© All Rights reserved @ LearnWithDash
Step-by-Step Solution
Step 1: Identify the Relevant Gravitational Potentials
Let the center-to-center distance between the two planets be 8R. Planet 1 has mass M and radius R. Planet 2 has mass 9M and radius 2R. A satellite of mass m is launched from the surface of Planet 1 (i.e., at a distance R from Planet 1's center).
At any distance x from Planet 1's center, the net gravitational potential due to the two planets (assuming both stay fixed in position) is
U(x) = -\,\frac{GMm}{x} \;-\; \frac{G(9\,M)\,m}{(8R - x)}.
Step 2: Locate the Point of Maximum Net Potential Between the Planets
The satellite must be able to cross the highest potential “hill” formed by the superposition of the two gravitational fields. To find where this maximum potential occurs, differentiate U(x) w.r.t. x and set it to zero:
\displaystyle \frac{d}{dx}\Bigl[-\frac{GMm}{x} \;-\; \frac{9\,GMm}{(8R - x)}\Bigr] \;=\; 0.
Carrying out the derivative and simplifying gives
-\frac{1}{x^{2}} + \frac{9}{(8R - x)^{2}} = 0,
which implies
\frac{9}{(8R - x)^{2}} = \frac{1}{x^{2}}.
Hence, (8R - x)/x = 3. Solving, 8R - x = 3x \implies x = 2R.
This distance x = 2R from Planet 1’s center is where the net potential is at its maximum value, U_{\text{max}}.
Step 3: Calculate the Value of the Maximum Potential, Umax
Substitute x = 2R into the potential expression:
U_{\text{max}} = -\,GMm \Bigl[\frac{1}{2R} \;+\; \frac{9}{(8R - 2R)}\Bigr]
\;=\; -\,GMm \Bigl[\frac{1}{2R} \;+\; \frac{9}{6R}\Bigr].
Next, simplify inside the brackets:
\frac{1}{2R} + \frac{9}{6R} = \frac{1}{2R} + \frac{3}{2R} = \frac{4}{2R} = \frac{2}{R}.
Therefore,
U_{\text{max}} = -\,\frac{2\,GMm}{R}.
Step 4: Write the Energy Conservation Condition at the Launch Point
Initially, the satellite is on the surface of Planet 1. The gravitational potential at this point ( x = R ) is:
U_{i} = -\,\frac{GMm}{R}\;-\;\frac{9\,GMm}{(8R - R)}
= -\,\frac{GMm}{R}\;-\;\frac{9\,GMm}{7R}
= -\,\frac{16\,GMm}{7R}.
Let the required minimum launch speed be v . Then the total initial energy is
E_{i} = \frac{1}{2}\,m\,v^{2} + U_{i}.
The satellite must at least reach the maximum potential point U_{\max} with zero kinetic energy there (i.e., it just makes it over the potential “hill”). Hence the total energy must equal U_{\max} :
\frac{1}{2}\,m\,v^{2} + \Bigl(-\frac{16\,GMm}{7R}\Bigr)
= -\,\frac{2\,GMm}{R}.
Step 5: Solve for the Required Speed v
Rearrange the above equation:
\displaystyle \frac{1}{2}\,m\,v^{2}
= -\,\frac{2\,GMm}{R} \;+\; \frac{16\,GMm}{7R}
= \frac{GMm}{R}\Bigl(-2 + \frac{16}{7}\Bigr).
Combine the terms inside the parentheses:
-2 + \tfrac{16}{7} = -\tfrac{14}{7} + \tfrac{16}{7} = \tfrac{2}{7}.
Hence,
\frac{1}{2}\,m\,v^{2} = \frac{2}{7}\,\frac{GMm}{R}
\quad\Longrightarrow\quad
v^{2} = \frac{4}{7}\,\frac{GM}{R}.
Therefore,
v = \sqrt{\frac{4}{7}\,\frac{GM}{R}}
= \sqrt{\frac{4}{7}}\;\sqrt{\frac{GM}{R}}
= \sqrt{\frac{a}{7}\,\frac{GM}{R}},
where comparing coefficients shows a = 4.
Final Answer
The value of “a” is 4.
