1. Initial circular orbit

The satellite is in a low circular orbit of radius approximately equal to Earth's radius: $r_0 = R_e$

For a circular orbit, speed is $v_c = \sqrt{\frac{GM}{R_e}}$ where $M$ is the mass of Earth.

2. Speed after firing rockets

Its speed is increased instantaneously in the same direction to $v = \sqrt{\frac{3}{2}}\,v_c$ So, $v^2 = \frac{3}{2}v_c^2 = \frac{3}{2}\cdot \frac{GM}{R_e}$

3. Nature of the new orbit

The velocity is increased tangentially, so the point where the impulse is given becomes the perigee (nearest point), since speed there is maximum.

Thus, $r_p = R_e$ Let the farthest distance be $R = r_a$.

4. Use conservation of mechanical energy

Total specific energy (energy per unit mass) just after the impulse is $\varepsilon = \frac{v^2}{2} - \frac{GM}{R_e}$ Substitute $v^2$: $\varepsilon = \frac{1}{2}\left(\frac{3}{2}\frac{GM}{R_e}\right) - \frac{GM}{R_e}$ $\varepsilon = \frac{3}{4}\frac{GM}{R_e} - \frac{GM}{R_e}$ $\varepsilon = -\frac{1}{4}\frac{GM}{R_e}$

For an elliptical orbit, $\varepsilon = -\frac{GM}{2a}$ where $a$ is the semi-major axis.

So, $-\frac{GM}{2a} = -\frac{1}{4}\frac{GM}{R_e}$ Cancel $-GM$: $\frac{1}{2a} = \frac{1}{4R_e}$ $2a = 4R_e$ $a = 2R_e$

5. Relate semi-major axis to nearest and farthest distances

For an ellipse, $a = \frac{r_p + r_a}{2}$ Thus, $2R_e = \frac{R_e + R}{2}$ $4R_e = R_e + R$ $R = 3R_e$

6. Check options

• A: $2R_e$ ❌
• B: $3R_e$ ✅
• C: $4R_e$ ❌
• D: $2.5R_e$ ❌

Therefore, the farthest distance from the center of Earth is $\boxed{3R_e}$