Editing Lagrange point (section)

==={{L1|nolink=yes}}===
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The location of L<sub>1</sub> is the solution to the following equation, gravitation providing the centripetal force:
<math display="block">\frac{M_1}{(R-r)^2}-\frac{M_2}{r^2}=\left(\frac{M_1}{M_1+M_2}R-r\right)\frac{M_1+M_2}{R^3}</math>
where ''r'' is the distance of the L<sub>1</sub> point from the smaller object, ''R'' is the distance between the two main objects, and ''M''<sub>1</sub> and ''M''<sub>2</sub> are the masses of the large and small object, respectively. The quantity in parentheses on the right is the distance of L<sub>1</sub> from the center of mass. The solution for ''r'' is the only [[Real number|real]] [[Zero of a function|root]] of the following [[quintic function]]

<math display="block">x^5 + (\mu - 3) x^4 + (3 - 2\mu) x^3 - (\mu) x^2 + (2\mu) x - \mu = 0</math>
where <math display="block"> \mu = \frac{M_2}{M_1+M_2} </math>
is the mass fraction of ''M<sub>2</sub>'' and <math display="block"> x = \frac{r}{R} </math>
is the normalized distance. If the mass of the smaller object (''M''<sub>2</sub>) is much smaller than the mass of the larger object (''M''<sub>1</sub>) then {{L1|nolink=yes}} and {{L2|nolink=yes}} are at approximately equal distances ''r'' from the smaller object, equal to the radius of the [[Hill sphere]], given by:
<math display="block">r \approx R \sqrt[3]{\frac{\mu}{3}}</math>

We may also write this as:
<math display="block">\frac{M_2}{r^3}\approx 3\frac{M_1}{R^3}</math>
Since the [[Tidal force|tidal]] effect of a body is proportional to its mass divided by the distance cubed, this means that the tidal effect of the smaller body at the L{{sub|1}} or at the L{{sub|2}} point is about three times of that body. We may also write:
<math display="block">\rho_2\left(\frac{d_2}{r}\right)^3\approx 3\rho_1\left(\frac{d_1}{R}\right)^3</math>
where ''ρ''{{sub|1}} and ''ρ''{{sub|2}} are the average densities of the two bodies and ''d''{{sub|1}} and ''d''{{sub|2}} are their diameters. The ratio of diameter to distance gives the angle subtended by the body, showing that viewed from these two Lagrange points, the apparent sizes of the two bodies will be similar, especially if the density of the smaller one is about thrice that of the larger, as in the case of the Earth and the Sun.

This distance can be described as being such that the [[orbital period]], corresponding to a circular orbit with this distance as radius around ''M''<sub>2</sub> in the absence of ''M''<sub>1</sub>, is that of ''M''<sub>2</sub> around ''M''<sub>1</sub>, divided by {{sqrt|3}} ≈ 1.73:
<math display="block">T_{s,M_2}(r) = \frac{T_{M_2,M_1}(R)}{\sqrt{3}}.</math>