A 90 kg body placed at \( 2R \) distance from the surface of the earth experiences gravitational pull of: (\( R \) = Radius of Earth, \( g = 10 \, \text{ms}^{-2} \)).
1. Determine Gravitational Acceleration at 2R above Earth's Surface: Gravitational acceleration \( g \) at height \( h \) above Earth's surface is calculated as: \[ g = g_s \left(1 + \frac{h}{R}\right)^{-2}, \] where \( g_s \) is surface gravitational acceleration and \( R \) is Earth's radius. 2. Set \( h = 2R \): Substituting \( h = 2R \) yields: \[ g = g_s \left(1 + \frac{2R}{R}\right)^{-2} = g_s (3)^{-2} = \frac{g_s}{9}. \] Using \( g_s = 10 \, \text{m/s}^2 \). 3. Compute Gravitational Force: The gravitational force \( F \) on the 90 kg object is: \[ F = mg = 90 \times \frac{g_s}{9} = 90 \times \frac{10}{9} = 100 \, \text{N}. \]
