Question

Two points \(A\) and \(B\) are \(16\,\text{km}\) from the surface of Earth. Point \(A\) is above the surface while point \(B\) is below the surface. Acceleration due to gravity at \(A\) and \(B\) are \(g_A\) and \(g_B\) respectively. Find \(\dfrac{g_A}{g_B}\). (Take radius of Earth \(R = 6400\,\text{km}\))

• A. \(1\)
• B. \(\frac{398}{399}\)
• C. \(\frac{399}{398}\)
• D. \(\frac12\)

Answer 1

The Correct Option is B

Concept: Above Earth's surface: \[ g_A = g\left(1-\frac{2h}{R}\right) \] Below Earth's surface: \[ g_B = g\left(1-\frac{d}{R}\right) \] Step 1: For point \(A\) \[ h=16\ \text{km} \] \[ g_A=g\left(1-\frac{2\times16}{6400}\right) \] \[ g_A=g\left(1-\frac{32}{6400}\right) \] \[ g_A=g\left(\frac{6368}{6400}\right) \] Step 2: For point \(B\) \[ d=16\ \text{km} \] \[ g_B=g\left(1-\frac{16}{6400}\right) \] \[ g_B=g\left(\frac{6384}{6400}\right) \] Step 3: Ratio \[ \frac{g_A}{g_B}= \frac{\frac{6368}{6400}}{\frac{6384}{6400}} \] \[ =\frac{6368}{6384} \] \[ =\frac{398}{399} \]