- Thank you received: 0
Precision of the solar eclipse test
21 years 10 months ago #4391
by jacques
Reply from was created by jacques
No response from anybody...
Maybe nobody that read this message have the book...
Maybe nobody that read this message have the book...
Please Log in or Create an account to join the conversation.
- tvanflandern
- Offline
- Platinum Member
Less
More
- Thank you received: 0
21 years 10 months ago #4838
by tvanflandern
Replied by tvanflandern on topic Reply from Tom Van Flandern
<BLOCKQUOTE id=quote><font size=2 face="Verdana, Arial, Helvetica" id=quote>quote:<hr height=1 noshade id=quote>Was the "gravity maximum" mesured with a gravimeter?<hr height=1 noshade id=quote></BLOCKQUOTE id=quote></font id=quote><font face="Verdana, Arial, Helvetica" size=2 id=quote>
No, it came from direct measures of orbital acceleration of Earth toward the Sun using radar ranging and spacecraft telemetry. -|Tom|-
No, it came from direct measures of orbital acceleration of Earth toward the Sun using radar ranging and spacecraft telemetry. -|Tom|-
Please Log in or Create an account to join the conversation.
21 years 10 months ago #4497
by jacques
Replied by jacques on topic Reply from
Thank you for your answer!
Is there any report of this experiment on the internet?
When, where and who did the experiment?
What eclipse(s) was it ?
Were you taking measurements during the eclipse or did you deduced the position from previous mesurements? (bouncing radar on planets and/or recording satellite data, while the eclipse took place)
Must need a lot of data to mesure such a small difference!
Thank you in advance for your answers.
Is there any report of this experiment on the internet?
When, where and who did the experiment?
What eclipse(s) was it ?
Were you taking measurements during the eclipse or did you deduced the position from previous mesurements? (bouncing radar on planets and/or recording satellite data, while the eclipse took place)
Must need a lot of data to mesure such a small difference!
Thank you in advance for your answers.
Please Log in or Create an account to join the conversation.
- tvanflandern
- Offline
- Platinum Member
Less
More
- Thank you received: 0
21 years 10 months ago #4512
by tvanflandern
Replied by tvanflandern on topic Reply from Tom Van Flandern
<BLOCKQUOTE id=quote><font size=2 face="Verdana, Arial, Helvetica" id=quote>quote:<hr height=1 noshade id=quote>[jacques]: Is there any report of this experiment on the internet? When, where and who did the experiment? What eclipse(s) was it ? Were you taking measurements during the eclipse or did you deduced the position from previous mesurements? (bouncing radar on planets and/or recording satellite data, while the eclipse took place)
Must need a lot of data to mesure such a small difference!<hr height=1 noshade id=quote></BLOCKQUOTE id=quote></font id=quote><font face="Verdana, Arial, Helvetica" size=2 id=quote>
Sorry, I thought you were asking about the difference between the direction of Earth's acceleration and the direction of sunlight. The eclipse experiment is completely different. A short description appears in "The speed of gravity ¡V What the experiments say", Phys.Lett.A, v. 250, #1-3, pp. 1-11 (1998/12/21). The preprint on the web site has fewer details. Here is the text from the publication. -|Tom|-
<b>The solar eclipse test</b>
Yet another manifestation of the difference between the propagation speeds of gravity and light can be seen in the case of solar eclipses (7). The Moon, being relatively nearby and sharing the Earth¡¦s 30 km/s orbital motion around the Sun, has relatively little aberration (0.7 arc seconds, due to the Moon¡¦s 1 km/s orbital speed around Earth). The Sun, as mentioned earlier, has an aberration of just over 20 arc seconds. It takes the Moon about 40 seconds of time to move 20 arc seconds on the sky relative to the Sun. Since the observed times of eclipses of the Sun by the Moon agree with predicted times to within a couple of seconds, we can use the orbits of the Sun and the Moon near times of maximum solar eclipse to compare the time of predicted gravitational maximum with the time of visible maximum eclipse.
In practice, the maximum gravitational perturbation by the Sun on the orbit of the Moon near eclipses may be taken as the time when the geometric lunar and solar longitudes are equal. For those interested in the technical details, the computed orbit of the Moon uses instantaneous force transmission between bodies; i.e., forces in numerical integrations are computed using true, instantaneous positions at a common instant of coordinate time. This is standard procedure in celestial mechanics. So I did an analysis to solve for any correction that might exist to the large perturbations of the Moon's orbit that depend upon the elongation angle D between the Moon and the Sun. The visible elongation angle is about 20 arc seconds larger than the computed elongation angle because light takes much longer to arrive from the Sun than from the Moon. So any transit delay for the gravitational force of the Sun would show up as a correction to the adopted value of D based on instantaneous force transmission. In particular, if gravity propagated at the speed of light, the correction to D as used in the computations would be 20 arc seconds. But an extensive analysis of lunar occultation timing data showed that the correction to D is zero to within an uncertainty of about one arc second. The elongation angle increases at the rate of an arc second every two seconds of time. So this result implies a difference between the time of optical and gravitational maxima at the time of eclipses of 40„b2 seconds. These results in turn imply that the propagation velocity of the force of the Sun acting on the Moon is at least 20 times the speed of light. Although this constraint is far less severe than some others, it uses three bodies (Sun, Moon, Earth) instead of two and changes in gravitational fields instead of static fields. Yet, it still constrains the speed of gravitational interaction to exceed the speed of light.
Must need a lot of data to mesure such a small difference!<hr height=1 noshade id=quote></BLOCKQUOTE id=quote></font id=quote><font face="Verdana, Arial, Helvetica" size=2 id=quote>
Sorry, I thought you were asking about the difference between the direction of Earth's acceleration and the direction of sunlight. The eclipse experiment is completely different. A short description appears in "The speed of gravity ¡V What the experiments say", Phys.Lett.A, v. 250, #1-3, pp. 1-11 (1998/12/21). The preprint on the web site has fewer details. Here is the text from the publication. -|Tom|-
<b>The solar eclipse test</b>
Yet another manifestation of the difference between the propagation speeds of gravity and light can be seen in the case of solar eclipses (7). The Moon, being relatively nearby and sharing the Earth¡¦s 30 km/s orbital motion around the Sun, has relatively little aberration (0.7 arc seconds, due to the Moon¡¦s 1 km/s orbital speed around Earth). The Sun, as mentioned earlier, has an aberration of just over 20 arc seconds. It takes the Moon about 40 seconds of time to move 20 arc seconds on the sky relative to the Sun. Since the observed times of eclipses of the Sun by the Moon agree with predicted times to within a couple of seconds, we can use the orbits of the Sun and the Moon near times of maximum solar eclipse to compare the time of predicted gravitational maximum with the time of visible maximum eclipse.
In practice, the maximum gravitational perturbation by the Sun on the orbit of the Moon near eclipses may be taken as the time when the geometric lunar and solar longitudes are equal. For those interested in the technical details, the computed orbit of the Moon uses instantaneous force transmission between bodies; i.e., forces in numerical integrations are computed using true, instantaneous positions at a common instant of coordinate time. This is standard procedure in celestial mechanics. So I did an analysis to solve for any correction that might exist to the large perturbations of the Moon's orbit that depend upon the elongation angle D between the Moon and the Sun. The visible elongation angle is about 20 arc seconds larger than the computed elongation angle because light takes much longer to arrive from the Sun than from the Moon. So any transit delay for the gravitational force of the Sun would show up as a correction to the adopted value of D based on instantaneous force transmission. In particular, if gravity propagated at the speed of light, the correction to D as used in the computations would be 20 arc seconds. But an extensive analysis of lunar occultation timing data showed that the correction to D is zero to within an uncertainty of about one arc second. The elongation angle increases at the rate of an arc second every two seconds of time. So this result implies a difference between the time of optical and gravitational maxima at the time of eclipses of 40„b2 seconds. These results in turn imply that the propagation velocity of the force of the Sun acting on the Moon is at least 20 times the speed of light. Although this constraint is far less severe than some others, it uses three bodies (Sun, Moon, Earth) instead of two and changes in gravitational fields instead of static fields. Yet, it still constrains the speed of gravitational interaction to exceed the speed of light.
Please Log in or Create an account to join the conversation.
21 years 10 months ago #4517
by jacques
Replied by jacques on topic Reply from
In a few word there is no mesurable difference in the computed position using instantanous effect of gravity and the observed position corrected by the time delay of light.
Is that the general idea?
Thank you
Is that the general idea?
Thank you
Please Log in or Create an account to join the conversation.
- tvanflandern
- Offline
- Platinum Member
Less
More
- Thank you received: 0
21 years 10 months ago #4533
by tvanflandern
Replied by tvanflandern on topic Reply from Tom Van Flandern
<BLOCKQUOTE id=quote><font size=2 face="Verdana, Arial, Helvetica" id=quote>quote:<hr height=1 noshade id=quote>[jacques]: In a few word there is no mesurable difference in the computed position using instantanous effect of gravity and the observed position corrected by the time delay of light. Is that the general idea?<hr height=1 noshade id=quote></BLOCKQUOTE id=quote></font id=quote><font face="Verdana, Arial, Helvetica" size=2 id=quote>
Yes. It's the same result as in the two-body problem, but extended to three bodies. The direction of gravitational acceleration toward the Sun, and the direction from which the Sun's light arrives, are not the same. The only way to explain that without a host of objectionable complications is if the propagation speed of light and the force of gravity are unequal. -|Tom|-
Yes. It's the same result as in the two-body problem, but extended to three bodies. The direction of gravitational acceleration toward the Sun, and the direction from which the Sun's light arrives, are not the same. The only way to explain that without a host of objectionable complications is if the propagation speed of light and the force of gravity are unequal. -|Tom|-
Please Log in or Create an account to join the conversation.
Time to create page: 0.391 seconds