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Step-by-Step Solution
Step 1: Identify the initial orbital velocity
For a satellite in a low, nearly circular orbit of radius $R_e$ (Earth’s radius),
the orbital speed $v_0$ is given by
$$
v_0 = \sqrt{\frac{GM}{R_e}},
$$
where $G$ is the universal gravitational constant and $M$ is the mass of the Earth.
Step 2: Determine the new velocity after firing the rockets
The problem states that the satellite’s speed is instantaneously increased by a factor of
$ \sqrt{\frac{3}{2}} $. Hence the new velocity $v$ is
$$
v = \sqrt{\frac{3}{2}} \, v_0
= \sqrt{\frac{3}{2}} \, \sqrt{\frac{GM}{R_e}}
= \sqrt{\frac{3\,GM}{2\,R_e}}.
$$
Step 3: Write the total mechanical energy at perigee
Just after the velocity is increased, the satellite is still at a distance $R_e$ from the center of the Earth (this point will become the perigee of the new elliptical orbit).
The total mechanical energy $E$ (per unit mass for simplicity) is:
$$
E = \frac{1}{2} {v}^2 \;-\; \frac{GM}{R_e}.
$$
Substituting the new speed $v$,
$$
E \;=\; \frac{1}{2} \left(\sqrt{\tfrac{3\,GM}{2\,R_e}}\right)^2
\;-\; \frac{GM}{R_e}
\;=\; \frac{1}{2} \cdot \frac{3\,GM}{2\,R_e}
\;-\; \frac{GM}{R_e}.
$$
Simplify this:
$$
E \;=\; \frac{3\,GM}{4\,R_e} \;-\; \frac{GM}{R_e}
\;=\; -\,\frac{GM}{R_e}\left(1 \;-\; \frac{3}{4}\right)
\;=\; -\,\frac{GM}{4\,R_e}.
$$
Step 4: Find the semi-major axis of the resulting elliptical orbit
For an elliptical orbit, the total energy $E$ (per unit mass) is related to the semi-major axis $a$ by:
$$
E = -\,\frac{GM}{2\,a}.
$$
Equating this to the value found in Step 3:
$$
-\,\frac{GM}{2\,a} \;=\; -\,\frac{GM}{4\,R_e}
\quad\Longrightarrow\quad
\frac{1}{2\,a} \;=\; \frac{1}{4\,R_e}
\quad\Longrightarrow\quad
a = 2\,R_e.
$$
Step 5: Determine the orbital eccentricity
The perigee $r_\text{p}$ of the elliptical orbit is given by $r_\text{p} = a (1 - e)$,
where $e$ is the orbital eccentricity.
Since the perigee is $R_e$,
$$
R_e = a(1 - e)
= 2R_e (1 - e)
\;\;\Longrightarrow\;\;
1 - e = \frac{1}{2}
\;\;\Longrightarrow\;\;
e = \frac{1}{2}.
$$
Step 6: Calculate the apogee distance
The apogee $r_\text{a}$ is given by $r_\text{a} = a(1 + e)$. Substituting $a = 2R_e$ and $e = \tfrac{1}{2}$:
$$
r_\text{a} = 2R_e \Bigl(1 + \tfrac{1}{2}\Bigr)
= 2R_e \times \frac{3}{2}
= 3R_e.
$$
Thus, the farthest distance from the center of the Earth that the satellite reaches is
$$
R = 3 R_e.
$$
Final Answer
The farthest distance from the Earth’s center is
$$
\boxed{3R_e}.
$$
Reference Image from the Provided Solution
