Earth's gravitational pull vs. the Sun's

Updated 26 Apr 26

To compare the Earth's gravitational pull with the Sun's, let's look at the Moon's orbit about the Earth, but to keep it simple, we'll ignore its mass. (Assuming a similar density to the Earth's, its mass would be about 1/50th of it, and this approximation is fine for our purposes).

The Moon's orbital period is 27.3 days or 27.5/365.25 years.

Square that and it's about 5.6 x 10⁻³.

From memory, it's about 240 thousand miles away. Again, from memory, the Sun is about 93 million miles away, so the radius of the Moon's orbit about the Earth is about 2.6 x 10⁻³ AUs. 

(Actually, I've done this back-to-front and imagine that the size of the AU was first determined in relation to the Earth-Moon distance, something that had been measured fairly accurately by the Ancient Greeks).

Cubing that is about 17.1 x 10⁻⁹.

The ratio of this number to the square of the orbital period is about 3 x 10⁻⁶. This indicates the strength of the gravitational field of the Earth (from Newtonian Gravitational theory). Let's call it Γₑ.

The equivalent figure for the Sun (let's call it Γₛ)  is 1 (see my earlier post on Kepler's third law). So, the gravitational pull of the Sun is Γₛ/Γₑ, i.e. about 330,000, times that of the Earth. We know from Newton that this is also the ratio of the Sun's mass to that of the Earth.