How are orbits made stable? I understand the concept of object staying in some orbit due to centrifugal force and gravity. 
However I do not understand how is orbit of a body like satellite or planet has perfect balance between gravitational pull and centrifugal force of revolution? 
because if the angular velocity is even a little bit more than required then the object will move away and gravitational pull will reduce with square of distance and so centripetal force will decrease leading the object to stray further.
so is everything in orbit is either slowly moving away or falling in?
or is it that only those bodies whose orbital velocity is just perfect stays in orbit with everything else spiraling in or out of it?
so are satellites put with very accurately calculated velocity to prevent them from falling in or moving away or somehow it balances itself? 
 A: First of all you should note that the orbit of such a satellite is stable orbit which means if you deviate it from the exact value of $r=r_0$ by small amount it will not go away and fall to the earth rather it will have a radial simple harmonic motion about $r=r_0$.This is because $r=r_0$ corresponds to the minimum of effective potential in which the satellite is bound.
This can be shown mathematically in following way.Let deviate it from $r=r_0$ by small amount so that the energy is given by $$E=\frac{1}{2}m\dot{r}^2+\frac{1}{2}mr^2\dot\theta^2-\frac{GMm}{r}=\frac{1}{2}m\dot{r}^2+\frac{L^2}{2mr^2}-\frac{K}{r}=\frac{1}{2}m\dot{r}^2+V_{eff}$$
If you expand $V_{eff}(r)$ about the minimum that is $r=r_0=\frac{L^2}{Km}$ you will get $$V_{eff}(r)=V_{eff}{(r_0)}+\frac{1}{2}k(r-r_0)^2_....$$
Where $k=V_{eff}''(r_0)=\frac{K^4m^3}{L^6}$.So the radial motion will be a simple harmonic oscillation about $r=r_0$ with frequency $$\omega=\sqrt{\frac{k}{m}}=\frac{mK^2}{L^3}$$
This will more clear if you just just try to plot $V_{eff}(r)$ vs $r$. About $r=r_0$ where $V_{eff}$ is minimum the potential can be approximated as that of simple harmonic oscillator for $r\sim r_0$.
A: In addition to the other answers, let me add an intuitive one:
Take the case of a body moving a bit too fast for circular orbit. As you say, it will move outward. But as it moves outward, it’s climbing against gravity. It’s kinetic energy goes down. It’s speed goes down. Eventually, the speed reduces to less than that needed for a circular orbit and it drops back down. 
This is the difference between circular orbits and elliptical ones. Elliptical orbits have more energy than a circular orbit at the lowest altitude, but less than that of a circular orbit at the highest. 
A: The reason why planets in our solar system have stable orbits is because, during the formation of the solar system, debris disk which consisted mostly of gas was orbiting the sun, During this period when protoplanets started forming they were interacting with this debris disk, due to this interactions(frictional forces) on the planets from the debris disk, Planets achieved a more or less a circular orbit. Afterwards our solar system continued to evolve and This debris disk disappeared(Asteroids, Comets were formed), From this point planets were fixed at an orbit they achieved.
The image bellow represents a star system with one planet orbiting also with debris disk.


how is orbit of a body like satellite or planet has perfect balance between gravitational pull and centrifugal force of revolution?

No it doesn't have a perfect balance. If that were the case all the orbits would have been perfectly circular which is not he case. at a given distance $r$ from the earths there is a specific orbital velocity at which an object has a circular orbit, If an object at that orbit accelerated to a larger velocity gravitational pull wouldn't increase just to keep that object in a circular path, gravitational pull stays the same Therefore that object will have an elliptical orbit. To explain this mathematically, For an object to be in a circular orbit it must have centripetal force equal to gravitational force:
$$\frac{mv^2}{r} = G\frac{Mm}{r^2} \Rightarrow v^2=G\frac{GM}{r}$$
$$v_{orbital}=\sqrt{\frac{GM}{r}}$$
From the equation we can see that if we increase the orbital speed the above equality doesn't hold between centripetal acceleration and gravitational attraction Thus it no longer has a circular orbit.
From the picture bellow we can see in red that if orbital velocity equals the above equation we calculated then it has a circular orbit, If it is less than or more it achieves an elliptical orbit.

A: In 2 dimensions, in any uniformly rotating frame of reference, there are two fictitous forces, the centrifugal force and the Coriolis force. The centrifugal force pulls directly away from the point of rotation is has the magnitude of $mr\omega^2$ where $m$ is the mass of the object; $r$ is the distance away from the center; and $\omega$ is the angular frequency. The Coriolis force goes in the direction 90° clockwise of the velocity in the rotating frame of reference if the frame of reference is rotating counterclockwise and 90° counterclockwise if the frame of reference is rotating clockwise and has magnitude $2mv$ where $v$ is the speed in the rotating frame of reference. In actuality, we live in the third dimension, not the second and for that reason, the force of gravity varies as the negative second power of distance. However, we can use the formulae for how to calculate centrifugal force and Coriolis force in the second dimension to figure out how to do so in the third dimension. The sun is so much more massive than Earth that we can neglect the gravitational effect of Earth on the sun. Suppose Earth has an exact circular orbit around the sun which it doesn't. Then we can take the uniformly rotating frame of reference where Earth and the sun are stationary. Then if Earth, has a tiny deviation from that orbit, than that will cause a Coriolis force to act on Earth in that frame of reference. It's the Coriolis force that prevents Earth from undergoing a runaway effect from the circular orbit.
