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.

(r + 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,

2rh L²
L

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:

r r_earth

We have

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 Earth.

Another way of seeing this by equations we already have established before is:

m_ac = Gm₁m₂/r²
mv²/r = Gm₁m₂/r²

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