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Derivation of orbital energy of satellite?
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Derivation of orbital energy of satellite?
Circular orbits arise whenever the gravitational force on a satellite equals the centripetal force needed to move it with uniform circular motion. Substitute this expression into the formula for kinetic energy. When U and K are combined, their total is half the gravitational potential energy.
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Derivation of orbital energy of satellite?
The satellites orbit around a central massive body in either a circular or elliptical manner. A satellite orbiting about the earth moves in a circular motion at a constant speed and at fixed height by moving with a tangential velocity that allows it to fall at the same rate at which the earth curves. The force of gravity acts in a direction perpendicular to the direction of motion of the satellite throughout the trajectory.As per the work-energy theorem, the initial total mechanical energy of the system plus the work done by any external force is equal to the final total mechanical energy.

Mathematically, KEi + PEi + Wext = KEf + PEf

For satellites, the force of gravity is the only external force and since gravity is considered as a conservative force, the term Wext is zero.

The equation can be simplified as KEi + PEi = KEf + PEf

In other words, the sum of the kinetic energy and the potential energy of the system is constant, while energy gets transformed between the kinetic energyand the potential energy.

Analysis for circular orbits

In circular motion about the Earth, a satellite remains at a fixed distance from the surface of the Earth at all the time. Since the tangential velocity is a function of the radius of the orbit, the velocity remains constant and so does the kinetic energy. Also, since the potential energy is dependent on the height of the object, which too remains constant in this case, thus, the potential energy too remains the constant throughout. So, the total mechanical energy i.e. KE + PE remains constant.

The motion of a satellite around the Earth is considered to be circular. In this section, we will derive the expression for the kinetic energy, potential energy and the total mechanical energy of an object orbiting in a circular path around the Earth.

For a satellite orbiting the earth, the tangential velocity can be given as:
V=√GM/r+h

Where M is the mass of the earth, R is the radius of the earth, h is the height from the surface of the earth where is an object is kept.

So, the kinetic energy of the satellite (mass m) in a circular orbit with speed v can be written as:

KE=1/2mvv =GmM/2(r+h)

AS per our assumption, the gravitational potential energy at infinity is considered to be zero, so, the potential energy at distance (R +h) from the centre of the earth can be written as:

PE=-GmM/(R+h)


The kinetic energy here is positive whereas the potential energy is negative. However, in magnitude, the kinetic energy is half the potential energy, so the total energy E is

E=KE+PE=GmM/2(R+h) - GmM/(R+h)=- GmM/2(R+h)

The total energy of a circularly orbiting satellite is thus negative, with the potential energy being negative but twice is the magnitude of the positive kinetic energy.

|E|=|KE|=1/2|PE|
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