# What is the acceleration due to Earth's rotational and orbital movements?

What are the accelerations due to earths spin and orbital motion around the sun? It must be negligible otherwise humans would have been feeling it, are there any instruments that can measure it? would a human feel the acceleration on mercury or that too still is too negligible? Are there any celestial places were rotational or orbital acceleration be significant enough to be felt?

PS : This was just a pondering about measurement, not an exercise or home work, was wondering if there are any instruments that are sensitive enough to measure it, or this is just something that can be computed but not measured.

• "otherwise humans would have been feeling it" The centripetal acceleration causing earth's orbital motion around is the sun's gravity. Neglecting tidal effects (which are very small in the scale of a human body), the acceleration is the same for each part of human's body, so a human cannot feel it.
– JiK
Commented Jun 10, 2015 at 12:37

The fictitious acceleration due to Earth's spin can be found using centrifugal acceleration,

$$a_{ca} = \omega^2 r,$$

where $\omega$ is Earth's angular velocity (roughly $7.292\cdot10^{-5}\ rad/s$) and $r$ the shortest distance between you and the axis of rotation of Earth. The direction of this acceleration will be aligned with the shortest distance between you and the axis of rotation, pointing outwards.

This acceleration is also the main cause why Earth and other planets are not a perfect spheres, but more like oblate spheroids. Therefore you could also attribute the lower gravitational force at the equator due to its bigger radius to centrifugal acceleration.

Under the assumption that Earth's gravity can still be approximated with Newton's law of universal gravitation (spherical symmetry of mass distribution), then the effect of Earth's spin would cause a difference in experienced acceleration between the poles and equator:

$$a_e = \frac{Gm_{\bigoplus}}{R_e^2} - \omega^2 R_e = 9.764\ m/s^2$$

$$a_p = \frac{Gm_{\bigoplus}}{R_p^2} = 9.864 m/s^2$$

which differs only by 1.0%, of which about 2/3 is due to the bigger radius and 1/3 due to the centrifugal acceleration.

The acceleration due to Earth's orbital motion around the Sun are defined by tidal forces, namely (the center of) the Earth is in free fall around the Sun, but the days side is slightly closer to the Sun and the night side is slightly further away from the Sun than Earth center. This effect will be most prominent around the equator (due to the axial tilt this will differ during a year). The most extreme difference would occur when the Earth is at is periapsis:

$$a_{day} = \frac{Gm_\bigodot}{(R_{pe}-R_e)^2} - \frac{Gm_\bigodot}{R_{pe}^2} = 5.3191\cdot10^{-7}\ m/s^2$$

$$a_{night} = \frac{Gm_\bigodot}{R_{pe}^2} - \frac{Gm_\bigodot}{(R_{pe}+R_e)^2} = 5.3184\cdot10^{-7}\ m/s^2$$

So you will feel ever so slightly lighter during the middle of the day and night than at sunrise and sunset. For comparison the acceleration difference from the tidal forces of our Moon is around $1.1\cdot10^{-6}\ m/s^2$. If these tides would coincide than that would only decrease the acceleration near the equator by about 0.0000167%.

The effect of the spin of a planet will be bigger the higher the angular velocity. For example Saturn is the planet in our solar system whose equator bulges. When using the same assumptions as for Earth then the difference between the acceleration at its "surface" at the poles and equator would be around 31%, about 3/5 due to the bigger radius and 2/5 due to the centrifugal acceleration, which should be noticeable.

The planet which experiences the biggest tidal forces, relative to its surface gravity would be Mercury, but would still only change the surface gravity by 0.00018%. If you want to find celestial bodies which experience bigger tidal forces you would have to look at moons, which orbit close to their planet. The nearest such moon would be the inner moon of Mars, named Phobos, which experiences a tidal force relative to its surface gravity of 24%. Or an even more extreme example the moon of Saturn, named Pan, which experiences a tidal force relative to its surface gravity of roughly 50-40%.

There is a (small) centripetal force imparted by Earth's rotation, but it's far too small to be noticeable by the human body. It's around 0.3% of gravity. So a highly precise scale can measure the difference at different latitudes, but you won't notice.

The answer detailing tidal forces is really the only force measurable in regards to Earth orbiting the sun, since the centripetal acceleration is basically counterbalanced by the pull of the sun's gravity (hence why we stay the same average distance from the sun over eons). Same deal with the solar system's rotation around the galaxy. The galaxy's movement towards the Great Attractor is linear (as far as we can tell) so there's no acceleration to note in that direction.

Left as an exercise to the reader is computing the centripetal force generated by the Earth's orbit and comparing it to the gravitational attraction between the Earth and the sun.