Basics of centripetal force Suppose an object is moving in a circular path. We know that the net force that is working on that object is the centripetal force towards the center. But the object should have gone closer towards the center in that case due to the radially inward force working on it, but it doesn't. Why does the object remain on the circular path instead of going closer towards the center?
For people who would be introducing centrifugal force in this case, i have a doubt on this too. Centrifugal is a pseudo force that only works when we are in the frame of the rotating object meaning we experience a pseudo force that pushes us radially outward. When we are in this frame, does centripetal and centrifugal both work on us?
But let us stay in ground frame as of now. Then what is the cause of the object not being pushed radially inward due to the effect of centripetal force? I am asking this question to clear out my doubts for strengthening my basic concept of physics. Hope the physics lovers will find this question relevant.
 A: I think this is the source of your confusion:
"...But the object should have gone closer towards the center...."
Never, ever, ever use relative words such as "closer" without asking yourself "Closer towards the center than what?"
And that's the answer to your question: The object has moved closer to the center than it would have been if it continued to move in a straight line.
Remember Newton's first law: Objects move with a constant velocity unless acted on by a force.  That is, they move at a constant speed and a constant direction (a straight line) unless a force moves them from that path. In the case of a circular (or elliptical) path, the centripetal force is continuously adjusting the object away from a straight-line path.
A: The centripetal force on an object moving in a circle radius $r$ and tangential velocity $v$ is $${\bf F}=\frac{mv^2}{r}{\bf \hat r}$$ where the force points to the center of rotation along the unit vector $\bf \hat r$.
The fact that the object moves with a tangential velocity $\bf v$ at all times is relevant, since while the direction of the force is toward the center, the object is moving in a direction at right angles to this force. In other words, $$\bf F\cdot v=\bf \hat r\cdot v=0$$ It may be "falling" toward the center but its distance from the center remains constant.
Therefore, the centripetal force does not move the object in an inward direction (decreasing the value of $r$), and hence there will be no displacement in the inward direction, and the object will maintain at the same distance while continually changing direction.
While objects that are for example, in circular orbits around the earth, like the ISS, this object is constantly in free-fall, but the average distance between this object and the earth's surface does not change.
You are also correct in your "suspicion" regarding the centrifugal force. While it does act away from the direction toward the center, it is simply a reaction (inertial force) to the inward centripetal force.
A: Since the workdone by the force acting radially inward  is always zero  in a circular motion, the kinetic energy of the object doesn't changes and hence doesn't pull inward.
A: A better way to understand is a stone tied to a string moving in a circle.
Think that the stone is intially moving with a velocity $v$. It tends to move along a straight line, if there is no external force acting on it. But the tension of the string is trying to pull it inwards. Therefore when it tries to move a little bit forward, the tension of the string pulls it a little bit toward the centre. Again while the stone tries to move forward (the new forward direction), the string tries to pull it inwards. This process happens throughout the motion within tiny amounts of period. The overall result is the stone seems to be moving in a circular path.
What happens if the string is unable to provide required centripetal force. It breaks. That is because: the stone is moving with a high velocity. So while the string tries to pull the stone inwards, the stone prefers to stay on its straight path because of the high momentum due to high speed. Thus it separates from the string and continues its journey along a straight line.

Then what is the cause of the object not being pushed radially inward due to the effect of centripetal force?

The simple answer is because it is moving, with a tangential velocity. You know what happens if the string was tied to a stationary stone and try to pull it.
Another point is that any other external force will cause slow down of the stone. But centripetal force doesn't because it is perpendicular to the velocity. This is more obvious because the work done by centripetal force is zero. When you make a stone move and stop making it move, it will slow down and finally stops. That is because external forces such as air resistance. If there are no air resistance, friction, or any other external force, you can move the stone in a circular path without any effort.
A: The object does fall towards the centre. It just misses...

*

*Imagine placing a satellite high up there and letting go. It will fall straight down and crash.

*Now push it slightly sideways while letting go so it has a small sideways speed to start with. I still falls down, but it also falls a bit sideways. It crashes on the ground slightly to the side from before.

*Now give an even greater sideways speed. It still crashes, but this time far to the side from the point that is directly underneath.

*And now give an even greater start speed, so large that the satellite flies so much sideways that it misses Earth. It still falls, but it falls besides Earth. And doesn't crash into Earth.

After missing Earth, the satellite flies away from Earth on the other side. Soon gravity will pull it back again. And the same thing will happen all over - it will miss Earth again. This continues forever; this is an elliptic orbit. With an even greater sideways start speed, the elliptic orbit becomes wider until it at some specific sideways speed exactly becomes as wide as it is tall - now it is a circular orbit.
The sideways speed needed for achieving an exactly circular orbit is found via the centripetal-acceleration formula:
$$a_c=\frac{v^2}{r}.$$
In this case the centripetal acceleration will be the gravitational acceleration at the orbit.
This was an explanation of why objects in circular motion don't fall inwards towards the centre. The answer is that they do fall. They fall constantly. They just miss the centre constantly as well. No need for centrifugal effects to explain this. You are correct that the so-called centrifugal force is a fictitious force that does not exist in the inertial frame - it is merely a force "invented" to explain the "swung outwards" tendency that we feel from our own perspective (from the rotational frame) when sitting in a turning car, in a spinning carousel etc.
A: The object would move closer to the centre if the centripetal force was increased. Likewise it would move further away from the centre if the centripetal force were decreased. The whole point about circular motion is that it is a state of equilibrium in which the force applied to the object is of the exact strength required to deviate its path from the straight ahead position without either allowing the object to spiral out or forcing it to spiral in.
A: Another way of analyzing the situation: imagine an object moving counterclockise at a velocity of $1$, starting at the point $(1,0)$ at time $0$. If it continues moving in a straight line, then at time $t$ it will be at point $(1,t)$, a distance of $\sqrt{1+t^2}$. The Taylor series for that is $1+\frac{t^2}2-\frac{t^4}{8}+\frac{t^6}{16}...$.
If we define $\Delta$ as the change in radial distance, $\Delta =  \sqrt{1+t^2}-1=1+\frac{t^2}2-\frac{t^4}{8}+\frac{t^6}{16}...-1=\frac{t^2}2-\frac{t^4}{8}+\frac{t^6}{16}...$
If we take the second derivative to get the acceleration, we get $\frac {d^2 \Delta}{dt^2}=1-\frac{3t^2}2+\frac {15t^4}{8}...$. If we add an acceleration of $1$ towards the center, the net acceleration is $-\frac{3t^2}2+\frac {15t^4}{8}...$
As we take the limit as $t \rightarrow 0$, this goes to zero.
The $1-\frac{3t^2}2+\frac {15t^4}{8}...$ is the centrifugal acceleration (where, here, "acceleration" means second derivative of the distance, not the second derivative of the location; since the direction of the distance is changing, those are two different things). Without any centripetal force, the distance from the center will increase, and in the rotating frame of reference, there is an apparent centrifugal force. When you include the centripetal force, the centripetal acceleration will cancel with the centrifugal acceleration, and the net instantaneous acceleration will be zero.
A: When going in a circle at constant speed, velocity changes. Velocity is speed and direction. Direction changing, is velocity changing. Changing velocity is acceleration. So, it takes acceleration to move in a circle. By $F=ma$, that takes force. That’s called centripetal force, the force to keep something revolving. In direction of velocity change: toward the center.
If you’re in the thing revolving, you feel a force pulling you away from the center, as if extra gravity. Thing circling is an “inertial reference frame” and is accelerating. The apparent force felt by things in the revolving frame is called centrifugal force and it is directly outward.
A: That's cause centripetal force is exactly the amount of twist needed for the momentum vector of the particle at an instant so that it sticks tangent to the path at the next.
This works for unit speed, and unit mass,  the force is exactly equal to the curvature of the path.
A: Already many useful answers, but...
I'll try to cover the topic of reference frames.
There are inertial (non-rotating) reference frames and rotating ones.
Whatever frame we are considering, as stubborn observers we want objects to obey the Newtonian rules, that without force they move at constant speed in a straight line (or stay at one point as a special case), and that any deviation from straight movement is caused by some force acting upon the object.
In an inertial frame, there's no problem, all objects follow the rules. If an object does not move in a straight line, there is a real force that makes it change its speed or direction. A circular path results if a force of appropriate magnitude constantly acts towards the center of the path.
But in a rotating frame, objects show a weird behavior like circling around the frame center or spiraling inwards and outwards (though, from an outside view the just stand still or move in straight lines). So, a (naive) observer on the rotating frame concludes that there must be forces pushing the objects around. As this effect is just caused by the weird observation environment, we call these forces pseudo-forces, and the centrifugal force is one special case.
Now, let's look at the circling object from a matched rotating frame, one that is centered at the center of the object path, and rotates with the same rate as the object. For an observer on that frame, the object does not move, meaning that there seems to be an equilibrium of forces. The observer already understands that (pseudo-) forces like the centrifugal force act upon all objects. So, to make the object "stand still", an opposite force must be acting on the object, the force called "centripetal". This is a real force, exerted e.g. by a string to the object or by gravitation (in the non-relativistic sense).
A: 
But the object should have gone closer towards the center in that case due to the radially inward force working on it

I'll take quite different approach here. Why you think body should move closer to the COM if acted by a constant centripetal force ? It does not follows. Take a look at first Newton law- it says that body is trying to be at rest or keeps a constant linear motion due to inertia. So somebody must introduce constant centripetal force only for forcing object to loose linear motion constantly, for changing it's tangential speed (speed is magnitude AND direction !). So by applying a centripetal force you constantly change object direction and as such force him to stay in the same orbit. If you want to pull object even closer to you (imagine rotating a toy over your head with a rope), then you need to apply even a greater centripetal force, you will be increasing it- for making an orbit shorter.
And final words - sometimes rotating objects do merge, check binary neutron stars. But reasons of this effect is not fully known and probably out of scope of Newtonian gravity. One of explanations of why does couple of neutron stars merge into one star is due to strong intensity of gravitational waves which these neutron stars emits. Due to emission of gravitational waves stars loose energy and because of that are forced into a lower orbits until full collapse happens. But as I said this needs General Relativity for explanation.
A: I will use some simple diagrams.
Imagine that the circular path is a series of very small diagonal trajectories (1). I will capture a small moment in time to see one of these diagonals (2).

The trajectory the object wants to do is the orange one (a). But the centripetal force (b) makes a change on the trajectory and produces the black vector (c).
If the deviation of the trajectory is the exact same length as the magenta component the trajectory is maintained as a circle.
But if you imprint more centripetal force than this magenta component (d) the object will start to "fall" to the center.
Another way to make the object fall to the center is by making the orange component shorter, for example, due to friction.

But I am using the term "falling" too loosely. I will add an example using gravity.
If I let fall a red object it will fall perpendicular to the surface of the planet (e).
But if I add a small tangential speed, the object will fall away from this spot. (f)
If I add more speed, it will now miss the planet (g). It still is "falling" at the same speed as the other cases, because the centripetal force is the same, but it is missing the crash. It is a matter of the vector components being the correct magnitude.

