Question

The ratio of the values of acceleration due to gravity at heights \( h_1 \) and \( h_2 \) from the surface of the earth is \( 16:9 \). If height \( h_1 = 2R_E \), then \( h_2 \) is:

• A. \( 4R_E \)
• B. \( 3R_E \)
• C. \( 5R_E \)
• D. \( 6R_E \)

Hint:

The acceleration due to gravity decreases with height, and the relationship involves the square of the factor \( \left( 1 + \frac{h}{R_E} \right) \).

Answer 1

The Correct Option is B

Step 1: Formula for Acceleration Due to Gravity at a Height The acceleration due to gravity at a height \( h \) from the surface of the Earth is given by: \[ g_h = g_0 \left( \frac{R_E}{R_E + h} \right)^2 \] where \( g_h \) is the acceleration due to gravity at height \( h \), \( g_0 \) is the acceleration due to gravity on the Earth's surface, and \( R_E \) is the Earth's radius. 

Step 2: Given Ratio of Gravity Values It is given that: \[ \frac{g_{h_1}}{g_{h_2}} = \frac{16}{9} \] and \( h_1 = 2R_E \), so using the formula: \[ g_{h_1} = g_0 \left( \frac{R_E}{R_E + 2R_E} \right)^2 = g_0 \left( \frac{R_E}{3R_E} \right)^2 = g_0 \left( \frac{1}{3} \right)^2 = g_0 \times \frac{1}{9} \] 

Step 3: Finding \( h_2 \) Let \( h_2 = xR_E \), then: \[ g_{h_2} = g_0 \left( \frac{R_E}{R_E + xR_E} \right)^2 = g_0 \left( \frac{1}{1 + x} \right)^2 \] Using the given ratio: \[ \frac{\frac{g_0}{9}}{\frac{g_0}{(1+x)^2}} = \frac{16}{9} \] \[ \frac{1}{9} \div \frac{1}{(1+x)^2} = \frac{16}{9} \] \[ (1+x)^2 = \frac{9}{16} \times 9 \] \[ (1+x)^2 = \frac{81}{16} \] \[ 1+x = \frac{9}{4} \] \[ x = \frac{9}{4} - 1 = \frac{5}{4} \] \[ h_2 = \frac{5}{4} R_E = 3R_E \] 

Step 4: Conclusion Thus, \( h_2 = 3R_E \), so the correct answer is option (B).