Evil Dr Ganymede wrote:First step is to multiply the Sun's mass by 3. Then press the 1/X button to invert that. Then multiply that number by the Earth's mass. This gives you the m/3M term in the brackets in
this equation.
Then multiply that number by the distance.
Finally, (I'm assuming you have a reasonably scientific calculator here), press the "x to the power y" button and then enter (1/3), then press "equals" to get the final answer. Or just press the "cube root" button if your calculator has that.
Or, if you have Excel, then use this formula in a cell to figure it out:
=(a*(m/(3*M)))^(1/3)
(where a, m, and M are the cells that have the orbital distance in metres, the mass of the Earth, and the mass of the Sun respectively).
You have to do the cube root before you multiply by a. So the calculator instructions should be
First step is to multiply the Sun's mass by 3. Then press the 1/X button to invert that. Then multiply that number by the Earth's mass.
Then press the "x to the power y" button and then enter (1/3), then press "equals" to get the final answer. Or just press the "cube root" button if your calculator has that.
Finally multiply that number by the distance.
or for Excel:
=a*(m/(3*M))^(1/3)
Other interesting Hill stuff:
This formula works for objects that orbit in the same plane as the priamary and secondary. But an object in a retrograde orbit can orbit well past the Hill radius as defined by this formula, in a weird triangluar orbit.
No object can exist beyond the Moon's orbit in a prograde orbit, not because of the Hill radius, but because the Moon's influence will either eject it, or cause it to collide with the Earth or the Moon.
Mars' moons, Phobos and Deimos have a hill sphere that is below their surfaces. Similar to the question posed in one of your links, "can an astronaut orbit the space shuttle?", the answer is no, nothing can orbit Phobos or Deimos.