|Now we will take a
look at objects such as satellites orbit around Earth.
Imagine throwing a baseball, the ball will eventually reach the ground.
The faster its initial velocity, the greater its distance will reach. If it's
speed is ever so increasing, the shape of its path will become ever so straighter, less
elongated. Finally, at one particular launch speed, the ball would glide out just
above planet's surface all the way to the other side without striking the ground.
Like the moon, it would wheel around the globe in a circular orbit until something,
(building, air friction, collisions) brought it down. Without anything to stop it,
it'll sail off in an orbit, never coming back to its starting point.
In other words, the ball is traveling so fast, that it's
parabolic flight (recall from projectile
motion section) follows the planet's curvature. It is still under the
effects of gravity, and continually drops toward the Earth's center, but since the earth
ground is round, and falls downward as the path, the ball never gets any closer to the
ground. Thus, forever orbiting the earth.
+ h)2 = r2 + L2
r2 + 2rh + h2 = r2 + L2
2rh + h2 = L2
In the diagram h seems relatively large
compare to r. However, in reality, h is tiny and h2 is even smaller in
comparison to 2rh, making it negligible. Hence,
Since h is the distance the object would fall,
and we learned this formula in vector kinematics:
d = vit + 1/2 at2
h = 0 + 1/2 (9.8) (1)2
h = 4.9 m
The ball would fall 4.9 m per second.
And we have already assumed the time interval of this occurrence is 1 second. Thus
L would be the distance the ball must travel in 1 second. With:
L 7.9 km.
This value means that the ball would have to
be fired with the initial velocity of 7.9 km/s to orbit the smallest orbit around the
Another way of seeing this by equations we
already have established before is:
mac = Gm1m2/r2
mv2/r = Gm1m2/r2