Question

A body (mass $m$) starts its motion from rest from a point distant $R_0$ ($R_0>R$) from the centre of the earth. The velocity acquired by the body when it reaches the surface of earth will be ($G =$ universal constant of gravitation, $M =$ mass of earth, $R =$ radius of earth)}

• A. 2GM \left( \frac{1}{R} - \frac{1}{R_0} \right)
• B. \left[ 2GM \left( \frac{1}{R} - \frac{1}{R_0} \right) \right]^{\frac{1}{2
• C. GM \left( \frac{1}{R} - \frac{1}{R_0} \right)
• D. 2GM \left[ \left( \frac{1}{R} - \frac{1}{R_0} \right) \right]^{\frac{1}{2

Hint:

In central force fields like gravity, loss in Potential Energy equals gain in Kinetic Energy.

Answer 1

The Correct Option is A

textbf{Step 1:} Apply the Law of Conservation of Mechanical Energy. The sum of kinetic and potential energy at distance $R_0$ equals the sum at distance $R$: \[ KE_i + PE_i = KE_f + PE_f \] textbf{Step 2:} Substitute the initial values (at rest at distance $R_0$): \[ 0 + \left( -\frac{GMm}{R_0} \right) = \frac{1}{2}mv^2 + \left( -\frac{GMm}{R} \right) \] textbf{Step 3:} Rearrange the equation to isolate the kinetic energy term: \[ \frac{1}{2}mv^2 = \frac{GMm}{R} - \frac{GMm}{R_0} \] textbf{Step 4:} Cancel $m$ and solve for $v^2$: \[ v^2 = 2GM \left( \frac{1}{R} - \frac{1}{R_0} \right) \] textbf{Step 5:} Take the square root to find the velocity $v$: \[ v = \left[ 2GM \left( \frac{1}{R} - \frac{1}{R_0} \right) \right]^{\frac{1}{2 \]