What happen to a spoon which is detached from the satellite? Suppose a spoon is a part of satellite but after detachment from the satellite Does it fall to the ground straight or does is take a parabolic path or any other path before coming to the surface of Earth
 A: A few assumptions before we get started:
1) The satellite that you refer to is traveling in an orbital path around the Earth, as opposed to some other type of motion (You did not specify whether the satellite was in orbital or sub-orbital flight.  If assumed incorrectly, please let me know with a comment.)
2) Relative to the satellite, the spoon is released from rest, along the path of the satellite.
As long as my assumptions in 1 and 2 are correct, than the spoon will continue traveling along the original path of the satellite, in orbital flight.
The  gravitational force of attraction between Earth and your satellite is given by $$F=\frac{m_{Earth}m_{satellite}}{r^2}$$, which I will here forward refer to as $m_E$ and $m_s$ for short.  For an orbiting satellite, the gravitational force of attraction between the Earth and the satellite acts as a centripetal force to hold the satellite in a circular orbit around the Earth.  The equation for this centripetal force is given by $$F=\frac{m_sv^2}{r}$$
Now before we go any further I want to point out that the Earth and satellite each orbit about the combined center of mass of the system.  However, for a typical satellite orbiting the Earth, this difference is so small, that for the purpose of your question, we can just say the satellite orbits the Earth.  Additionally, your satellite will actually travel in an ellipse, not a circle.  However, the eccentricity for an orbiting satellite is close to that of a circle, so again, for the purpose of this question, we can simplify the situation by saying that the satellite orbits in a circle about the Earth.
Anyway, because it is the force of gravity that acts as a centripetal force holding the satellite in orbit, we can apply Newton's Synthesis and set these two equations equal to each other $$\frac{m_Em_s}{r^2}=\frac{m_sv^2}{r}$$  This simplifies to $$\frac{m_E}{r}=v^2$$
Now the reason why I went through this whole derivation is just to show you that the orbital velocity of a satellite does not depend upon its mass.  The $m_s$ cancelled out of the final equation.  What this means is that when the spoon is released from the satellite, it will continue to travel at the same velocity it had before, along the orbital path, due to inertia.  Even if you threw the spoon away from the satellite, the new orbit of the spoon would not be too much different, because the velocity you would have provided to the spoon by throwing it is still very small in comparison to the orbital velocity it already had.
A: The spoon will keep moving until orbital decay takes control and it reaches the earth over several orbital periods.
To explain why, imagine a spoon hung to a satellite traveling at a constant velocity (ignoring any accelerations, for ease of understand)  of $x$ and therefore all its components must be travelling at the exact same velocity to make sure their relative velocity is $0$ m/s, that said the spoon must be also travelling at $x$ velocity but as orbital decay slowly pulls the object by slowing it down it falls to earth gradually. 
Furthermore, space rocks and other objects could disturb the object by slowing it down or pushing it towards earth and such or destroying it in process. 
The key effect here you might want to see is orbital decay: http://en.wikipedia.org/wiki/Orbital_decay
