Originally posted by twhitehead
I stand corrected. However the definition is not infact what is used in science. In science we measure the gravitational force exerted in releation to a given object. If you do not include a point of reference then objects on one side of the earth would have to have negative weight compared to objects on the other side as the force of gravity acting on th ght have to take into account any other known forces other than gravity that may be acting).
Your post is very confusing. The point of reference is arbitrary and does not change the magnitude of the gravitational force, only its direction. I'm not arguing about the direction, I'm arguing that the vector sum is unlikely to be zero.
Also, it is not necessary to measure the velocity of the Earth in relation to the rest of the universe in order to calculate the gravitational attraction between the Earth and the rest of the universe, nor do we need to account for any other forces. Newton gave us a handy-dandy formula for calculating gravitational attraction all by itself.
To that end, I've made some quick calculations using some simplifying assumptions. I got the masses of the Sun and all the planets in the solar system, and calculated an "average" distance from the Sun by taking the mean of the aphelion and the perihelion (all data provided by Wikipedia). Then I assumed that all the planets lay in a straight line - this turned out to be a sketchy assumption based on the current state of the solar system as seen here:
http://www.fourmilab.ch/solar/
But it ends up not mattering much, since the Sun, the Earth, and Jupiter all do lie on a straight line (more or less), and they turn out to be the only important bodies for this calculation.
If positive is towards the Sun, the vector sum of all the gravitational attractions between the Earth and the rest of the bodies in the solar system turns out to be 3.33E+22 N. If you only take into account the Sun, the Earth, and Jupiter, it turns out to be 3.33+22 N again. I also checked the effect of Alpha Centauri, which is about 4.5 light years away, but it only contributes 4.66E+11 N which doesn't change the answer (no matter which direction it pulls in).
If you divide the force by the mass of the Earth, you get 5.58E-3 m/s^2, which is very small, especially compared to the standard acceleration due to gravity at the Earth's surface of 9.81 m/s^2. Fine measurements (especially in orbit) would probably have to take this into account, but for everyday purposes we can neglect it without much error.