A satellite is moving in a low nearly circular orbit around the earth. Its radius is roughly equal to that of the earth's radius $$R_e$$. By firing rockets attached to it, its speed is instantaneously increased in the direction of its motion so that it becomes $$\sqrt{\frac{3}{2}}$$ times larger. Due to this the farthest distance from the centre of the earth that the satellite reaches is $$R$$. Value of $$R$$ is:

We start with the satellite in a very low, almost circular orbit whose radius is practically the earth’s radius itself. Hence the initial orbital radius is $$r_1 = R_e$$.

For a circular orbit, the standard formula for orbital speed is

$$v_c = \sqrt{\frac{GM}{r}},$$

where $$G$$ is the universal gravitational constant and $$M$$ is the mass of the earth. Substituting $$r = R_e$$ gives the initial speed

$$v_1 = \sqrt{\frac{GM}{R_e}}.$$

Immediately after the rockets are fired, the speed is increased tangentially to

$$v_2 = \sqrt{\frac{3}{2}}\;v_1 = \sqrt{\frac{3}{2}}\;\sqrt{\frac{GM}{R_e}} = \sqrt{\frac{3GM}{2R_e}}.$$

The moment after the impulse, the satellite is still at the same position $$r = R_e$$ but now with the higher speed $$v_2$$. Its subsequent path is no longer circular; it becomes an ellipse with the present point as the perigee (closest point). We shall find the apogee distance $$R$$ using conservation laws.

1. Specific mechanical energy (total energy per unit mass)
The formula is

$$\varepsilon = \frac{v^2}{2} - \frac{GM}{r}.$$

At perigee just after firing, we substitute $$v = v_2$$ and $$r = R_e$$:

$$\varepsilon = \frac{1}{2}\left(\frac{3GM}{2R_e}\right) - \frac{GM}{R_e} = \frac{3GM}{4R_e} - \frac{GM}{R_e} = \frac{3GM}{4R_e} - \frac{4GM}{4R_e} = -\frac{GM}{4R_e}.$$

For an elliptical orbit, the same energy is also expressed as

$$\varepsilon = -\frac{GM}{2a},$$

where $$a$$ is the semi-major axis. Equating the two expressions, we get

$$-\frac{GM}{2a} = -\frac{GM}{4R_e} \;\;\Longrightarrow\;\; 2a = 4R_e \;\;\Longrightarrow\;\; a = 2R_e.$$

2. Relation between perigee, apogee, and semi-major axis
For any ellipse,

$$r_p + r_a = 2a,$$

where $$r_p$$ is the perigee distance and $$r_a$$ is the apogee distance. Here

$$r_p = R_e,\qquad r_a = R.$$

Substituting these values and $$a = 2R_e$$, we obtain

$$R_e + R = 2a = 4R_e \;\;\Longrightarrow\;\; R = 4R_e - R_e = 3R_e.$$

So the farthest distance from the earth’s centre reached by the satellite is three times the earth’s radius.

Hence, the correct answer is Option C.