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Q5.5(c):
Answer the following : An artificial satellite orbiting the earth in very thin atmosphere loses its energy gradually due to dissipation against atmospheric resistance, however small. Why then does its speed increase progressively as it comes closer and closer to the earth ?

Solution :

Step 1: Energy Analysis of an Orbiting Satellite

The Total Mechanical Energy ($E$) of a satellite of mass $m$ orbiting a planet of mass $M$ at a distance $r$ from the center is given by the sum of its kinetic energy ($K$) and gravitational potential energy ($U$):

$E = K + U = \frac{1}{2}mv^2 - \frac{GMm}{r}$

For a stable circular orbit, the centripetal force is provided by the gravitational pull [Per Newton’s Law of Universal Gravitation]:

$\frac{mv^2}{r} = \frac{GMm}{r^2} \implies mv^2 = \frac{GMm}{r} \implies K = \frac{GMm}{2r}$

Substituting this into the energy equation:

$E = \frac{GMm}{2r} - \frac{GMm}{r} = -\frac{GMm}{2r}$

Step 2: The Effect of Atmospheric Dissipation

When the satellite encounters atmospheric drag, it loses total mechanical energy ($E$ decreases, becoming more negative). [Per the Law of Conservation of Energy]. Since $E = -\frac{GMm}{2r}$, for $E$ to decrease (become more negative), the denominator $r$ must decrease. This confirms that the satellite spirals inward toward the Earth.

Step 3: Calculating Velocity Change

From the virial theorem relation derived in Step 1, the orbital velocity $v$ is given by:

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

As the satellite loses energy due to friction, its orbital radius $r$ decreases. Examining the inverse relationship between $v$ and $r$:

If $r \to r - \Delta r$ (where $\Delta r > 0$), then $v \to \sqrt{\frac{GM}{r - \Delta r}}$.

Since the denominator decreases, the magnitude of the velocity $v$ must increase. [By Algebraic Inequality Logic].

Step 4: Summary of Energy Conversion

The dissipative force (atmospheric drag) does negative work on the system, reducing the total mechanical energy. However, the loss in potential energy ($\Delta U$) is twice the loss in total energy ($\Delta E$), which necessitates an increase in kinetic energy ($\Delta K$).

Earth Start (r1) End (r2) Spiral trajectory: r decreases, v increases

Final Solution: The satellite loses total mechanical energy due to atmospheric drag, causing its orbital radius ($r$) to decrease. Because orbital velocity is inversely proportional to the square root of the radius ($v \propto r^{-1/2}$), the decrease in $r$ results in a net increase in kinetic energy and orbital speed. The loss of potential energy is so significant that it compensates for the energy dissipated as heat, leading to an acceleration of the satellite as it descends.


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