- Thank you received: 0
Requiem for Relativity
- Joe Keller
- Offline
- Platinum Member
Less
More
13 years 10 months ago #24049
by Joe Keller
Replied by Joe Keller on topic Reply from
An empirical relation between Earth's mass, Earth's radius, Earth's semimajor orbital axis, the speed of light, and the Golden Ratio
Sconzo (Astronomical Journal 67:19-21, 1962) noted that if one chooses the unit of length to be Earth's radius, and chooses the unit of mass to be Earth's mass, then the gravitational constant will be exactly one, if the unit of time is 806.832 sec. That is, the (average) acceleration of gravity at Earth's surface, is one Earth radius per 807sec, per 807sec.
The time for light to travel one Astronomical Unit ( = semimajor axis of Earth's orbit) is 1.4959787*10^13 / (2.9979246*10^10) = 499.0048sec. The Golden Ratio is (1+sqr(5))/2 = 1.618034.
806.832/499.0048 = 1.616882
The ratio, 1.000712, of the Golden Ratio, to this number, differs from one, by about the present relative uncertainty of the gravitational constant. So, the numbers are the same, as nearly as they have been determined.
Sconzo (Astronomical Journal 67:19-21, 1962) noted that if one chooses the unit of length to be Earth's radius, and chooses the unit of mass to be Earth's mass, then the gravitational constant will be exactly one, if the unit of time is 806.832 sec. That is, the (average) acceleration of gravity at Earth's surface, is one Earth radius per 807sec, per 807sec.
The time for light to travel one Astronomical Unit ( = semimajor axis of Earth's orbit) is 1.4959787*10^13 / (2.9979246*10^10) = 499.0048sec. The Golden Ratio is (1+sqr(5))/2 = 1.618034.
806.832/499.0048 = 1.616882
The ratio, 1.000712, of the Golden Ratio, to this number, differs from one, by about the present relative uncertainty of the gravitational constant. So, the numbers are the same, as nearly as they have been determined.
Please Log in or Create an account to join the conversation.
13 years 10 months ago #21032
by Jim
Replied by Jim on topic Reply from
Dr Je, It seems dreadfully complex to set the gravity constant at one-at least to me. What is to be gained by this project?
Please Log in or Create an account to join the conversation.
- Joe Keller
- Offline
- Platinum Member
Less
More
- Thank you received: 0
13 years 10 months ago #24022
by Joe Keller
Replied by Joe Keller on topic Reply from
<blockquote id="quote"><font size="2" face="Verdana, Arial, Helvetica" id="quote">quote:<hr height="1" noshade id="quote"><i>Originally posted by Jim</i>
<br />...complex to set the gravity constant at one...<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">
Here is an equivalent way of saying what I said in my previous post:
The acceleration of gravity at the surface of the Planet (not counting centrifugal force due to rotation) times T squared (where T is the light time from the Sun to the Planet, at its semimajor axis, i.e. approx. its average distance from the Sun) times the "Golden Ratio" ( 1.618034 = (1+sqrt(5))/2 ) to some whole power, equals the radius of the Planet. Like the NASA "Fact Sheets", I use the "volumetric average" radius, i.e. the geometric mean of the three semi-axes of the Planet's ellipsoidal shape (two of which are equal). Using the current figures on NASA's Fact Sheets, I find that all the rocky planets, except Venus, and spherical minor planets, on which data are available, conform to this law.
I approximated Ceres as an ellipsoid, used the larger of the two diameters (customary for asteroids) given by NASA as the equatorial, then found the volumetric average radius myself. For Charon, only an equatorial radius was given, so I used it as the average, since for Pluto the equatorial and polar radii are given as equal. Other minor planets are not very round and/or lack moons and have unknown compositions, so their surface gravity is unknown or undefined. The gas giant planets have surface accelerations which NASA gives arbitrarily as that at a depth of one bar pressure in their atmosphere.
The formula is:
(surf. accel.)*(Sun light time)^2 * 1.618...^n = radius of Planet
Here is a list of the values of "n" found:
Mercury 5.968
Venus 3.441
Earth 1.994
Mars 0.951
Jupiter -1.812
Saturn -2.932
Uranus -7.209
Neptune -9.604
Pluto -10.89
Charon -11.04
Ceres 2.818
For the outer planets (Jupiter & beyond) "n" is negative.
For the minor planets Pluto & Ceres the last digits above are not significant, and for Charon the next to last digit is hardly significant. For these planets, NASA gives the masses or surface accelerations only to two, or not quite three, decimal places. Agreement is excellent for Earth; good for Mercury, Mars, and Saturn; and at least fair for Jupiter, Uranus, Pluto, Charon and Ceres.
<br />...complex to set the gravity constant at one...<hr height="1" noshade id="quote"></blockquote id="quote"></font id="quote">
Here is an equivalent way of saying what I said in my previous post:
The acceleration of gravity at the surface of the Planet (not counting centrifugal force due to rotation) times T squared (where T is the light time from the Sun to the Planet, at its semimajor axis, i.e. approx. its average distance from the Sun) times the "Golden Ratio" ( 1.618034 = (1+sqrt(5))/2 ) to some whole power, equals the radius of the Planet. Like the NASA "Fact Sheets", I use the "volumetric average" radius, i.e. the geometric mean of the three semi-axes of the Planet's ellipsoidal shape (two of which are equal). Using the current figures on NASA's Fact Sheets, I find that all the rocky planets, except Venus, and spherical minor planets, on which data are available, conform to this law.
I approximated Ceres as an ellipsoid, used the larger of the two diameters (customary for asteroids) given by NASA as the equatorial, then found the volumetric average radius myself. For Charon, only an equatorial radius was given, so I used it as the average, since for Pluto the equatorial and polar radii are given as equal. Other minor planets are not very round and/or lack moons and have unknown compositions, so their surface gravity is unknown or undefined. The gas giant planets have surface accelerations which NASA gives arbitrarily as that at a depth of one bar pressure in their atmosphere.
The formula is:
(surf. accel.)*(Sun light time)^2 * 1.618...^n = radius of Planet
Here is a list of the values of "n" found:
Mercury 5.968
Venus 3.441
Earth 1.994
Mars 0.951
Jupiter -1.812
Saturn -2.932
Uranus -7.209
Neptune -9.604
Pluto -10.89
Charon -11.04
Ceres 2.818
For the outer planets (Jupiter & beyond) "n" is negative.
For the minor planets Pluto & Ceres the last digits above are not significant, and for Charon the next to last digit is hardly significant. For these planets, NASA gives the masses or surface accelerations only to two, or not quite three, decimal places. Agreement is excellent for Earth; good for Mercury, Mars, and Saturn; and at least fair for Jupiter, Uranus, Pluto, Charon and Ceres.
Please Log in or Create an account to join the conversation.
13 years 10 months ago #21048
by Jim
Replied by Jim on topic Reply from
Dr Joe, I'm not able to determine what "n" stands for-can you explain what it is and what units it has? It looks to me it could be a calculated ratio different for every body in the system. Your latest post sidesteps my question regarding setting the gravity constant at unity.
Please Log in or Create an account to join the conversation.
- Joe Keller
- Offline
- Platinum Member
Less
More
- Thank you received: 0
13 years 10 months ago #24023
by Joe Keller
Replied by Joe Keller on topic Reply from
Luna also conforms to the relationship given in my previous post. Using the NASA Fact Sheet figure for Earth's volumetric average radius, but the figures in the 2008 Astronomical Almanac, p. E4, for Earth's mass and for Luna's mass and radius,
mLuna / mEarth / (rLuna / rEarth )^3 = 0.60650,
so if Earth is changed to Luna, in the formula, an extra factor of 1/0.60650 = 1.6488 is needed, that is,
n = 1.994 + log(1.6488)/log(1.618034) = 3.033
so the exponent, n, for Luna, is 3, vs. n=2 for Earth and n=1 for Mars.
Furthermore, the Sun itself conforms fairly well:
(rSun /1AU)^2 * mSun/mEarth / (rSun/rEarth)^3 = 1/1.618...^25.1605
so the exponent, n, for the Sun, is 1.994+25.1605 = 27.15.
mLuna / mEarth / (rLuna / rEarth )^3 = 0.60650,
so if Earth is changed to Luna, in the formula, an extra factor of 1/0.60650 = 1.6488 is needed, that is,
n = 1.994 + log(1.6488)/log(1.618034) = 3.033
so the exponent, n, for Luna, is 3, vs. n=2 for Earth and n=1 for Mars.
Furthermore, the Sun itself conforms fairly well:
(rSun /1AU)^2 * mSun/mEarth / (rSun/rEarth)^3 = 1/1.618...^25.1605
so the exponent, n, for the Sun, is 1.994+25.1605 = 27.15.
Please Log in or Create an account to join the conversation.
13 years 10 months ago #21053
by Stoat
Replied by Stoat on topic Reply from Robert Turner
First thoughts on this Joe, how do you think it relates to the fission v collision models of planetary formation?
Please Log in or Create an account to join the conversation.
Time to create page: 0.402 seconds