Attraction Forces During Total Eclipse: Sun, Moon, and Earth Explained

  • Thread starter Thread starter zarmewa
  • Start date Start date
  • Tags Tags
    Astronomy
AI Thread Summary
During a total eclipse, the gravitational forces acting on the Moon from the Sun and Earth are analyzed using the formula F = GMm/r². The calculations show that the net force on the Moon is directed towards the Sun, yet it continues to revolve around the Earth due to the Earth's gravitational influence and their mutual acceleration towards the Sun. The discussion highlights that while the Sun's gravitational pull is stronger, the Moon's orbit around the Earth is maintained because it is gravitationally bound to the Earth. Additionally, the concept of escape velocity is examined, but it is clarified that the Moon's gravitational relationship with the Earth is the primary factor in its orbit. Ultimately, the Moon's orbit is a result of the balance of gravitational forces and relative motion within the Earth-Sun system.
zarmewa
Messages
44
Reaction score
0
Here is the summary of attraction forces (F = GMm/r2) during Total Eclipse when the Moon “M” is between Sun “S” and Earth “E”.
Sun (F1 = GMm/r2) Moon (F2= GMm/r2) Earth

(F1 = GMm/r2) = S - M = 4.1984 x 1020
(F2 = GMm/r2) = M - E = 2.2 x 10^20

Net force on the Moon= F3 = F1-F2= 4.1984 x 1020 Minus 2.2 x 1020 = 1.998 x 1020 towards Sun

At this point why Earth force the moon to revolve around its centre when the net force on the moon is much greater towards the sun ? Explain please. Pls also check the calculation. If we consider the Sun - Earth Force
(F4= GMm/r2) = S - E= 3.67 x 1022 , then

Net Force on the Moon = F3+F4=1.998 x 1020 Plus 3.67 x 1022 = 3.68 x 1022 towards Sun.
Please also note that force of attraction F = GMm/r2 between sun and moon in any case (perigee, apogee, average) is much greater than between moon and Earth F = GMm/r2 (perigee, apogee, average). So technically it should revolve around the sun in a separate orbit not earth. So why moon revolves around earth?

Here is the other Forum answer but I disagree because law of gravitation can not applied to the common center of gravity of two masses.
“Both Earth and moon are constantly revolving around the sun. They are also revolving around their common center of gravity as they move around the sun, but they constantly move around the sun.

Their speed of revolution and the vast amount of angular momentum keeps them from simply dropping like proverbial rocks into the sun.

Think of astronauts aboard the Space Station. They are weightless, yet they are just a hundred miles or so farther from the center of the Earth than you and me. Why? Why don't they fall?

The answer is that they ARE falling. But as fast as they fall, their momentum has carried them forward so that they endlessly fall around the earth, not into it. The same with the Earth and moon, relative to the sun, during a total eclipse and at all other times”.

So what do you suggest??
 
Astronomy news on Phys.org
The answer from another forum you received seems correct, whether or not you agree.

The part you seem to be missing in your calculation is the fact that the Earth is falling towards the sun at approximately the same speed as the moon is. So you're right, the moon is accelerating towards the sun, but so is the Earth.
 
As Jack said in post #2, you are ignoring that the Earth is also accelerating toward the Sun. What I suggest you do is account for the Earth's acceleration. In other words, you should be the acceleration of the Moon with respect to the Earth.

At the instant of a total eclipse, the Moon's acceleration with respect to the Earth is directed toward the Earth and is equal to

a_m =<br /> \frac{G(M_e+M_m)}{{r_m}^2}<br /> - \frac{G M_s}{{r_e}^2}\left(\frac 1{(1-r_m/r_e)^2}-1\right)<br /> \approx<br /> \frac{G(M_e+M_m)}{{r_m}^2}<br /> - 2\frac{G M_s}{{r_e}^2}\frac {r_m}{r_e}<br />

where Ms, Me, and Mm are the masses of the Sun, Earth, and Moon, re is the distance between the Sun and the Earth, and rm is the distance between the Earth and the Moon.

When looked at in this light, the Sun's gravitational influence is small (about 1/90th) compared to that of the Earth.
 
Sun--------(gm, wrt s=GMs/Rsm^2 )-------Moon-----(gm, wrt e=GMe/Rem2)-----Earth
gm, wrt s >>>> gm, wrt e , gm = accelaration of moon
Since no one agree to this, therefore would I be wrong to represent the same problem with escape velocity of moon Vem wrt to both sun and Earth when it is in between Earth and sun in total eclipse.

Sun-------(Vem, wrt s=2GMs/Rsm)1/2 -----Moon---(Vem, wrt e=2GMe/Rem)1/2---Earth

Vem, wrt s=(2GMs/Rsm)1/2 >>> Vem, wrt e=(2GMe/Rem)1/2
Vem, wrt s >>>> Vem, wrt e
Please check all the calculation and then you will know how many time it is greater. So Earth has to use extra force to escape moon from the sun gravity. So am I wrong to say that moon can not escape from sun at this point towards earth.
 
So what? You are using the wrong metric. Nobody is claiming that the Moon, along with the Earth, and all of its artificial satellites, are not orbiting the Sun, or that they are not orbiting the Milky Way, or even some larger structure in space. The Moon is orbiting the Earth by virtue of a simple test: Is the Moon gravitationally bound to the Earth? That the Moon is also gravitationally bound to the Sun, or to the Milky Way, or something even larger, is irrelevant to the question of whether it is gravitationally bound to the Earth.
 
Publication: Redox-driven mineral and organic associations in Jezero Crater, Mars Article: NASA Says Mars Rover Discovered Potential Biosignature Last Year Press conference The ~100 authors don't find a good way this could have formed without life, but also can't rule it out. Now that they have shared their findings with the larger community someone else might find an explanation - or maybe it was actually made by life.
TL;DR Summary: In 3 years, the Square Kilometre Array (SKA) telescope (or rather, a system of telescopes) should be put into operation. In case of failure to detect alien signals, it will further expand the radius of the so-called silence (or rather, radio silence) of the Universe. Is there any sense in this or is blissful ignorance better? In 3 years, the Square Kilometre Array (SKA) telescope (or rather, a system of telescopes) should be put into operation. In case of failure to detect...
Thread 'Could gamma-ray bursts have an intragalactic origin?'
This is indirectly evidenced by a map of the distribution of gamma-ray bursts in the night sky, made in the form of an elongated globe. And also the weakening of gamma radiation by the disk and the center of the Milky Way, which leads to anisotropy in the possibilities of observing gamma-ray bursts. My line of reasoning is as follows: 1. Gamma radiation should be absorbed to some extent by dust and other components of the interstellar medium. As a result, with an extragalactic origin, fewer...
Back
Top