Question

If \(R_E\) be the radius of Earth, then the ratio between the acceleration due to gravity at a depth 'r' below and a height 'r' above the earth surface is : (Given : \(r<R_E\))

• A. \(1 + \frac{r}{R_E} + \frac{r^2}{R_E^2} + \frac{r^3}{R_E^3}\)
• B. \(1 + \frac{r}{R_E} - \frac{r^2}{R_E^2} - \frac{r^3}{R_E^3}\)
• C. \(1 - \frac{r}{R_E} - \frac{r^2}{R_E^2} - \frac{r^3}{R_E^3}\)
• D. \(1 + \frac{r}{R_E} - \frac{r^2}{R_E^2} + \frac{r^3}{R_E^3}\)

Hint:

Note that the approximation \(g_h = g(1 - 2r/R_E)\) is only valid when \(r \ll R_E\). Since the options include higher order terms (\(r^2, r^3\)), you must use the exact formula for \(g_h\) and expand the expression fully.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
Acceleration due to gravity varies with height and depth. We need to find the ratio of \(g_{depth}\) to \(g_{height}\) for a displacement \(r\).
Step 2: Key Formula or Approach:
1. At depth \(r\): \(g_d = g \left( 1 - \frac{r}{R_E} \right)\)
2. At height \(r\): \(g_h = \frac{g}{(1 + r/R_E)^2} = g \left( 1 + \frac{r}{R_E} \right)^{-2}\)
Step 3: Detailed Explanation:
We need to find the ratio \(\frac{g_d}{g_h}\):
\[ \text{Ratio} = \frac{g(1 - \frac{r}{R_E})}{g(1 + \frac{r}{R_E})^{-2}} = \left( 1 - \frac{r}{R_E} \right) \left( 1 + \frac{r}{R_E} \right)^2 \]
Expand the squared term:
\[ \text{Ratio} = \left( 1 - \frac{r}{R_E} \right) \left( 1 + \frac{2r}{R_E} + \frac{r^2}{R_E^2} \right) \]
Multiply the terms:
\[ \text{Ratio} = 1\left( 1 + \frac{2r}{R_E} + \frac{r^2}{R_E^2} \right) - \frac{r}{R_E} \left( 1 + \frac{2r}{R_E} + \frac{r^2}{R_E^2} \right) \]
\[ \text{Ratio} = 1 + \frac{2r}{R_E} + \frac{r^2}{R_E^2} - \frac{r}{R_E} - \frac{2r^2}{R_E^2} - \frac{r^3}{R_E^3} \]
Group like terms:
\[ \text{Ratio} = 1 + \left( \frac{2r}{R_E} - \frac{r}{R_E} \right) + \left( \frac{r^2}{R_E^2} - \frac{2r^2}{R_E^2} \right) - \frac{r^3}{R_E^3} \]
\[ \text{Ratio} = 1 + \frac{r}{R_E} - \frac{r^2}{R_E^2} - \frac{r^3}{R_E^3} \]
Step 4: Final Answer:
The ratio is \(1 + \frac{r}{R_E} - \frac{r^2}{R_E^2} - \frac{r^3}{R_E^3}\).