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How to Calculate the Force of Gravity on the Earth’s Surface

Discussion started by Adam Rangihana 8 years ago

Refer: http://www.dummies.com/how-to/content/how-to-calculate-the-force-of-gravity-on-the-earth.html

How to Calculate the Force of Gravity on the Earth’s Surface

Starting with the physics equation for the force of gravity, you can plug in the mass and radius of the Earth to calculate the force of gravity near the surface of the Earth.

The equation for the force of gravity is

and it holds true no matter how far apart two masses are.

The gravitational force between a mass and the Earth is the object’s weight. Mass is considered a measure of an object’s inertia, and its weight is the force exerted on the object in a gravitational field. On the surface of the Earth, the two forces are related by the acceleration due to gravity: F_g = mg. Kilograms and slugs are units of mass; newtons and pounds are units of weight.

You can use Newton’s law of gravitation to get the acceleration due to gravity,g, on the surface of the Earth just by knowing the gravitational constant G, the radius of the Earth, and the mass of the Earth. The force on an object of massm₁ near the surface of the Earth is

F = m₁g

This force is provided by gravity between the object and the Earth, according to Newton’s gravity formula, and so you can write

The radius of the Earth, r_e, is about 6.38 × 10⁶ meters, and the mass of the Earth is 5.98 × 10²⁴ kilograms. Putting in the numbers, you have

Dividing both sides by m₁ gives you the acceleration due to gravity:

Newton’s law of gravitation gives you the acceleration due to gravity near the surface of the Earth: 9.8 meters/second².

Of course, you can measure g by letting an apple drop and timing it, but what fun is that when you can calculate it in a roundabout way that requires you to first measure the mass of the Earth?

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Part of the problem is the word time and how it is misused. Time can mean, among other things, the reading on a clock at a particular moment or the difference in two such readings. The former should be more correctly called a clock reading while the latter should be more correctly called a duration.

Another part of the problem is that authors use the same symbol, usually just tt, to represent both of these quantities! While tt is fine for a clock reading, a better symbol for duration is ΔtΔt.

I fault the authors of physics textbooks, especially introductory textbooks, for perpetuating this obfuscation and its resulting confusion to students.

Anyway, there is hardly ever (I can't think of one as I write this) a case in physics when we need to substitute a clock reading into a kinematic or dynamic equation. In every case I can think of, what we really deal with is duration ΔtΔt, and there are two reasons for this. The first is that the numbers on a clock, although astronomical in origin, are otherwise completely arbitrary and thus have no inherent physical significance. The second is that the difference between two clock readings does indeed have great physical significance. It is the latter, duration, that appears in the definitions of velocity, acceleration, and any other physical quantity that involves "time" (note the quotes).

Now, to specifically address your question about the meaning of acceleration, consider that the correct articulation of the definition of average acceleration is the change in an object's velocity compared to the duration through which that change in velocity happens. That is far more physically accurate than the sloppy and overly simple distance over time. I always make sure my students learn the more correct definition; it prevents so many errors.

Next, look at the unit for acceleration. It must be the unit of velocity (m/sm/s) compared to the unit of duration (ss). This can correctly be written as the obtuse (and nearly devoid of physical meaning) combination m/s2m/s2 but this hides the meaning so clearly articulated by the better definition given above. So, write the unit of acceleration as m/ssm/ss and read it from the bottom up. An acceleration of, say, 3m/ss3m/ss would be read and interpreted as

"For every second of duration, the object's velocity changes by 3m/s3m/s."

Remember that acceleration is a vector quantity so there could, in general, be both a change in magnitude and a change in direction. Any such arbitrary change can be decomposed into a change that is parallel (or antiparallel) to the original acceleration and a change that is perpendicular to the original acceleration. My point is we should also quote a direction, but I didn't do that here.

Your point about the meaning of time squared is a great question and one that should be brought up by more students. Time squared means nothing. Duration squared, on the other hand, is physically meaningful and is the basis of our physical understanding of motion.

You bring up another issue, and that is the term distance. It too has been somewhat abused, but I'll save that for another time if anyone asks.

I really wish the community would hold textbook authors accountable for this serious abuse of terminology and notation.

8 years ago

Inertia: - a property of matter by which it continues in its existing state of rest or uniform motion in a straight line, unless that state is changed by an external force.
"the power required to overcome friction and the inertia of the moving parts"

8 years ago

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