Suppose that a planet were discovered between the sun and Mercury, with a circular orbit of radius equal to 2/3 of the average orbit radius of Mercury. (Such a planet was once postulated, in part to explain the precession of Mercury's orbit. It was even given the name Vulcan, although we now have no evidence that it actually exists. Mercury's precession has been explained by general relativity.) The orbital period of Mercury is 88.0 days.
What would be the orbital period of such a planet?
Physics Problem: Orbital Period of Vulcan?
2008-04-02 7:17 pm
回答 (2)
2008-04-02 9:12 pm
✔ 最佳答案
The physics is this...anything in orbit has two major forces acting on it...centripetal force F, which comes from gravity, and centrifugal force C, which is the reaction to centripetal. The two forces are in balance (F = C) because the orbiter of mass m is neither falling into the attracting body of mass M nor is it flying away. Thus in general, we can write F = GmM/R^2 = mW^2 R = C and right away we can see that GM/R^3 = W^2 = (1/T)^2 and T^2/R^3 = 1/GM = constant for a given attracting mass M like the Sun. Thus, if there is another body with radius r = 2R/3 for example and period t, we have T^2/R^3 = 1/GM = t^2/r^3
Thus, t^2 = T^2 (r/R)^3 and t = T sqrt((r/R)^3) = 88*sqrt((2/3)^3) = 47.9 days for Vulcan orbiting at r = 2R/3; where T = 88 day orbit for Mercury orbiting at R.
2008-04-02 7:27 pm
What's that formula?
D1<cubed> D2<cubed>
---------------- = ----------------
P1<squared> P2<squared>
Let's say D1 is the distance of Mercurey to the Sun.
Let's say P1 is the orbital period.
Let's say D2 would be the body in question.
D1<cubed> D2<cubed>
---------------- = ----------------
P1<squared> P2<squared>
Let's say D1 is the distance of Mercurey to the Sun.
Let's say P1 is the orbital period.
Let's say D2 would be the body in question.
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