How do you calculate a planet's period based on K and radius?
Here's a question I need help solving:
The value of K for planets moving around our sun is 3.35 x 10^18 m^3/s^2. If Neptune's orbit has an average radius of 4.5 x 10^12 m, calculate the period of Neptune.
Can someone show me how to solve this? I have no idea what formula to use.
- Dr SpockLv 61 decade agoFavorite Answer
Knowing SOMETHING that appears in an equation is USELESS unless you know what the equation is, and how to use it.
If you are being asked this question, you were presumably introduced to the "Kepler constant, K" by the VERY equation in which it is needed, as a convenient constant normalizing an empirically observed relationship (later put on a sound theoretical basis by Newton, with a subtle modification):
P^2 = K a^3,
where P is the period of a planet, and a is NOT the "average [orbital] radius," but rather the SEMI-MAJOR AXIS of the planet's orbit.
[For mathematical reasons, the average orbital radius is NOT the semi-major axis. Kepler's relationship actually holds for the semi-major axes of planetary orbits. Note also that you should NOT talk of a "planet's radius" in this context, since that just means something to do with the planet's OWN size, that is half of the planet's physical diameter. Instead, say "planet's orbital radius."]
Now that you have the proper equation, just substitute your values of K and a into it!
Live long and prosper.
****** IMPORTANT CAUTIONARY REMARK! ******
I have just realized from the size and the physical dimensions of your quoted "K" that SOMEONE is setting you up for serious problems should you ever take an astronomy course in a university:
YOUR K is inverted from the classical representation of Kepler's relationship!
I have just evaluated the classical K, using Earth's own orbital information. I find that
Kepler's K = 2.9746... x 10^(-19) s^2 / m^3,
or in other words, YOUR K (let's call it K*) would be given by
K* = 3.362 ... x 10^18 m^3 / s^2.,
which is, as you will notice, essentially the value you quoted.
Unfortunately your K* would be the normalizing constant that would have to appear in the following, inverted and historically unconventional form of Kepler's 3rd Law,
a^3 = K* P^2.
(While you COULD write the relationship that way, it's unconventional, and simply at odds with the more physical way to think about the functional relationship it implies. For example, Kepler's 3rd Law is usually expressed in words as "The SQUARE of the Period is proportional to the CUBE of the semi-major axis." That's like saying that the PERIOD is a function of the SIZE of the orbit --- a much more physical way of looking at it than declaring that the size of the orbit is a function of its period! In other words, when you're thinking about an orbit as a problem in physics, the most imortant thing to know about an orbit (after learning that it's an ellipse) is its size. As a problem in orbital physics, its period is worked out physically FROM its size, rather than the other way round. Of course, once one HAS the relationship P^2 = K a^3, the values of P or a can be related to one another in either direction, that is by (i) evaluating P if a is given, or (ii) evaluating a if P is given. Naturally, one doesn't have to WRITE the formal relationship in an unconventional way in order to do that!)
If you don't believe me about the standard way in which K is defined and used, simply look at any good university level astronomy text.
I really cannot understand how some educator can have cavalierly expressed Kepler's 3rd Law in a way that conflicts with its usual expression. Perhaps it comes from a reluctance to have a constant with a large negative exponent! (I can't imagine any other rationalization.) It has to have been the act of a pedagogical educator who simply doesn't give a damn about the way in which practicing astronomers usually write the relationship. As I said, it's setting you up for future misunderstandings if you try to carry something half-remembered from this over into your university studies.
Of course, the actual physical value of K is not needed in many problems. Kepler's relationship is used in effect as a RATIO in the solar system, with the Earth's semi-major axis (by definition, 1 A.U.) taken as a size scale, and the Earth's year as the period scale. With that normalization,
K = 1 numerically, its units being (year)^2 / (A.U.)^3. That is,
K = (1 year)^2 / (1 A.U.)^3.
Analogously, once one has learned how K depends upon the parent star's mass (through Newton's derivation), the relationship can be scaled to apply to other systems, such as the 270 or so extrasolar planetary systems discovered in the last decade or so. That enables one to appreciate the physical size of extrasolar planet's orbits in terms of the orbits we're familiar with in the solar system.Source(s): Teaching university level astronomy and astrophysics for almost four decades.
- RickBLv 71 decade ago
Sounds like you have no clue what "K" means. This is an expression of Kepler's 3rd Law, which says that the cube of the radius is proportional to the square of the period. In math terms, this law written as:
R^3 = (K)(T²)
where "R" is the radius and "T" is the period.
They give you "R" and "K" in the problem. Just solve for "T".