Kepler's third law doesn't give earth's orbital period! Why? 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

 A: 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. 
A: 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.
A: 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. 
