Question

The ratio of the accelerations due to gravity at heights 1280 km and 3200 km above the surface of the earth is: (Radius of the earth = 6400 km)

• A. \( 25:16 \)
• B. \( 5:2 \)
• C. \( 1:1 \)
• D. \( 25:4 \)

Hint:

Use \( g' = g \left( \frac{R}{R+h} \right)^2 \) to find acceleration at a given height. - Always compute the ratio carefully using proper fraction division.

Answer 1

The Correct Option is A

Step 1: Use gravity formula at height \( h \)
The acceleration due to gravity at height \( h \) is: \[ g' = g \left( \frac{R}{R+h} \right)^2. \] For \( h_1 = 1280 \) km: \[ g_1 = g \left( \frac{6400}{6400+1280} \right)^2 = g \left( \frac{6400}{7680} \right)^2. \] \[ g_1 = g \left( \frac{5}{6} \right)^2 = g \times \frac{25}{36}. \] For \( h_2 = 3200 \) km: \[ g_2 = g \left( \frac{6400}{6400+3200} \right)^2 = g \left( \frac{6400}{9600} \right)^2. \] \[ g_2 = g \left( \frac{2}{3} \right)^2 = g \times \frac{4}{9}. \] Step 2: Compute ratio \( g_1:g_2 \)
\[ \frac{g_1}{g_2} = \frac{25}{36} \div \frac{4}{9} = \frac{25}{36} \times \frac{9}{4} = \frac{25 \times 9}{36 \times 4} = \frac{225}{144} = \frac{25}{16}. \] Thus, the correct answer is \( \boxed{25:16} \).