# Astronomy Question

## Main Question or Discussion Point

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??

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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.

D H
Staff Emeritus
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

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

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.

D H
Staff Emeritus