Suppose two planets (spherical in shape) of radii $$R$$ and $$2R$$, but mass $$M$$ and $$9M$$ respectively have a centre to centre separation $$8R$$ as shown in the figure. A satellite of mass $$m$$ is projected from the surface of the planet of mass $$M$$ directly towards the centre of the second planet. The minimum speed $$v$$ required for the satellite to reach the surface of the second planet is $$\sqrt{\frac{a}{7} \frac{GM}{R}}$$, then the value of $$a$$ is
[Given: The two planets are fixed in their position]

We need to determine the value of the parameter $$a$$ representing the minimum projection speed required for a satellite of mass $$m$$ to travel from the surface of the first planet to the surface of the second planet.

1. Identify the Neutral Point (Null Point)

From the system geometry:

• Planet 1: Mass $$M_1 = M$$, Radius $$R_1 = R$$
• Planet 2: Mass $$M_2 = 9M$$, Radius $$R_2 = 2R$$
• Center-to-Center Distance: $$d = 8R$$

For the satellite to reach the second planet, it must be launched with enough kinetic energy to cross the neutral point ($$r_0$$) where the gravitational pull of both planets perfectly balances out. Let $$r_0$$ be the distance of this point from the center of the first planet:

$$\frac{G M m}{r_0^2} = \frac{G (9M) m}{(8R - r_0)^2}$$

Taking the square root of both sides:

$$\frac{1}{r_0} = \frac{3}{8R - r_0}$$

$$8R - r_0 = 3r_0 \implies 4r_0 = 8R \implies r_0 = 2R$$

Thus, the neutral point lies at a distance of $$2R$$ from the center of the first planet, and $$6R$$ from the center of the second planet.

2. Apply Principle of Conservation of Energy

We compare the mechanical energy of the satellite at its launch position (the surface of the first planet, at distance $$R$$ from its center) to its mechanical energy at the neutral point (at distance $$2R$$ from the first center), where its velocity momentarily drops to zero ($$v_f = 0$$):

Energy at the Launch Surface ($$E_i$$):

$$E_i = \frac{1}{2}mv^2 - \frac{GMm}{R} - \frac{G(9M)m}{7R}$$

Energy at the Neutral Point ($$E_f$$):

$$E_f = 0 - \frac{GMm}{2R} - \frac{G(9M)m}{6R} = -\frac{GMm}{2R} - \frac{3GMm}{2R} = -\frac{2GMm}{R}$$

Equating $$E_i = E_f$$ and canceling out the satellite mass $$m$$:

$$\frac{1}{2}v^2 - \frac{GM}{R} - \frac{9GM}{7R} = -\frac{2GM}{R}$$

$$\frac{1}{2}v^2 = \frac{GM}{R} + \frac{9GM}{7R} - \frac{2GM}{R}$$

$$\frac{1}{2}v^2 = \frac{9GM}{7R} - \frac{GM}{R}$$

$$\frac{1}{2}v^2 = \frac{2GM}{7R}$$

$$v^2 = \frac{4GM}{7R} \implies v = \sqrt{\frac{4GM}{7R}}$$

3. Compare with the Given Expression

The problem states that the required minimum velocity is written in the form:

$$v = \sqrt{\frac{a}{7}\frac{GM}{R}}$$

Comparing our calculated result with this formula, we find:

$$a = 4$$

Conclusion

The value of $$a$$ required for the satellite to successfully reach the second planet's surface is 4.