Re: Gravity

[lenvanzanten]

It looks like the experts in their online calculations have yet to figure out how to compute gravity and/or centrifugal force.

Let us take a 3000 lb rocket at 200 miles up traveling 18.000 mph

Example 1: 3000-lb x 26,400-ft/sec2 : 22,176,000-ft/r = 94,286-lb/cg

Then that same rocket at the equator.

Example 2: 3000-lb x 1521-ft/sec2 : 20,929,920-ft/r = 331.6 lb/cg

The online calcs specify 2930.4 lb for example 1 and 10.3 lb for example 2.

On the ground we specify G/force at 3000 + 331.6 to 3331.6 lb, and 200 miles up at 18.000 mph at 94,286 lb.

If then we simply lift that rocket straight up to 200 miles above earth to travel with the same speed as the earth’s rotation (1000 mph) what would be its weight in G force?

In order to be correct we must determine what the figure of G is 4200 miles from earth’s center. Assuming the rocket is balanced in orbit, and assuming g at about 90% down from 32.14 to 31.43, the 94286 : 31.43 = 3000 lb. The new weight of the rocket 200 ft up therefore is 2933.7 lb.

Then: 2933.7 lb x 1521 ft/sq : 22176000 ft/r = 306 lb centrifugal force, this added to its basic weight, G/force comes to 3239.7 lb of G/force on the rocket with only 306 lb of centrifugal holding it up, which by law of equal action to reaction cannot possibly stay up there.

The scientists with their online calcs are however telling us that it will stay up there by no more than 2930.4 lb of G/force, as g and c (for centrifugal) must be equal. But how can they possibly come up with a c and g of less than its own weight at a velocity 18 times faster than sitting still on earth and/or 200 miles up? Why accelerate it 18000 mph? Whatever happen to the square of the force in linear motion? Is 18000 mph so minute in force, while at a mere 60 mph a car into a brick wall will be smashed?

These fellows rewrote the law as follows:

"The gravitational acceleration of a mass in uniform circular motion is proportional to the square of its speed and inversely to the radius of its path, and once again inversely to its inertial gravity."

For that is how they came to their eroneous figures.

The link to the calc is (on line) (calctool)

The above weight, speed, and radius are for example, actual velocity of space shuttle 17,532 mph

[CraigD]

Whoa! This post appears to show some severe confusion about basic mechanics!

Let us take a 3000 lb rocket at 200 miles up traveling 18.000 mph

Example 1: 3000-lb x 26,400-ft/sec2 : 22,176,000-ft/r = 94,286-lb/cg

First, there is a confusion of units and term. The Pound (lb) is a unit of force, not of mass. Force = Mass * Acceleration, so the mass of a body with a weight of 3000 lb at the Earth’s surface has a mass of about 3000 lb / 32.2 ft/s/s = 93.3 slugs (these English traditional units are so awkward and unusual, it’s a good idea to use the more common SI system when discussing science, so I’ll convert to them as I go).

 

Next g-force, or “gees”, is not a force, but a unit of perceived acceleration, such that 1 g is the acceleration experienced by an accelerometer on the Earth’s surface.

 

The g-force experienced by a body in a circular path around the Earth is the difference between its acceleration due to gravity and its centripetal acceleration.

 

Gravitational acceleration is given by the formula

[math]a_{g} = \frac{G m_E}{r^2}[/math]

where [math]m_E[/math] is the mass of the Earth, [math]r[/math] the distance between the center of the Earth and the body, and [math]G[/math] the gravitational constant.

 

Centripetal acceleration is given by the formula

[math]a_C = \frac{v^2}{r}[/math]

where [math]v[/math] is the body’s speed.

 

So, for the above example, the g-force on a body circling the earth at a height above the surface of 200 miles and a speed of 18000 miles/hour is

[math]a = \frac{G m_E}{r^2} - \frac{v^2}{r}[/math]

where [math]r = 200 \,\mbox{miles} + r_{Earth} \dot= 21981646 \,\mbox{ft} \dot= 6700006 \,\mbox{m}[/math],

 

so [math]a = \left( \frac{6.67428 \times 10^{-11} \cdot 5.9736 \times 10^{24}}{6700006^2} - \frac{8046.7^2}{6700006} \right) \div 9.8 \dot= -.07984681 \,\mbox{gees}[/math]

That is, a person in a vehicle following the path would experience an upward force about 8% as strong as the downward force he experiences on the surface.

It looks like the experts in their online calculations have yet to figure out how to compute gravity and/or centrifugal force.

…

The link to the calc is (on line) (calctool)

I’m unable to figure out what URL you’re referring to, lenvanzanten, but from your post, I suspect your understanding of basic mechanics isn’t good enough to understand many specialized online calculators. Until you’ve mastered the basic skills necessary to reproduce the calculations I did above, I’d recommend avoiding labor-saving aids like online calculators, and doing all of your calculations in longhand on paper (computer rendering tools, like the hypography’s LaTeX feature I used above, can be handy, but unless you’re very proficient with them, can be a distraction).

[lenvanzanten]

I put this on to see what responds there would be, but you seem to be more confused than the persons constructing the calculators.

[Zythryn]

I put this on to see what responds there would be, but you seem to be more confused than the persons constructing the calculators.

 

Len, being critical at best, and insulting at worst, without offering any critic of the actual post you are being critical of is, well... not constructive.

 

If you feel Craig's post is in error please feel free to share with us where. I think he went to great length to explain. Please return the common courtesy.

[Pyrotex]

Whoa! ...That is, a person in a vehicle following the path would experience an upward force about 8% as strong as the downward force he experiences on the surface. ....

Sorry Craig, but you seem to have derived an incorrect value.

 

If the radius of the Earth is 4,000 miles, and the force of gravity at its surface is 9.8 m/sec^2;

 

Then going up to an orbit of 200 miles only changes the radius to 4,200 -- an increase of only 5%.

 

This would mean the force of gravity (at 4,200 miles) should be around 8.9 m/sec^2.

Or 91% of the pull felt at the surface.

:lol:

[modest]

I figured out the problem. Lenvanzanten is referring to this calculator:

CalcTool: Centrifugal force calculator

He input these parameters to find centrifugal force:

m = 3,000 lbs

v = 26,400 ft/s

r = 22,176,000 ft

and got this answer:

2930.49 lbf

To check that answer he used the formula for centrifugal force:

[math]F_{centrifugal}=m\frac{v^2}{r}[/math]

inputting the parameters above gives:

[math]F_{centrifugal}=(3,000 \ lbs ) \frac{(26400 \ ft/s)^2}{(22,176,000 \ ft)} = 94,286 \ lbs \ ft / s^2[/math]

He then figured 2930.49 lbf is not equal to 94,286 lb ft/s^2 so the calculator must be wrong. The problem which Lenvanzanten didn't realize is that an "lbf" is not equal to "lb ft/s^2". Pound force (lbf) is defined as "The pound-force is approximately equal to the gravitational force exerted on a mass of one avoirdupois pound on the surface of Earth". This means 1 lbf = 32.174 lb ft/s^2. Therefore:

[math]F_{centrifugal}=94,286 \ lbs \ ft / s^2 = 2930.5 \ lbf[/math]

The calculator is most certainly correct. As Craig said "these English traditional units are so awkward and unusual, it’s a good idea to use the more common SI system when discussing science".

 

Using SI units such as kilograms, meters, and seconds would give consistent answers such as newtons and would avoid the kind of trouble you, lenvanzanten, ran into.

 

I put this on to see what responds there would be, but you seem to be more confused than the persons constructing the calculators.

Neither Craig nor the people who made the calculator were confused. You should read Craig's post again and try harder to make sense of it because it's spot on and it will help you.

 

~modest

[belovelife]

Len, being critical at best, and insulting at worst, without offering any critic of the actual post you are being critical of is, well... not constructive.

 

If you feel Craig's post is in error please feel free to share with us where. I think he went to great length to explain. Please return the common courtesy.

 

i concur

[lenvanzanten]

Much thanks Modest for your reply, that makes good sense. I did not really think the calcs were wrong, but why they did so.

In my book (as also in the textbooks) however whatever one makes of it, the term acceleration should only be used where there is in fact and indeed an increase in velocity, a change in speed not a change in direction when speed remains constant. And don't get me wrong I know all the arguments given.

[lenvanzanten]

QUOTE: Next g-force, or “gees”, is not a force, but a unit of perceived acceleration, such that 1 g is the acceleration experienced by an accelerometer on the Earth’s surface.

 

 

Sorry CraigD I did not mean to be rough but a bit return of your own medicine,

Tell me then on the above g is very much a force standing still on the ground, as much as it is under acceleration. The value is used for both and even when no gravitational direction is present such as a car making a turn in the horizontal plane. It applies since the ever present g is by and in the inertia of all substance, (Therein by the way lies also the secret to how and what gravity is, but a lenghty subject)

[CraigD]

Sorry Craig, but you seem to have derived an incorrect value.

 

If the radius of the Earth is 4,000 miles, and the force of gravity at its surface is 9.8 m/sec^2;

 

Then going up to an orbit of 200 miles only changes the radius to 4,200 -- an increase of only 5%.

 

This would mean the force of gravity (at 4,200 miles) should be around 8.9 m/sec^2.

Or 91% of the pull felt at the surface.

:)

You’re correct, Pyro.

 

Your calculation, however, gives the acceleration experienced by a body at rest relative to the center of the Earth at a distance of 4200 miles from it. Mine was for the acceleration of a body at the same distance traveling in a circle with a speed of 1800 MPH (8046.7 m/s).

 

Replacing the [math]v=8046.7[/math] in

[math]a = \frac{G m_E}{r^2} - \frac{v^2}{r}[/math]

…

[math]a = \left( \frac{6.67428 \times 10^{-11} \cdot 5.9736 \times 10^{24}}{6700006^2} - \frac{8046.7^2}{6700006} \right) \div 9.8 \dot= -.07985\,\mbox{gees}[/math]

gives

[math]a = \left( \frac{6.67428 \times 10^{-11} \cdot 5.9736 \times 10^{24}}{6700006^2} - \frac{0}{6700006} \right) \div 9.8 \dot= .9063\,\mbox{gees} = 8.88 \,\mbox{m/s/s}[/math]

Which agrees with your calculation. :thumbs_up

In my book (as also in the textbooks) however whatever one makes of it, the term acceleration should only be used where there is in fact and indeed an increase in velocity, a change in speed not a change in direction when speed remains constant.

What textbook are you referencing, lenvanzanten? If it actually states what you say, I recommend trying another, as it’s in bad agreement with the usual definitions of these terms, which have been well defined for the past few centuries.

 

As the linked wikipedia article states, acceleration “is the change in velocity over time”. Velocity is a vector quantity, so consists of a direction and a magnitude. The magnitude of velocity is called speed. A body has nonzero acceleration if either its direction of speed changes.

 

This is an important concept in mechanics. Without it, important concepts such as centripetal acceleration can’t exist, and ones ability to answer many questions of practical importance is greatly impaired. In short, without understanding mechanics, one can’t be even an engineer, let alone a physicist.

Tell me then on the above g is very much a force standing still on the ground, as much as it is under acceleration.

Please reread post #2, being sure to follow its link to the wikipedia article on “g-force”.

 

It’s important to understand that, despite having the word “force” in it, “g-force” is not a force, but an acceleration, and the word “gee” and unit abbreviation “g” is simply a unit of acceleration, equal to about 9.8 m/s/s. There is nothing complicated or specialized about the concept or unit. They’re simply useful for everyday applications, and so are widely used.

 

When all is said, mechanics should allow us to answer questions about our everyday physical universe. By learning mechanics, and other branches of physics and other scientific disciplines, we become better able to answer such questions, and to understand one another. Claiming that “experts” are “confused” isn’t very useful, and isn’t the purpose of science, of these science forums.

 

Lenvanzanten, it’s important that when you make claim in your posts at hypography, you back them up with links or references. Notice that I’ve done that in my posts in this thread. Be sure you’ve read and understand the site rules, noting that backing up your claims is rule #1. Hypography is all about sharing ideas and learning in the process, and its rules are meant to help this happen. If you don’t follow them, your threads will be moved to the strange claims forum, and, after being gives a reasonable chance to begin following them, your posting privileges will be temporarily or permanently suspended. Please follow the rules.

[belovelife]

Hey I can validate craigd's claims, most of my stuff is in the strange claim forums, and I don't even insult other posters. And the people here are intelligent, so it is a backward step to insult here.

I would say to try and defend your claims as best you can, and if you still feel you are right, then build up a case for yourself. That is my "vector". (lol)

[lenvanzanten]

I have heard enough, the erro of man is held viable while truth and reality are considered strange. I therefore am quiting the forum. This however I am sure of; that current foundations of science will be replaced by a new foundation, one that inhibits truth with reality.

I will mention a new law of gravity which does not require reference, it being indisputable in itself.

; "The gravitational force upon a mass in uniform circular motion is; its weight times the value of g at the radius thereof."

[modest]

I have heard enough, the erro of man is held viable while truth and reality are considered strange. I therefore am quiting the forum.

But, despite what they say—I've seen far stranger things than truth :eek:

 

:eek2:

I've seen a rich man beg, I've seen a good man sin, I've seen a tough man cry.

I've seen a loser win, And a sad man grin, I heard an honest man lie.

I've seen the good side of bad and the downside of up and everything between.

I licked the silver spoon, drank from the golden cup and smoked the finest green.

I stroked the fattest dimes at least a couple of times before I broke their heart.

Y'know where it ends, yo—usually depends on where you start.

:doh:

 

This however I am sure of; that current foundations of science will be replaced by a new foundation, one that inhibits truth with reality.

Change is the only constant :hihi:

 

I will mention a new law of gravity which does not require reference, it being indisputable in itself.

; "The gravitational force upon a mass in uniform circular motion is; its weight times the value of g at the radius thereof."

And I thought "the gravitational force acting upon a mass" was equal to it's weight... Perhaps you meant to say mass. That I would agree with.

 

Force of graivty on object = weight = mass times acceleration of gravity :cocktail:

 

The weight of an object is defined as the force of gravity on the object and may be calculated as the mass times the acceleration of gravity, w = mg.

 

Mass, Weight, Density

 

~modest

[CraigD]

I’m sorry you’re leaving hypography, lenvanzanten. If you a change of heart leads you to feel that you can follow its simple rules, including the supporting of ones claims, and would like to share your enthusiasm for science with like minded people, please feel free to return.

This however I am sure of; that current foundations of science will be replaced by a new foundation, one that inhibits truth with reality.

Please take this as well-intended advice. Before you can effectively question the “current foundations of science”, you must understand them. You appear to lack an understanding of introductory level physics. If you wish to be taken seriously, gain this understanding. If you are able to take classes in it in school, do. If not, self-educations is possible, in which case sites like hypography can help.

 

Key to mastering any formal discipline is understanding the concepts behind its terms. In you last post, you write

"The gravitational force upon a mass in uniform circular motion is; its weight times the value of g at the radius thereof."

Here, in describing a force as a weight times an acceleration, you make a common and serious mistake involving units.

 

A force is a mass times an acceleration. It’s units, therefore, are mass/acceleration = mass/velocity/time = mass/distance/time/time. In SI units, force is in kg/m/s/s, a unit given the sort name Newton, abbreviated N. In English units, its in slugs/feet/s/s, a unit given the short name pound, abbreviated lb. or #.

 

A weight is a force, not a mass. Unfortunately, the English unit pound is commonly, and incorrectly, used as a unit of mass – a good reason to avoid using English units as much as possible. In the US, unfortunately, despite efforts by legislators and teachers, we continued to use English units in everyday life.

 

Lenvanzanten, if you take nothing away from your experience at hypography than an understanding of the difference between mass and weight, your time has not been wasted.