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# Physics/Earth's sphere of influence

Question
How does one utilize the 2/5 in the formula for calculating a planet's sphere of influence. I found the formula at http://en.wikipedia.org/wiki/Sphere_of_influence_(Astrodyamics) I understand a, (m/M) but not how to apply the 2/5 above and to the right of the (m/M). 9.27 e 8 meters is listed as the Earth's sphere of influence but it seems neither I nor my coworkers understand the 2/5 part of the formula.

For the Earth, specifically, the Hill sphere is more accurately what you want.  http://en.wikipedia.org/wiki/Hill_sphere  It's also better derived, but it's an approximation.  It basically can be approximated at a minimum by the radius of the L1 point of the Earth orbiting the Sun (see here: http://en.wikipedia.org/wiki/Lagrangian_point ).  There are many descriptors for it, but the wikipedia is accurate.  The technical solution is indeed (read the details) the solution to a quintic function, so I can see why using 2/5=0.4 as an approximation will get you close...but someone on the Hill sphere page and the Lagrangian point page gave the perhaps more accurate approximation of R(m/3M)^1/3.  Keep in mind that the Hill sphere is not a sphere, but rather an oblate spheroid with a complex and asymmetric shape in three dimensions.  This is especially true for the Earth, with our relatively large moon.  These are approximations for a kinda-spherical volume of space which changes as the Earth orbits the Sun.

Excellent question, by the way.

Physics

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#### Dr. Stephen O. Nelson

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