If total energy of an earth satellite is zero, it means that

The correct option is: The satellite moves away from the orbit along a parabolic path.

To understand why this is the case, let us analyze the total energy of a satellite orbiting the Earth. The total mechanical energy ($E$) of a satellite is the sum of its kinetic energy ($K$) and its gravitational potential energy ($U$):

$E=K+U$

Let $m$ be the mass of the satellite, $M$ be the mass of the Earth, $r$ be the distance of the satellite from the center of the Earth, and $v$ be its velocity. The expressions for kinetic and potential energy are:

$K=\frac{1}{2}m{v}^2$

$U=-\frac{GMm}{r}$

Here, $G$ is the universal gravitational constant. Therefore, the total energy of the satellite is given by:

$E=\frac{1}{2}m{v}^2-\frac{GMm}{r}$

The trajectory and bound nature of the satellite's orbit depend on the value of its total energy ($E$):

1. Bound Orbit ($E<0$): When the total energy is negative, the kinetic energy is insufficient to overcome the gravitational pull of the Earth. The satellite remains bound in a closed circular or elliptical orbit.
2. Hyperbolic Escape Orbit ($E>0$): When the total energy is positive, the kinetic energy exceeds the escape energy. The satellite escapes the gravitational field of the Earth along a hyperbolic path.
3. Parabolic Escape Orbit ($E=0$): When the total energy is exactly zero, the kinetic energy is just enough to allow the satellite to escape to infinity. Let us set the total energy to zero to find the required velocity:

$\frac{1}{2}m{v}^2-\frac{GMm}{r}=0$

Solving for $v$, we get:

$v=\sqrt{\frac{2GM}{r}}$

This critical speed is precisely the escape velocity of the satellite. A body moving at escape velocity from a gravitational source follows a parabolic trajectory, allowing it to move away from its orbit and escape the Earth's gravitational field completely with zero net kinetic energy at an infinite distance.