How can you get the density of a planet given only the period. (probably universal constant G is in the equation…F=Gm1m2/d^2, a=4pi^2d/T^2)
Anyways a sample question is: find the density of planet x given that the period of orbit of an object skimming the surface is about 2 hours.
Hints: Gm1/d^2 is also a. That is, a = F/m2
You still have to know the formula for the volume of a sphere…
Unless there is some trick to this, you can’t.
A 1 million kilogram asteroid (iron) or a 1 million kilogram comet (water), both with vastly differing densities, will possess the same orbital period if their orbital radius is the same. It is equivalent to say that the gravitational attraction between the sun and either body is the same, which results in the same orbital velocity being required to maintain orbital equilibrium.
Unless we’re just talking about what the equations reduce or convert to…
I guess the trick is that m1 = density times the volume of the sphere, of radius d–notice that the object is skimming the surface.
Short answer: You can’t. You need more information.
The trick is you’d have to figure out the mass of the planet, but you can’t from the orbital period. The mass of the planet gets cancelled out in the math because the gravitational force is proportional to mass and so is the acceleration (what you can determine from period.) This is why Kepler’s third law of planetary motion makes no reference to the mass of the planet:
From Kepler’s Three Laws of Planetary Motion
It goes like this:
F = [sup]Gm[sub]s[/sub]m[sub]p[/sub][/sup]/[sub]d[sup]2[/sup][/sub] = m[sub]p[/sub]a
Let m[sub]p[/sub] be the planetary mass, and m[sub]s[/sub] be the stellar mass.
So, the acceleration of the planet is
a = [sup]Gm[sub]s[/sub][/sup]/[sub]d[sup]2[/sup][/sub]
which is independent of the planetary mass. Since acceleration in an orbit is directly tied to orbital period, the orbital period is independent of planetary mass.
If you can measure how much the star is moving, you can determine the mass of the planet - but you can’t do it from the period alone. A more massive planet will move the star more and more - this is how we are able to detect planets, by the motions of their stars. If you use this trick to find the planetary mass AND find the radius of the planet, you can find the density. However, the period of orbit is useless in this endeavour.
The title of this thread is misleading. If you read the OP carefully, you will see that the problem gives the period of a satellite orbiting the planet at the planet’s surface, and asks us to find the density of the planet. If we can assume the planet is a sphere, this problem is soluble. In solving it, I used the equations
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mv[sup]2[/sup]/r = GMm/r[sup]2[/sup] [centripetal force equals gravitational force]
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v=2pir/T [speed is circumphrence over period]
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V =[sup]4[/sup]/[sub]3[/sub]pir[sup]2[/sup] [volume of a sphere]
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D=M/V [density is mass over volume]
Of course #3 should have r[sup]3[/sup], not r[sup]2[/sup]
Yep. The sample question clears it up.
This problem is soluble?
Yeah, soluble means “possible to solve or explain” (AHD) as well as “easily dissolved”. Or is that what you meant?
A safe assumption since anything in a solar system that is too small to have enough gravity to remain spherical would not be called a planet. (gee, that was an awkward sentence)
The devil is in the details though. The earth is not spherical, due to its rotation (and a few odd mountain
ranges). To first or second approximation, it is spherical. Almost as much as a billiard ball.