Problem 71 - A Complete Resource Book in Physics for JEE Main

Question

The change in the value of \(g\) at a height \(h\) above the surface of the earth is the same as at a depth \(d\) below the surface of earth. When both \(d\) and \(h\) are much smaller than the radius of earth, then which one of the following is correct? (A) \(d=\frac{h}{2}\) (B) \(d=\frac{3 h}{2}\) (C) \(d=2 h\) (D) \(d=h\)

Step-by-Step Solution

Answer

The relationship between the depth \(d\) and the height \(h\) is that they are equal when the change in the gravitational acceleration is the same. Thus, the correct option is: (D) \(d = h\)

Step 1: Newton's Law of Universal Gravitation

Let's first recall the formula for gravitational acceleration. In general, the acceleration due to gravity at a distance r from the center of a massive body (such as Earth) is given by: \[g(r) = \frac{G M}{r^2}\] where G is the gravitational constant, and M is the mass of the body.

Step 2: Gravitational Acceleration at Height h and Depth d

In our problem, we can consider the radius of Earth as r, and so, the height above the surface and the depth below will change the overall r value. For a height h above the surface, the distance from the center of the earth is: \[r_h = r + h\] and for a depth d below the surface, the distance from the center of the earth is: \[r_d = r - d\] Then, using the formula for gravitational acceleration, we can express g(h) and g(d) as follows: \[g(h) = \frac{G M}{(r + h)^2} \quad \text{and} \quad g(d) = \frac{G M}{(r - d)^2}\]

Step 3: Change in Gravitational Acceleration

We are asked to find the relationship between h and d when the change in the value of g is the same at both heights. We can represent this as: \[\Delta g_h = \Delta g_d\] or \[|g - g(h)| = |g - g(d)|\] where g represents the gravitational acceleration at the surface of the Earth: \[g = \frac{G M}{r^2}\]

Step 4: Simplifying the Equation

Now we can substitute the expressions for g(h) and g(d) and simplify: \[|\frac{G M}{r^2} - \frac{G M}{(r + h)^2}| = |\frac{G M}{r^2} - \frac{G M}{(r - d)^2}|\] First, we can cancel G and M: \[|\frac{1}{r^2} - \frac{1}{(r + h)^2}| = |\frac{1}{r^2} - \frac{1}{(r - d)^2}|\] Now let's cross-multiply and simplify: \[|(r - d)^2 - r^2| = |(r + h)^2 - r^2|\] Solving both sides: \[|r^2 - 2dr + d^2 - r^2| = |r^2 + 2rh + h^2 - r^2|\] Which simplifies to: \[2dr - d^2 = 2rh + h^2\]

Step 5: Solving for the Relationship between h and d

Now we can solve for d in terms of h: \[2dr - d^2 = 2rh + h^2\] \[d(2r - d) = h(2r + h)\] Since d and h are much smaller than r, we can safely ignore the d^2 and h^2 terms, so we are left with \[2dr = 2rh\] Divide both sides by 2r: \[d = h\] So, the relationship between the depth d and the height h is that they are equal when the change in the gravitational acceleration is the same. Thus, the correct option is: (D) \(d = h\)