Visualizing the Setup
Imagine you are standing on the surface of the Earth, which has a radius of RE. The problem asks us to explore how the acceleration due to gravity changes as we move away from the surface in two opposite directions.
First, we dig a tunnel and go down to a depth r. Second, we take a rocket and fly up to a height r. Our mission is to find the exact ratio of the gravity at this depth, gd, to the gravity at this height, gh.
The Master Equations
To solve this, we need to recall our fundamental formulas for the variation of gravity.
For a point inside the Earth at a depth r, the gravity decreases linearly because the mass of the outer shell no longer contributes to the net gravitational pull. The exact formula is:
For a point outside the Earth at a height r, gravity follows the inverse-square law. The exact formula is:
The Algebraic Execution
Now, we set up the ratio ghgd. When we divide the two expressions, the surface gravity g cancels out perfectly. The squared term in the denominator of gh flips up to the numerator, giving us:
ghgd=(1−REr)(1+REr)2
Let's expand the squared bracket first using the standard identity (a+b)2=a2+2ab+b2:
(1+REr)2=1+RE2r+RE2r2
Next, we carefully multiply this expanded quadratic expression by the linear term (1−REr):
ghgd=(1−REr)(1+RE2r+RE2r2)
Distributing the terms, we get:
=1+RE2r+RE2r2−REr−RE22r2−RE3r3
Finally, combining the like terms yields our elegant final answer:
ghgd=1+REr−RE2r2−RE3r3
The Catch
Why Not Approximate?
You might be wondering, "Why didn't we just use the binomial approximation gh≈g(1−RE2r)?"
That is a brilliant question! The approximation is only valid when r≪RE. However, if you look closely at the options provided in the question, they explicitly contain higher-order terms like r2 and r3. This is a massive hint from the examiner that they want the exact algebraic expansion, not an approximation. Always let the options guide your mathematical approach!