# Tag Info

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I generally regard NASA as authoritative, and they report the orbital parameters on their Earth Fact Sheet. I note that they disagree with Wikipedia about the aphelion though they agree on the perhelion, semi-major axis and eccentricity: NASA Wikipedia Aphelion 152.10 151.93 Perhelion 147.09 147.095 Semi-major 149.60 ...

19

Are there any exact data about Earth's orbit? No. There are always measurement errors. There are however very good estimates. The best estimates come from three competing organizations, the Jet Propulsion Laboratory (the Development Ephemeris models), the Russian Institute for Applied Astronomy (the Ephemerides of the Planets and Moon), and the IMCCE ...

8

The explanation is Birkhoff's theorem, which states that the Schwarzschild solution is the unique spherically symmetric vacuum solution in general relativity. An immediate result of this is that, just as in Newtonian gravity, a spherical shell does not contribute to the gravity experienced by an object within it. If this were otherwise it would suggest the ...

8

Your diagram is not quite to scale, and the errors are important. Notice that only the hemisphere of the moon which points toward the sun is illuminated, rather than what your drawing shows. This has the following implications. 1) When the moon is new, it rises and sets at the same time as the sun, and is not (mostly) visible at night. The extreme example ...

7

Here's a slightly more accurate diagram: It's still not quite to scale — the Moon is actually a lot further away from the Earth than shown here — but it should suffice to demonstrate that the moon can indeed be seen from the night side of the Earth even when it's nearly between the Earth and the Sun. Note how, in the orientation shown in the ...

6

The geometry of spacetime is described by a function called the metric tensor. If you're starting to learn GR then any moment you'll encounter the Schwarzschild metric that describes the geometry outside a sphrically symmetric body. When you go inside the body the geometry is described by the (less well known) Schwarzschild interior metric. The exact form ...

4

To two significant figures, the acceleration due to gravity is $g=9.8\:\mathrm{m/s^2}$ everywhere on Earth (at sea level). That is to say, if you use e.g. a pendulum to measure $g$ to two sig figs, you will get this value no matter where you are. In a sense, this is the precision to which the Earth is well-approximated by a uniform sphere of matter. The ...

2

The elements you give describe an idealised orbit that does not exist in reality. Those numbers are parameters to an approximate model. Earth's closest distance to the sun is different each and every year, by a lot (about 20,000 km in fact). Are there any exact data about Earth's orbit? There are certainly far better models than the 6-parameter ...

2

You are right that it should not be considered an inertial frame for many types of problems. This is how you end up with fictitious forces to account for (such as the Coriolis effect). However, this only has practical effect for larger scale problems. For the types of problems generally considered in physics class, the inertial frame approximation will ...

1

Strictly speaking, $g$ (even more strictly speaking, $g_0$ or $g_n$) is a constant. It is exactly 9.80665 m/s$^2$, by definition. There are many places in science and engineering where it is very useful to have an exact (albeit arbitrary) defined constant for gravitation on the surface of the Earth. That said, gravitation on the surface of the Earth does ...

1

This Q boils down to How bright is Earthshine? According to the usual source Oceans reflect the least amount of light, roughly 10%. Land reflects anywhere from 10–25% of the Sun's light, and clouds reflect around 50%. So the amount of sunlight reflected (i.e. albedo) depends on what part of the earth is facing your observer and how cloudy it is. ...

1

This depends on the reflectivity of the objects and the size in the sky. The visual albedo of the moon when near (but not at) full, is about $0.12$. Earth's is around $0.39$. Being in low-earth orbit, the actual amount will vary based on the terrain and atmosphere. If it's overcast below, the value could be much higher. You can assume that a given area ...

1

In this answer we assume a spherically symmetric spacetime and no cosmological constant $\Lambda=0$. I) Birkhoff's theorem (BT) only works for vacuum branches of a spherically symmetric spacetime, i.e. in regions without any matter, cf. e.g. this Phys.SE post. Therefore BT would apply to a hollow planet, cf. e.g. this Phys.SE post. The Newtonian shell ...

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