Calculate the Period of a Planet.

  • Thread starter Thread starter destinee
  • Start date Start date
  • Tags Tags
    Period Planet
Click For Summary
SUMMARY

The discussion focuses on calculating the orbital period of a planet using Kepler's 3rd Law, specifically the formula T²/R³. The initial calculation attempted by a user was incorrect due to neglecting to square the values appropriately. The correct approach involves using the proportional relationship T1²/R1³ = T2²/R2³, where T represents the period and R the radius in astronomical units (AU). The discussion also emphasizes that while Kepler's law generally holds, it assumes the planet's mass is negligible compared to the star's mass.

PREREQUISITES
  • Understanding of Kepler's 3rd Law of planetary motion
  • Basic knowledge of orbital mechanics
  • Familiarity with astronomical units (AU)
  • Ability to perform algebraic manipulations involving squares and cubes
NEXT STEPS
  • Study the application of Kepler's 3rd Law in various planetary systems
  • Learn about the implications of mass ratios in orbital mechanics
  • Explore the mathematical derivation of Kepler's laws
  • Investigate the effects of density and mass on planetary orbits
USEFUL FOR

Astronomy students, astrophysicists, educators teaching orbital mechanics, and anyone interested in the dynamics of planetary motion.

destinee
Messages
1
Reaction score
0
Consider a planet with a period of orbit to be 0.241 years and an average radius of 0.39 astronomical units. Calculate the period of a planet with an average radius of 1.4 astronomical units.



If someone can answer and explain this question to me, I would be very thankful. I know it has something to do with Kepler's 3rd law (T²/R³), but I do not know how to apply it.The teacher counted my answer wrong. See my work below:



P1 = 0.241 years with a radius of 0.39 AU
P2 = ? years with a radius of 1.4 AU
So set 0.241 over 0.39 equal to X (? years) over 1.4

You get
0.3374=0.39x than divide
x = 0.865 year orbit
 
Physics news on Phys.org
0.865 years ISN'T correct, You have forgotten to square each part, use this:

Using Kepplers 3rd Law:

P12/P22 = R12/R22

However, this is a poor question, it assumes these planets will have the same density/mass which is pretty much never the case. may want to mention this to your teacher!
 
Kepler's 3rd law is a bit more than "T2/R3". For a planetary system with a big sun and relatively tiny planets, it is more like "T2/R3=k", where k is approximately the same for all tiny planets.

You used a simple proportion--that's what got you in trouble. Find k and try again using the big sun equation.
 
"The square of the orbital period of a planet is directly proportional to the cube of the semi-major axis of its orbit."

So for one planet, T_1^2 \propto a_1^3 For another, T_2^2 \propto a_2^3 Thus \frac {T_1^2} {a_1^3} = \frac {T_2^2} {a_2^3} Can you see how you could use this?

However, this is a poor question, it assumes these planets will have the same density/mass which is pretty much never the case.
No, the question is OK. The law works as long as the planet's mass is much less than the mass of the star, which is pretty much always the case.
 
voko said:
No, the question is OK. The law works as long as the planet's mass is much less than the mass of the star, which is pretty much always the case.

Really? Surely if one planet is extremely light and the other is heavy by comparison but still less than the sun then the force on the heavier planet and therefore it's orbital velocity would be much larger?

I'm not disagreeing, it's just this was the impression I was always under: that if there is a large difference in the relative mass of the two planets then the law can become skewed? Or is this just one of those assumptions we make because it pretty much never occurs? :S
 
Bowenwww said:
Really? Surely if one planet is extremely light and the other is heavy by comparison but still less than the sun then the force on the heavier planet and therefore it's orbital velocity would be much larger?

The coefficient of proportionality has, as a factor, \frac 1 {M + m}, where M is the mass of the star, and m of the planet. Typically M >> m, so that factor is almost exactly \frac 1 M no matter what planet you choose. Take our Sun, Mercury and Jupiter, and see how this factor differs.

I'm not disagreeing, it's just this was the impression I was always under: that if there is a large difference in the relative mass of the two planets then the law can become skewed? Or is this just one of those assumptions we make because it pretty much never occurs? :S

As you can see from the above, of importance is the ratio planet to star, not planet to planet.
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
3K
Replies
5
Views
2K
  • · Replies 13 ·
Replies
13
Views
8K
  • · Replies 7 ·
Replies
7
Views
4K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 12 ·
Replies
12
Views
5K
  • · Replies 12 ·
Replies
12
Views
4K
Replies
2
Views
1K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 11 ·
Replies
11
Views
3K