2.8 Game 2302-0380 motion -- variable velocity and acceleration  (Page 3/17)

Let's examine the validity of these two assumptions.

Newton's law of universal gravitation

This law states that every point mass in the universe attracts every other point mass with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. (Separately ithas been shown that large spherically symmetrical masses attract and are attracted as if all their mass were concentrated at their centers.)

Effect of gravity versus altitude

An object at the surface of the earth is approximately 3963 miles from the center of the earth. An object that is six miles above the surface of the earth(typical flying altitude for a passenger plane) is approximately 3969 miles from the center of the earth.

Therefore, (if I didn't make an arithmetic error) the gravitational attraction between the earth and that object changes by less than 0.5 percentwhen the object is transported from the surface of the earth to a position that is six miles above the surface of the earth.

For practical purposes, therefore, we can assume that the acceleration of gravity is constant up to at least 32000 feet above the surface of the earth.

Acceleration is independent of mass

Newton's law, as applied to the gravitational attraction of the earth, expressed in algebraic terms, looks something like the following :

f =E*m/d^2

where

• f is the attractive force between the earth and another object
• E is the mass of the earth
• m is the mass of the other object
• d is the distance between the center of mass of the earth and the center of mass of the other object

Force equals mass times acceleration

It is a well-known principle of physics that an object that is free to move will move and will accelerate when subjected to a force. The acceleration of the object will beproportional to the force and inversely proportional to its mass. In other words,

a = f/m

where

• a is the acceleration in units such as meters/sec^2
• f is force in units such as kilograms*meters/sec^2 (newtons)
• m is mass in units such as kilograms

By multiplying both sides of the equation by m, we get a more common presentation of this relationship, which is

f = m*a

where the symbols mean the same as listed above .

A ratio of two different forces of gravity

Now let's use the equation from above to form a ratio of the forces exerted on two different masses by the earth, assuming thatboth masses are the same distance from the center of the earth.

f1/f2 =(E*m1/d^2)/(E*m2/d^2)

Cancel like terms to simplify

If we cancel like terms from the numerator and denominator of the expression on the right, we can simplify the ratio to

f1/f2 = m1/m2

Replacing the forces on the left by the expression from above , we get

m1*a1/m2*a2 = m1/m2

Multiplying both sides by m2/m1 we get

a1/a2 = 1

or

a1 = a2

This shows that the acceleration resulting from the gravitational force exerted on two objects that are equally distant from the center of mass of theearth is the same regardless of the differences in mass of the two objects.