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Asking a question here is quite intimidating for me -- while I love Physics, my high-school understanding only allows me to go so far... I've been trying to solve this problem for the last couple of days without success, so I thought that perhaps someone here wouldn't mind helping me to better understand it.

Lets say, our origin location is London, defined by the WGS84 coordinates 51.507222 and -0.1275. What I want to do, is calculate the magnitude (?) and direction of the gravitic force exerted by the Sun and Moon bodies (only, for the sake of simplicity) on the origin location for a given date. Using pyEphem:

origin = ephem.Observer()
origin.lon, = '51.507222', '-0.1275', origin.pressure = '2012/05/27 23:00:00', 0 # 0 ignores atmosferic refraction

moon = ephem.Moon(origin)
sun = ephem.Sun(origin)

print moon.distance * ephem.meters_per_au # 388989920.985 m (from Earth)
print sun.distance * ephem.meters_per_au  # 1.51606987131e+11 m (from Earth)

Assuming $1.988435 * 10^{30} kg$, $5.9721986 * 10^{24} kg$ and $7.3459 * 10^{22} kg$ as the mass of the Sun, Earth and Moon, respectively, I'm now able to calculate the gravitational force for each body individually:

$$ F(Earth|Sun) = G {5.9721986 * 10^{24} * 1.988435 * 10^{30} \over (1.51606987131 * 10^{11})^2} = 3.448 * 10^{22} N $$

$$ F(Earth|Moon) = G {5.9721986 * 10^{24} * 7.3459 * 10^{22} \over (3.88989920985 * 10^{8})^2} = 1.935 * 10^{20} N $$

Combining Gravitational Force Vectors in a 1-D Plane

Now, consider the following scenarios ($\#1$ and $\#2$) in the image, where the circles represent the Sun, Moon and Earth and the little black square represents a (small) object of $100$ kg at our origin location:

hypothetical 2-D scenarios

PS: Sorry for the imagery but I don't know how to adequately explain this using the proper jargon...

I'm not completely sure if this is the case, but my intuition tells me that the object will be slightly heavier in the first scenario and slightly lighter in the other. However, if I try to work out the math:

$$ F(Earth|Obj) = G {5.9721986 * 10^{24} * 100 \over (6.3675 * 10^{6})^2} = 983 N $$

$$ F_{\#1} = F(Earth|Obj) + F(Earth|Moon) + F(Earth|Sun) = 3.467 * 10^{22} N $$

$$ F_{\#2} = F(Earth|Obj) - F(Earth|Moon) - F(Earth|Sun) = -3.467 * 10^{22} N $$

Clearly, I'm doing something wrong here. I suspect the right answer is:

$$ F(Sun|Obj) = G {1.988435 * 10^{30} * 100 \over (1.51606987131 * 10^{11})^2} = 0.5773 N $$

$$ F(Moon|Obj) = G {7.3459 * 10^{22} * 100 \over (3.88989920985 * 10^{8})^2} = 0.00324 N $$

$$ F_{\#1} = F(Earth|Obj) + F(Moon|Obj) + F(Sun|Obj) = 983.58054 N $$

$$ F_{\#2} = F(Earth|Obj) - F(Moon|Obj) - F(Sun|Obj) = 982.41946 N $$

Which looks way more realistic, the mass difference on the object would only be of $≈ ±59$ grams. However, I'm a bit confused, since the mass of the object is part of the mass of the Earth, and this Wikipedia entry states that: "large spherically symmetrical masses attract and are attracted as if all their mass were concentrated at their centers" and "r is the distance between the centers of the masses".

Combined Gravitational Force Vectors in a Equatorial Coordinate System?

Continuing with the pyEphem code, I can get the Apparent Topocentric Position of both the Sun and the Moon, as seen from our chosen date and origin location (pressure = 0 ignores atmospheric refraction):

print moon.ra, moon.dec # 9:43:33.65, 7:47:56.6
print sun.ra, sun.dec   # 4:20:51.51, 21:28:53.9

Converting these values to degrees yields:

  • Moon: $145.89º$ Right Ascension, $+7.80º$ Declination
  • Sun: $65.21º$ Right Ascension, $+21.48º$ Declination

Where the Right Ascension is $Φ$ (azimuthal) and the Declination is $Θ$ (polar); I believe $R$ (radial) should either be the distance from the origin location to the center of mass $X$ (but then I wouldn't be able to calculate the resultant force, would I?) or the result of $F(Body|Obj)$ -- would this make sense?

So, converting the above spherical coordinates into $(X, Y, Z)$ Cartesian coordinates gives:

$$ X_{Earth} = F(Earth|Obj) * cos(0) * sin(0) = 0 $$

$$ Y_{Earth} = F(Earth|Obj) * sin(0) * sin(0) = 0 $$

$$ Z_{Earth} = F(Earth|Obj) * cos(0) = 983 $$

$$ X_{Sun} = F(Sun|Obj) * cos(21.48) * sin(65.21) = 0.48802 $$

$$ Y_{Sun} = F(Sun|Obj) * sin(21.48) * sin(65.21) = 0.19185 $$

$$ Z_{Sun} = F(Sun|Obj) * cos(65.21) = 0.24189 $$

$$ X_{Moon} = F(Moon|Obj) * cos(7.80) * sin(145.89) = 0.00180 $$

$$ Y_{Moon} = F(Moon|Obj) * sin(7.80) * sin(145.89) = 0.00025 $$

$$ Z_{Moon} = F(Moon|Obj) * cos(145.89) = -0.00268 $$

$$ (0, 0, 983) + (0.48802, 0.19185, 0.24189) + (0.00180, 0.00025, -0.00268) $$

Using the above vectors as input on WolframAlpha yields a vector length of $983.239$ (Newtons?) -- is this my resultant force? This would mean that my $100$ kg object has gained (or lost?) $262$ grams, which is more than my exaggerated 1-dimensional scenario would allow. What might I be doing wrong?

I realize that this probably isn't the right way to tackle this, I also noticed that there is a vector equation for Newton's Law of Universal Gravitation, but honestly I don't even understand the notation it uses.

I hope this isn't too much of a dumb question... And please let me know if I should explain it better.

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up vote 1 down vote accepted

Not to worry, you've explained your question fine. However, your lack of familiarity with vectors is probably making this a lot more complicated than it could be.

Anyway, to get to the point: the correct way to calculate the force in the one-dimensional case is

$$\sum F = F(\text{Earth|Obj}) + F(\text{Moon|Obj}) + F(\text{Sun|Obj})$$

It's not correct to assume that the object is part of the Earth in this problem, because you're calculating the force on the object itself. In fact, it's meaningless to add up forces that act on different things. So in your first (incorrect) calculation, where you add one force on the object to two other forces on the Earth, you're absolutely going to get a nonsensical result.

When you get to doing this in the 3D case, you do the same thing except with vectors:

$$\sum \vec F = \vec F(\text{Earth|Obj}) + \vec F(\text{Moon|Obj}) + \vec F(\text{Sun|Obj})$$

This just means that you need to use a copy of this equation for each component, $x$, $y$, and $z$.

$$\begin{align}\sum F_x &= F_x(\text{Earth|Obj}) + F_x(\text{Moon|Obj}) + F_x(\text{Sun|Obj}) \\ \sum F_y &= F_y(\text{Earth|Obj}) + F_y(\text{Moon|Obj}) + F_y(\text{Sun|Obj}) \\ \sum F_z &= F_z(\text{Earth|Obj}) + F_z(\text{Moon|Obj}) + F_z(\text{Sun|Obj})\end{align}$$

This you've done correctly. You just made a mistake in calculating the equivalent mass change: note that $983.239\text{ N}$ is between the two extremes you calculated in the 1D case, so the equivalent mass it gives should be less than $59\text{ g}$ different from $100\text{ kg}$. In fact it corresponds to a difference in equivalent mass of about $14\text{ g}$, not $262\text{ g}$.

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I don't know how I got to the 262g, I probably wasn't paying the due attention at the time, but I'm very happy to know that my reasoning was correct given that I picked almost everything I know about vectors from Youtube videos. Thank you so much for your answer! =) – Alix Axel May 28 '12 at 18:07
Sure, no problem. I do also want to mention that the mass of the object does not change, only the weight. – David Z May 28 '12 at 18:13

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