# Kepler's third law doesn't give earth's orbital period! Why? [closed]

I tried to calculate earth's orbital period using Kepler's third law, but I found 365.2075 days for the orbital period instead of 365.256363004 which is the correct value. I checked everything, and I couldn't find what's the problem. I used these values for my calculation:

• Semi-major axis, a: 149,598,261 km
• Gravitational constant, G: 6.67*10-11 N·(m/kg)2
• Solar mass, M: 1.9891*1030 kg
-

## closed as off-topic by David Z♦Apr 11 '14 at 1:26

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – David Z
If this question can be reworded to fit the rules in the help center, please edit the question.

accuracy of 1 part in $10^4$ when one of the constants was accurate to 1 part in $10^3$ is not bad –  ratchet freak Apr 10 '14 at 13:11
Where did you get 365.256363 days? According to britannica.com/EBchecked/topic/553076/solar-year, a tropical year (Vernal Equinox to Vernal Equinox) is 365d 5h 48m 46s, which works out to 365.242199 days. That agrees with the figure I've always heard, of 365.2422 days. –  Phil Perry Apr 10 '14 at 14:12
@PhilPerry from here link. but according to this link link, tropical year is different from orbital period. it's 20 minutes shorter. –  Mostafa Apr 10 '14 at 14:23
In any case, it's probably a good idea to specify exactly which year you're using. And as @ratchetfreak pointed out, you got a good result considering the precision of the data. –  Phil Perry Apr 10 '14 at 15:49
@PhilPerry Funny, I always heard 365.2425 + some further decimals. Wikipedia actually gives several different numbers and the reasons for each –  Izkata Apr 10 '14 at 21:03

You've used the gravitational constant with only three significant digits. So it's no surprise that your answer isn't accurate to five significant digits. Instead of $G$ and $M_\odot$ separately, you should use the product $GM_\odot$, known as the standard gravitational parameter. Its value is known very accurately: in the link, you'll find $$GM_\odot = 132\,712\,440\,018\;\text{km}^3\text{s}^{-2}$$ We could even include the value for the Earth: $$GM_\oplus = 398\,600\;\text{km}^3\text{s}^{-2}$$ so we get $$T = 2\pi\sqrt{\frac{a^3}{G(M_\odot+M_\oplus)}} = 2\pi\sqrt{\frac{(149\,598\,261)^3}{132\,712\,838\,618}} = 31\,558\,272\;\text{s}=365.2578\;\text{d},$$ which is very close to the actual value. As remarked in the other answers, the remaining small difference is mainly due to planetary perturbations.

-
Why should we include $M_\oplus$ in the above calculation? –  Prahar Apr 10 '14 at 14:14
@Prahar Because the Earth-Sun system is a two-body problem: the Earth also pulls on the Sun, so they orbit around each other, and you need both masses. Naturally, the effect of the Earth is very small. –  Pulsar Apr 10 '14 at 14:25
Ah! Yes. Thanks. –  Prahar Apr 10 '14 at 15:30
To be precise: ignoring it gives you an error of about 1E-7, which is about the magnitude of errors that we're worrying about here. As you see, we still have a difference of 365.2564 - 365.2578. –  MSalters Apr 10 '14 at 15:45

To expand on Prahar's answer, let me run some numbers to try and convince you this is reasonable. Your answer is correct to within one part in 104: $$\frac{365.256363004}{365.2075}\approx 1.000133795.$$ The main perturbing influence on Earth's orbit is the gravitational pull of Jupiter, whose mass is about 1000 times smaller than the Sun, and which orbits at about 5 A.U. from the Sun. The latter means that its gravitational influence on the Earth is reduced by something between $$\frac{1}{4^2}=\frac{1}{16}\text{ and }\frac{1}{6^2}=\frac{1}{36}$$ compared to an equal-mass body at 1 A.U. from Earth. This puts the perturbation at something like the 10-4 ballpark, so you should not expect substantially better accuracy from a model that ignores it.

-
I'm not convinced. While the effect is numerically of the right order, the gravitational force exerted by Jupiter on Earth is a vector. During the course of a year, this angle between this vector and the Suns gravitational pull makes almost a full 360 degree sweep (almost since Jupiter itself moves ~30 degrees). So, in first-order approximation the average pull is zero. Of course, second-order effects aren't - but those aren't in the 1E-4 ballpark. –  MSalters Apr 10 '14 at 15:53
An error of one part in $10^4$ is quite reasonable when one of the inputs has only three digits of accuracy. That is much larger than the planetary perturbations. –  Ross Millikan Apr 10 '14 at 16:23
@MSalters unless you are in a resonance regime where the contributions add up over many cycles. In general you can't tell whether that's the case and you need to explicitly check that the per-cycle contributions do cancel out before you rule out contributions at first order. –  Emilio Pisanty Apr 10 '14 at 16:58
@EmilioPisanty: Even with resonance the majority of contributions cancel out - that too is a second order effect. Another way to view it is that Jupiter trails earth for 6.5 months and then leads for 6.5 months. it the trailing phase, the gravitational pull of Jupiter is opposite our orbital velocity, in the leading phase the angle between them is <90 degrees. In first order, these cancel - any remaining term by definition is second order. –  MSalters Apr 10 '14 at 17:19
I'm going to agree with Pulsar's answer: the proper two-body approach gives an error of .0014 day/year. Your first-order 1E-4 influence would increase the error to about .0365 day/year - clearly that's nonsense. But a second order effect, a magnitude smaller, is still possible. –  MSalters Apr 10 '14 at 17:40

Kepler's 3rd law assumes that the Earth travels in a perfect ellipse with the only gravitational force on it being from the Sun. Further, Kepler's laws are derived from Newtonian gravitation. In reality, the orbit of the Earth is affected by the gravitational pull of other planets, and by the effects of General Relativity and is therefore not quite elliptical.

The value you have obtained is as close as you can get to the correct value under the aforementioned assumptions. If you incorporate the other effects, then you get closer to the real value.

-
These points may be true, but I don't think they are relevant considering the precision of the data. –  Octopus Apr 10 '14 at 18:43
GR might not be relevant, but the effect of the other planets definitely are. –  Prahar Apr 10 '14 at 19:14