Question

Suppose two planets (spherical in shape) of radii R and 2 R, but mass M and 9 M respectively have a centre to centre separation 8 R 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 $\sqrt{\frac{aGM}{7R}}$ then the value of 'a' is _________. [Given : The two planets are fixed in their position] 

 

Hint:

For minimum projection speed problems involving two massive bodies, the key is to find the "null point" where gravitational forces balance. The minimum energy required is just enough to get the object to this point, after which it will be captured by the other body's gravity.

Answer 1

Correct Answer: 4

To find the minimum speed, the satellite must just be able to reach the null point (where the gravitational forces of the two planets cancel out) with zero kinetic energy. After this point, the gravity of the second planet will pull it in.
Let the null point be at a distance x from the center of the first planet (mass M). The distance from the second planet (mass 9M) will be (8R - x).
At the null point, the net gravitational force is zero:
$\frac{G M m}{x^2} = \frac{G (9M) m}{(8R - x)^2}$
$\frac{1}{x^2} = \frac{9}{(8R - x)^2} \implies \frac{1}{x} = \frac{3}{8R - x}$
$8R - x = 3x \implies 4x = 8R \implies x = 2R$.
Now, we apply the principle of conservation of energy. The initial energy at the surface of the first planet must equal the final energy at the null point.
Initial Energy ($E_i$) at the surface of the first planet:
$E_i = K.E._{initial} + P.E._{initial} = \frac{1}{2}mv^2 - \frac{GMm}{R} - \frac{G(9M)m}{(8R-R)} = \frac{1}{2}mv^2 - \frac{GMm}{R} - \frac{9GMm}{7R}$
Final Energy ($E_f$) at the null point (with minimum speed, so $v_{final}=0$):
$E_f = K.E._{final} + P.E._{final} = 0 - \frac{GMm}{x} - \frac{G(9M)m}{(8R-x)} = -\frac{GMm}{2R} - \frac{9GMm}{6R} = -\frac{GMm}{2R} - \frac{3GMm}{2R} = -\frac{4GMm}{2R} = -\frac{2GMm}{R}$
By conservation of energy, $E_i = E_f$:
$\frac{1}{2}mv^2 - \frac{GMm}{R} - \frac{9GMm}{7R} = -\frac{2GMm}{R}$
Divide by m and rearrange:
$\frac{1}{2}v^2 = \frac{GM}{R} + \frac{9GM}{7R} - \frac{2GM}{R} = GM \left( \frac{1}{R} + \frac{9}{7R} - \frac{2}{R} \right)$
$\frac{1}{2}v^2 = \frac{GM}{R} \left( \frac{7+9-14}{7} \right) = \frac{GM}{R} \left( \frac{2}{7} \right)$
$v^2 = \frac{4GM}{7R} \implies v = \sqrt{\frac{4GM}{7R}}$
Comparing this with the given expression $v = \sqrt{\frac{aGM}{7R}}$, we get $a=4$.