about the gravity?

Question

what does acceleration of the gravity means & how do we calculate it???and caculate the acceleration of the earth gravity

Answer 1

It's acceleration "due" to gravity, not "of" gravity. Gravity does not accelerate. Gravity is a force, like when you pull (a mechanical force) a door open.

I presume you've heard of F = ma? Your pull on the door exerts a force (F) on the mass (m) of that door. In which case, the door starts to move (accelerate...a).

OK, then, gravity exerts a force on things; but that force is downward...towards the center of Earth (and other planets too). One interesting note...when we talk about the force due to gravity, we often call it "weight." So, for consistency, we rewrite F = ma as W = mg; where W is weight in Newtons or pounds and g is that acceleration due to the force (weight) of gravity.

In general, acceleration is just the change in velocity per unit time. So to measure any accleration, we simply measure how long an accelerating something took to cover a known distance. For example, you have a souped up Mustang (1959 model) with turbo charger. You want to know how fast that hummer can accelerate from a standing stop over some distance. What do you do?

There is an equation you should learn: s = ut + 1/2 at^2; where s = distance traveled in time (t), at acceleration (a), starting with initial velocity (u). Suppose you want to find out how fast your souped up Mustang accelerated over a distance of s = 88 feet with initial velocity = u = 0. Then s = 1/2 at^2; so that 2s/t^2 = a.

You know the distance, all you have to do now is time (t) how long it took to cover that distance. A good stop watch should do the trick. If it took 2 sec, for example, your acceleration over the 88 feet would be 2*88/2^2 = 44 feet per second per second (ft/sec^2).

You can do the same thing with g, the acceleration due to gravity. But instead of a Mustang on the straight and level, you drop a golf ball from a known height (h). Using the same logic as the Mustang problem, you time the fall (t) and plug that into the same equation. Thus, h = ut + 1/2 gt^2 and g = 2h/t^2.

What is important to note is that for a given radius from the center of the Earth, g = constant. For example, on the Earth's surface g = 9.81 m/sec^2 or 32.2 ft/sec^2 no matter what size object you drop (e.g., a bowling ball will drop with the same acceleration as your golf ball) as long as air drag forces are insignificant (like in a vacuum). Thus, on Earth's surface, we can always calculate the mass of something by W = mg; so that m = W/g = W/32.2 and W is the weight as measured in pounds. (mass is in "slugs" when using pounds)

Answer 2

Gravity is the attraction of two objects that have mass. Because the earth it wildly more massive than anything else we deal with, we only notice the force of gravity from the earth.

The force is related to the mass of the objects by the square of the distance between them. The means that the force increases as the objects get closer and therefore they move closer even faster and that is acceleration.

The acceleration due to gravity at the earths surface is estimated at about 10 meters per second per second. You can measure this by dropping an object from various heights and measuring the difference in the average velocity of the fall.

Answer 3

Things held up and let go, fall. They don't fall at a constant speed, but rather speed up. That speed up is called the "acceleration due to gravity."

That falling things accelerated was appreciated by an English Abbott who did experiments with his monks, dropping things like cannon balls from towers of various heghts. Things couldn't be taken much further, because everything happened so fast.

However, Galileo later reasoned that at a slope of angle theta, the effective acceleration down the slope would only be (g sin theta) instead of the full downward acceleration g. By effectively "slowing down gravity" and doing lots of experiments with rolling objects, he was able to test and confirm his ideas on accelerated motion, and the various relationshipe he produced. Even more importantly, he understood the motion of pendulums of various lengths, and the dependence of their periods on their length and on the value of g, leading to the best way at the time of measuring 'g.'

[That relationship, still used with many sophisticated additional experimental details and refinements to measure ' g ' is:

P = 2 pi sqrt (L/g),

where ' L ' is the length of the pendulum.]

Later, Isaac Newton applied his new understanding of mechanics to three empirical "laws" of planetary motion found by Johannes Kepler. With their help, and a further test involving the Moon's motion, Newton realized that there was a gravitational force of attraction between any two masses, proportional to the product of their masses, and inversely proportional to the square of the distance between them.

Newton worked entirely in proportionalities, but later the "Gravitational Constant," ' G,' was introduced, so that we now write the following expression for the attraction between two bodies:

F = G m1 m2 / r^2.

The value of ' G ' was first estimated by a British Astronomer Royal, cunningly "weighing" the Earth by astronomical measurements of the different angle of the "vertical" at two cleverly chosen places. He compared those two "verticals" on either side of a Scottish mountain whose mass he could estimate from its shape, size, and density (using trial samples obtained from boring into it or going deep into caves, etc.) Cavendish later invented analogous, ingenious laboratory methods that refined the value of ' G ' significantly. (These latter methods did not depend on measuring small angles in the deviation of the apparent "vertical" over small distances. Rather, he very carefully measured the controlled small angle "twisting" of suspended arms separating two known masses as another known attracting mass was moved from one side of the apparatus to the other side. Check 'Cavendish' on the web.)

If YOU or any other object is considered to be the mass m2, the acceleration towards the Earth will be:

g = G ME / RE^2,

where ME and RE are the mass and radius of the Earth itself.

Thus, from the value of ' g ' determined by careful pendulum experiments, ' G ' from refined versions of the Cavendish experiment, and RE from what are effectively surveying methods, we can learn the mass of the Earth very accurately.

And that effectively brings things full circle. Note that logically, we don't CALCULATE ' g,' we observe what it is. We have to USE that value of ' g ' to learn what ME IS. Thus, when we write:

g = G ME / RE^2,

we are NOT "calculating the acceleration of gravity due to the Earth," we are simply writing an equation that tells us the value of ' ME '. So in a sense, the answer to your question is that we don't "calculate" the acceleration of gravity here on Earth, we MEASURE it by interpreting sophisticated pendulum experiments.

Live long and prosper.

Answer 4

first,it is accleration due to gravity and not accl of gravity.
it is the accl which a body attains when it is thrown towards the earth. it has a fixed value of 9.8

accl due to gravity = G*mass of earth/square of earth's radius

G is univresal gravitaton constant.

Answer 5

Were you awake in class? The first half of question 1 is self explanatory. Use a dictionary. The other 2 use formulae created by Sir Isaac Newton found in your text book.