this makes gravity all but impossible since any object on the surface of a globe (ie. Earth) would be held fast against the surface.  Anything pushing against gravity would encounter massive (no pun intended) resistance (like blood flow, plant growth, etc) to the point where no life could form.   Nor could objects be buoyant.  An object floating on the ocean surface is essentially accelerating away from the center of the earth (until it finds equilibrium at the ocean surface).  The gravity of the Earth is far greater then the total outward thrust of the buoyant object (ie. air inflated beach ball).  In a nutshell, buoyancy would be completely overwhelmed by gravity.  As well, the lift experience by commercial airplanes would also be insufficient to overcome gravity.

However, no matter how reasonable this line of thinking is, many gravity apologists will simply drag out their favorite solution:

Einstein – The grand-daddy of excuses

If we push aside Newton for the moment a look at what Einstein proposed, we are actually in a less favorable possible (if you believe in gravity).  Firstly, Space-Time needs to bend or be distorted to create this magical affect.  Looking at the area around an object (like a planet), we see it is spherical.  The so-called gravity “well” needs to encompass the entire planet not just a portion underneath.  I talked about this in a previous post.  The point being, the distortion of space time is not like this:

It needs to be an gravity sphere.  So what about the distortion around objects on the earths surface?  Is this not what causes gravity?  Does space-time wrap around a cube or a oddly shaped stone?

The explanation is somewhat specious:

Bodies with spatial extent

If the bodies in question have spatial extent (rather than being theoretical point masses), then the gravitational force between them is calculated by summing the contributions of the notional point masses which constitute the bodies. In the limit, as the component point masses become “infinitely small”, this entails integrating the force (in vector form, see below) over the extents of the two bodies.  

In this way it can be shown that an object with a spherically-symmetric distribution of mass exerts the same gravitational attraction on external bodies as if all the object’s mass were concentrated at a point at its centre.[2] (This is not generally true for non-spherically-symmetrical bodies.)

https://en.wikipedia.org/wiki/Newton’s_law_of_universal_gravitation

Right.  This explanation effectively reduces all objects to single point masses and runs gravitational vectors between two bodies.  It’s really a rather grotesque idea.  Then the gravity within a body is nullified since all the internal objects of a “single mass” are counted as one:

  • The portion of the mass that is located at radii r < r0 causes the same force at r0 as if all of the mass enclosed within a sphere of radius r0 was concentrated at the center of the mass distribution (as noted above).

  • The portion of the mass that is located at radii r > r0 exerts no net gravitational force at the distance r0 from the center. That is, the individual gravitational forces exerted by the elements of the sphere out there, on the point at r0, cancel each other out.

As a consequence, for example, within a shell of uniform thickness and density there is no net gravitational acceleration anywhere within the hollow sphere.

So now we have a hollow uniform body and only the surface itself has gravity.  So if all the gravitational forces are cancelled out as per the explanation above, then all the gravitational forces most somehow come from the surface.  How is that even possible? Geometrically, the math flattens out the sphere by pushing everything to the surface so the concept of a larger mass having greater density and therefore greater gravity is expunged and we are left with flat plane – in essence.

So does the math coincide with reality?  If the math says all the gravity is on the surface, then is it really on the surface?  I mean really only on the surface.  If yes, then how could a surface, no matter how big, generate a sufficiently potent gravity field as to warp space-time?

Does Gravity make sense?

If you actually think about how gravity is supposed to work, you might find that there are significant issues with the theory.  I’m just pointing out some factors that might be furiously refuted since without gravity (as it is proposed by the majority of scientists), the universe would be a completely different place.  So I’m moving into a territory that is highly guarded and sanctified.

If we grant that there is a force “pulling” towards the center with a force of 9.8 Newtons/kg at all points on the ‘globe’ simultaneously *and* a centrifugal force that pushes away from the center at such insignificant values, then magical traits would have to be applied to the centripetal force. To put this into perspective, if you are standing on a disk (r=10m) that spins at 10 m/s and a person with a mass of 72kg stands on the edge of this disk, they would feel 720N of outward force (centrifugal). The only way that the centrifugal force can be ‘balanced’ is by having the person hold onto the disk with a greater force than 720N. In other words, the centripetal force. If this person stands close to the center of this disk [cos(89)X10] then then would only have to hold on with 0.2N.

http://www.calctool.org/CALC/phys/newtonian/centrifugal

So what is the difference between gravity and centripetal force? Of course centripetal force is valid if the object is attached to the center of a spinning globe by a rope or some kind of real tensor (like a tree rooted in the ground). However, centripetal force can not be applied to an object that is not tethered to the center of a spinning globe unless you are applying some kind of magical grappling force to it (I hope not). So for objects like people or cars or dogs, you can only apply gravity and the centrifugal force (and no the dog is not tied to a tree by a leash). So we are left with a tiny centrifugal force and gravity.

A 72kg person at the equator will have 705N of force due to gravity on them at all times (9.8 N/kg * 72kg). The centrifugal force at the equator is only 2.4N or an effective force of 702.6N. At the 89th parallel it is effectively null. So the centripetal and centrifugal forces are inconsequential to our experience on this ‘globe’.

What we are now dealing with is the 702.6N-705N of force that is pulling down on the average human. So 9.8N of force is being applied to every atom (and supposedly a lot of empty space) in that body so that the person in question ‘weighs’ 72kg. However, the force of gravity on an object reduces at the square of the distance from the source. So the hydrogen atoms at the top of my head weigh less than the hydrogen atoms in my feet – unless I’m lying down or standing on my head.

https://en.wikipedia.org/?title=Inverse-square_law

To get a real measurement of gravity we need to get the source value of gravity on Earth at its core. As far as I can tell, there is no clear answer as to how gravity functions below the surface. Some theorize that it decrease as you leave the surface and head towards the core. But how can that be? This means that the surface of the Earth is the center point of mass. So how can we get the actual source value for gravity? Without that, no one can give a real answer as to what the value for gravity on the earth’s surface is. The only answer I have is to reverse the formula when heading towards the core of the earth. At 4000 miles from the earth’s core gravity is 9.8 N/kg. The surface area is A = 4πr^2 @ 4000 miles = 200960000. So the energy or intensity decreases (divided by 4) as the distance r is doubled (or increases and inverted going in the other direction). At 2000 miles then the intensity of gravity would be 9.8 N/kg * 4 = 39.2 N/kg. At 1000 miles it would be 156.8 N/kg. 500 miles would be 624 N/kg; 250 miles 2496 N/kg…etc..etc.

Therefore, 8000 miles from the core of the earth would be 2.45 N/kg which makes the idea of weightlessness at 200 miles above the surface a little suspect. I don’t see how an object like the ISS (370,131kg) is able to ‘float’ at a orbit of only 4200 miles from the earth’s core. It would still be have an effective ‘weight’ very similar to what it would have on the earth’s surface (about 2.5% less ‘weight’). Even at 8000 miles the ISS would ‘weigh’ 92,532kg and would still fall back to earth. Until the object approaches the Moon at approximately 110,000 miles, the gravity of earth will always be greater. However, gravity due to the sun will begin to pull at a greater force than both the moon and the earth at similar distance. With this in mind, I don’t see how space travel is possible without massive amounts of fuel and propulsion.

As an aside, you can’t really have an effective gravity ‘at sea level’ since there can’t be a ‘sea level’ on a globe due to the roundness of the water (which really makes no sense at all). And if we are on a spinning globe, there seems to be a magical grappling effect on every water molecule via the centripetal force that keeps the water in a spherical shape. This then puts the concept of orbiting objects into question since the centripetal force is the only thing keeping it in a circular orbit (or is it pear shaped…I can’t keep up) and the centripetal/centrifugal force is a fraction of the ‘pull’ of gravity, then all objects must fall. The counter argument that since the object is in the ‘vacuum’ of space and a force perpendicular to the pull of gravity (ie. Newton’s first law) which keeps it in orbit is nonsensical since the force of gravity is greater than the centrifugal force and weightlessness can’t be within such proximity to the earth’s surface. A constant pull is being applied perpendicular to all orbiting objects therefore they cannot ‘float’ in ‘space’ without the addition of acceleration (ie rockets or other forms of propulsion). And finally, the question of *how* a planet stays in an orbit without additional forces is beyond me. What is supplying the additional acceleration? The gravitational forces of the sun would have long ago exhausted any capacity to stay in an orbit.