Re-entry

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hoi.polloi
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Re: Re-entry

Unread post by hoi.polloi »

Dear Heiwa, and others, what do you make of the proposition that the Earth's gravity has been poorly calculated, that escape velocity might not be what they claim it is, and that Earth's gravity has a stronger force when you are closer and a weaker force when you are further, both than have been calculated previously?

As per this article:
Gravitational Bias and Escape Velocity

Author: Greg Alexander
Date: 7th May 2003.

When using the equation describing the acceleration due to gravity, g = GM/r², it is always assumed for a body like the Earth that the gravitational field acts from a point at the exact centre of the mass such that when stood at its surface the distance between you and that point is always exactly equal to the radius of the Earth. However knowing that the inverse square law applies to any gravitational field, wouldn’t it be the case that the mass directly beneath your feet would have a disproportionately greater effect that that on the other side of the planet? As the relation between the field intensity and distance isn’t linear, surely it can’t be valid to assume that one can summate all the gravitational forces produced by all the particles within the Earth’s mass into a single force acting from an isolated, central point?

The following example would appear to demonstrate quite clearly that this indeed is not a valid assumption :

If we divide the Earth’s mass into 100 smaller units taking the form of spheres (these spheres needn’t necessarily have to fit alongside each other without overlapping, and assuming, for the sake of the argument, that the density of the Earth is uniform throughout), we obtain the following proportions for such a sphere; a mass of 5.98 x 10exp22 kg, and a radius of 1,372.6 km. Stood at the surface of the Earth directly over one such sphere the observer would be exactly 1,372.6 km from its mathematical centre and experience a ‘g’ of 2.13 m/sec² as a consequence. Placing a second theoretical sphere at the Earth’s exact centre a further ‘g’ is produced of 0.098 m/sec² where ‘r’ is the radius of the Earth. (It can be noted that 0.098 is exactly 100 times less than 9.8 m/sec²). A third theoretical sphere is placed on exactly the opposite side of the Earth to the first such that the distance between its centre and the observer is 11,383.4 km. This would produce a ‘g’ of 0.0309 m/sec².

From this example it is apparent that the sum of the g’s produced by the first and third spheres, 2.13 + 0.0309 = 2.16 m/sec², is not at all equal to twice the ‘g’ produced by the second centrally located sphere of 0.196 m/sec². In fact there is a definite bias of over a 1,000%!

If such a bias does occur on the Earth’s surface as a result of the non-finite proportions of its mass which is spread out within a substantial volume of space, then this must reflect upon the relevant accuracy of the equation g = GM/r² (when applied at the surface). When the value of the universal gravitational constant G (6.7 x 10exp-11 N m² kg-1), the mass of the Earth in kilograms (5.98 x 10exp24 kg) and the Earth’s radius in metres (6,378,000 m) are put into the equation in the relevant places you always end up with g = 9.8 m/sec². But how much of this value (which is measurable experimentally) is the result of such a gravitational bias?

If the 9.8 m/sec² acceleration at the Earth’s surface is, to a significant part, the result of such a bias, one consequence of this would be that the calculated value of ‘g’ would decrease far more rapidly as you left the Earth’s surface, say, in a rocket. This in turn would inevitably effect the Earth’s escape velocity which would inevitably decrease. However currently it is calculated as being 11.18 km/sec.

Is it possible that this value for the Earth’s escape velocity is wrong? Although such a suggestion may sound slightly outrageous there is at least some evidence that such may be the case. On the 10th of August 1972 a meteorite was seen to enter the Earth’s atmosphere and travel in a horizontal direction over Utah and Montana, eventually leaving the atmosphere somewhere over Alberta. It apparently approached the Earth at some 10 km/sec and was accelerated to 15 km/sec by the Earth’s gravity.

Straight forward enough however one can’t help making the following observation: it is an indisputable fact that just as any object leaving the Earth’s gravitational field needs to travel at the escape velocity, any object entering it would be accelerated by the same amount. In the case of the meteorite described above, whose trajectory was apparently well documented, it had accelerated by only some 5 km/sec, far less than 11.18 km/sec one might have expected.

A further consequence of such a gravitational bias is that the calculated g’s on the surface of other bodies such as the Moon or Mars, may actually be different. Because the mass immediately beneath the observer’s feet has a far greater effect, this would mean that as the body in question became smaller and the mass less, the acceleration due to gravity at its surface would not decrease in the same proportion but significantly less than this. As a result one might expect that the g’s on the surface of the Moon and Mars would be greater than the currently quoted values.

Also apparently affected by this bias are the celestial mechanics of the Solar System. Because the value of ‘g’ tails off quicker as you leave the surface of the gravitating body in question, its gravitational profile is also changed completely and tends to be weaker at a distance than previously calculated. Any body in orbit about it would by necessity have to move in a little closer to retain the same period of orbit. This would equally affect the orbits of the Moon about the Earth as well as each of the planets around the Sun.

Of course we have all been told that the distance between the Earth, the Moon and the Sun, and all the planets have already been calculated precisely but how are we to reconcile this with the above observations?
(My bolds.) http://www.webspawner.com/users/gravibias

This is by the same writer who claims to have spotted the ISS in a telescope and seen it is physically little more than a round ball of some kind sitting in place for the fake space station. (See: ISS thread)
Heiwa
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Re: Re-entry

Unread post by Heiwa »

Physicists and engineers like to simplify things, e.g. a 3D mass is represented by a point with no dimensions (i.e. the density of the point mass is big to say the least), etc, etc. :rolleyes:
When GG released a ball of iron from the roof of the Pisa Tower and noted that it dropped down being attracted by the ground (Earth) below – an early re-entry test - he probably missed that the (mass of the) Pisa Tower also attracted the ball like a magnet sideways and that the ball of iron also attracted the ground below upwards. B)
It is thus simpler to reduce the ball and Earth to points with masses that attract each other – gravity - and forget the Tower at the side.
Gravity is a mystery force. :blink:
Moses (later Ramses III) allegedly was handed some stone tablets from the sky 3 200 years ago on a mountain top (early media fakery?) and it was a re-entry miracle … because when Moses, aka Ramses III, later dropped one tablet (with point mass m) on his toes (from height h) it really hurt. You don’t fool around with gravity force F = mg because when it has displaced distance h it applies energy or work done on your toe = mgh. ^_^
NASA, SpaceX, Sir Branson & Co, pls take note. :P
Last edited by Heiwa on Fri Oct 12, 2012 4:15 pm, edited 2 times in total.
agraposo
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Re: Re-entry

Unread post by agraposo »

Dear Hoi.

It is safe to ignore these articles, as it is impossible to deduce something with their poor mathematical style. For example, in this article

http://www.webspawner.com/users/gjalex

the writer is saying this:
The equation for kinetic energy is derived from the equation Work Done = Force x Time as follows:

K.E. = W.D. = ∫ F.dx = ∫ m dv/dt . dx = ∫ m dx/dt . dv

= ∫ m v dv = ½ mv²,

However if Work Done is actually equal to Force x Time this would change the existing equation for kinetic energy as follows:

K.E. = W.D. = ∫ F.dt = ∫ m dv/dt . dt = ∫ m dv = mv.
so he says that energy (mv²) is equal momentum (mv), which are different physical magnitudes. It's like measuring time in meters.
hoi.polloi
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Re: Re-entry

Unread post by hoi.polloi »

Interesting to the formula being bad, but the explanation is good. For instance, it should be true that work is worth more than just force exerted. Work should be worth force exerted within a time frame. Does the original formula for work anticipate this?
Fred54
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Re: Re-entry

Unread post by Fred54 »

Work is force through a distance. Power is what you have when you introduce time. Units for work are Btus, calories, foot-pounds, newton-meters, ergs, joules, and so on. Units for power are Btus/min, horsepower, ft-lbs/sec, watts, etc. If you lift a box car with a wind up alarm clock and large series of gears and pulleys it may take you 10 years to lift it one foot if you put a small lawn mower engine in a similar apparatus it may take you 10 minutes to also lift it one foot the total amount of work done is the same in each case. It is power that determines the rate at which work can be done. One final note a watt is power since it is a joule/sec but a killowatt-hour is work since you have a joule per second multiplied by time which in effect cancels the time unit and leaves joules.
hoi.polloi
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Re: Re-entry

Unread post by hoi.polloi »

Aha. So he is just describing power. That's "funny" that he overlooked that.
lux
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Re: Re-entry

Unread post by lux »

You might want to take a look at the book Gravitational Force of the Sun by Pari Spolter for an alternative view of gravity. My math is not up to vouching for all the equations it contains but I found it an interesting read.
Heiwa
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Re: Re-entry

Unread post by Heiwa »

lux, have your really read Pari's book? :rolleyes:
lux
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Re: Re-entry

Unread post by lux »

Yep.
Fred54
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Re: Re-entry

Unread post by Fred54 »

@ hoi
I have always wanted to use the wind up alarm clock analogy but the opportunity has never arose until now. It makes it pretty crystal clear what each definition comprises yes/no?
hoi.polloi
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Re: Re-entry

Unread post by hoi.polloi »

Thanks for the clear and curt replies.

Please elaborate, Fred54!
Fred54
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Re: Re-entry

Unread post by Fred54 »

@hoi
Elaborate on the gravitation? I will have to think on that for a bit. I posted in my welcome that I am a mechanical engineer not a physicist. Here is the difference. There was a giant science hall and in the middle of it was suspended a 17 pound (8kg) bowling ball attached to the ceiling by a 50 ft (17m) rope. An engineer walked over to the bowling ball and pulled it sideways until he got the bowling ball at the exact height of his nose, held his nose next to it and let the ball go. It swung out and returned and at the exact moment it stopped he grabbed it. A physicist saw what he did and said "I can do that" took the bowling ball placed it on his nose just like the engineer and gave it a shove...... :rolleyes:

Like I said I will have to think on this, right now I am still trying to shoot large holes in the mars lander and I haven't even begun to read the tons of other things that are contained in this forum, many of which I have spent a major amount of time investigating on my own.
hoi.polloi
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Re: Re-entry

Unread post by hoi.polloi »

Alright. That is an amusing story. Eh, but I meant this:
I have always wanted to use the wind up alarm clock analogy but the opportunity has never arose until now. It makes it pretty crystal clear what each definition comprises yes/no?
What did you mean by the wind up alarm clock?
Fred54
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Re: Re-entry

Unread post by Fred54 »

OMG I am sorry I must be really old now here is a picture
Image

There is a spring in the clock and you wind a key on the back so that it will work, no batteries, no wall plug and the alarm is the 2 bells on the top and there is a little striker between that hits them.
So my analogy is using that as a primary source of energy to lift the box car and of course you would have to wind it every day as well
hence 10 years vs 10 minutes
Clocks have been mechanical devices since their invention it is only in the last 50 years that they have become electrical.
Heiwa
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Re: Re-entry

Unread post by Heiwa »

Next re-entry is Oct. 28 for the SpaceX Dragon module:
http://www.cbsnews.com/8301-205_162-575 ... e-station/
... Again using the robot arm, ISS station commander Sunita Williams and Japanese astronaut Akihiko Hoshide plan to un-berth the capsule Oct. 28 for re-entry and splashdown in the Pacific Ocean off the southern coast of California where recovery crews will be standing by.

The Russian Soyuz spacecraft that ferry crews to and from the space station can only carry a few hundred pounds of small items back to Earth. All other station vehicles -- unmanned Russian Progress supply ships and European and Japanese cargo craft -- burn up during re-entry.
I am glad to see that all other station vehicles, without heat shields, I assume, burn up during re-entry, so that they will not be in the way. :lol:

So the only question that remains is: Can a heat shield absorb enough friction heat to prevent capsule burning up, while turbulence decelerate the capsule to a suitable speed when parachutes can be deployed?
Last edited by Heiwa on Tue Oct 16, 2012 3:09 pm, edited 1 time in total.
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