Troubleshooting Kepler's Third Law with Halley's Comet

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ehrenfest
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Homework Statement


When I plug in all of the parameters for Halley's comet (from Wikipedia) into Kepler's third law a get a semimajor axis of 38.56 AU when it should be about 17? Can someone else try it and see if I am crazy?

Homework Equations


The Attempt at a Solution

 
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mass of halley's comet = negligable
mass of the sun
G
T = 76 years
 
OK here are the details:

38.5654 = (T^2/(4 pi^2) * G * (Ms))^(1/3)/(1.4*10^11)

where T is the period in seconds, Ms = 1.991*10^31 and G = 6.674 * 10^(-11)
what am I doing wrong?
 
ehrenfest said:
OK here are the details:

38.5654 = (T^2/(4 pi^2) * G * (Ms))^(1/3)/(1.4*10^11)

where T is the period in seconds, Ms = 1.991*10^31 and G = 6.674 * 10^(-11)
what am I doing wrong?

The mass of the Sun is 1.99*10^30 kg... (Your result for a is off by very nearly the cube root of 10.)
 
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Dick said:
You beat me! I just figured that out. But, ehrenfest, for solar orbits if you work in AU and years, the constant proportionality k in R^3=k*T^2, is one.

My training's largely in astrophysics, so I have the solar mass by heart. I would usually take the proportionality approach myself as well, though...
 
Ahh! 30 minutes of frustration because my short-term memory is not good enough to look at a computer screen and then write down a two-digit number without botching a digit!

Thanks guys.
 
Dick said:
Funny, my training is in cosmology, so I know it's like to ten the fifty some proton masses. And fifty plus what I forget. Good job.

Close enough... ;-) When I was an undergraduate, cosmology was called "the science where you're happy when your order of magnitude is right to an order of magnitude". Nowadays we speak of "precision" cosmology -- what an age we live in...
 
ehrenfest said:
30 minutes of frustration...

Everybody makes copying errors (when they're not making *sign* errors), so I know just how you feel...