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Physics/Acceleration due to gravity.

Question
Dear Dr Jeffery

en.wikipedia.org/wiki/Acceleration_due_to_gravity

The value of Acceleration due to gravity is taken as 9.8 m/s square.

How this value is derived and computed and why this number is not any other number viz 6, 8, 12, 15 etc ?.

Thanks & Regards,
Prashant S Akerkar

Hello again Prashant!  I hope all is well.

In this instance, it is pure experimentation that gives us the number and (oddly) astrophysics that gives us the answer.

The method of measuring it is simply to put an object into a vacuum tube (with a depth of d) and then observe the rate of acceleration by seeing how long it takes to hit bottom.  The general equations of motion for uniform acceleration will give you this:  d = (a*t^2)/2.  This turns in to a = 2d/t^2. So, just from a practical perspective acceleration due to gravity is 9.81 m/s^2

Now the "why": By observing celestial bodies (http://web.stanford.edu/~buzzt/gravity.html) , it was observed that a mass-mass attractive force was observed. To this effect, the force due to gravity is described by Newton's universal law of gravitation: F = G * m1 * m2 / r^2.  This is where G is a universal constant (observed), m1 is the mass of the first object (in our case, the dropped object) and m2 is the mass of the second object (in this case, the earth) and r is the distance between the centers of the masses. Since the earth's mass doesn't change much, m2 is constant and since the distance doesn't change much (since the center of the earth is about the same distance everywhere on the surface), r is constant. Where does this leave us?

It means that the force do to gravity is m1 times some constant equal to G*m2/r^2... or F = m1 * (G*m2/r^2).

Now lets look at the equation for acceleration of a mass due to an applied force:  F = m * a
Force equals mass times acceleration.  Since the mass in question is our object (m1) we can plug that in to have F = m1 * a.    Combining this with the force due to gravity equation we come to:

F = m1 * (G*m2/r^2) = m1 * a

Removing m1 from both sides then gives

G*m2/r^2 = a

Note that our final equation does not care about m1 after all... hence all objects at the surface of the earth (when no other forces are present) accelerate at a constant rate that is equal to G*m2/r^2.

I hope this helps!

Take care,
Jeff

Physics

Volunteer

Dr. Jeffery Raymond

Expertise

Materials chemistry. Materials science. Spectroscopy. Polymer science. Physical Chemistry. General Physics. Technical writing. General Applied Mathematics. Nanomaterials. Optoelectronic Behavior. Science Policy.

Experience

Teaching: General Inorganic Chemistry I & II, Organic Chemistry I & II, Physical Chemistry I, Polymeric Materials, General Physics I, Calculus I & II
My prior experience includes the United States Army and three years as a development chemist in industry. Currently I am the Assistant Director of the Laboratory for Synthetic Biological Interactions. All told, 13 years of experience in research, development and science education.

Organizations
Texas A&M University, American Chemical Society, POLY-ACS, SPIE

Publications
Journal of the American Chemical Society, Nanoletters, Journal of Physical Chemistry C, Journal of Physical Chemistry Letters, Ultramicroscopy Proceedings of SPIE, Proceedings of MRS, Polymer News, Chemical and Engineering News, Nano Letters, Small, Chemistry.org, Angewandte

Education/Credentials
PhD Macromolecular Science and Engineering (Photophysics/Nanomaterials Concentration), MS Materials Science, BS Chemistry and Physics, Graduate Certificate in Science Policy, AAS Chemical Technology, AAS Engineering Technology