The Sun's gravitational pull at the Earth's surface

We know from an earlier post that the acceleration due to a massive object is:

i. 4π²k/d²

where k is the Kepler constant for the massive object (yes, I'm avoiding G and gravitational mass) and d is the distance from that object.

Hence at the Earth's surface the ratio of the gravitational acceleration due to the Sun to that due to the Earth is:

ii. (kₛ.dₑ²)/(kₑ.dₛ²).

Where kₛ and kₑ are the Kepler constants for the Sun and Earth respectively and dₑ is the distance from centre of the Earth, i.e. one Earth radius, and dₛ is the distance from the Sun, i.e. 1 AU.

We know also that:

iii. kₛ/kₑ ~ 10⁶/3 &

iv. dₑ/dₛ ~ 1/24000

Hence, at the surface of the Earth, the ratio of the gravitational acceleration due to the Sun to that due to the Earth is about

10⁶/(3 x 24000²) ~ 0.0006.

This tiny, tiny acceleration, less than one-thousandth that we experience due to the Earth, is enough to keep the Earth in its orbit about the Sun, year in, year out!

At Saturn, ~ 9.6 AU from the Sun, the acceleration due to the Sun is about 100 times smaller than at the Earth and yet, it's sufficient to keep Saturn orbiting too!

Gravity is such a puny force and yet it permeates the cosmos!