Question

An object of mass \( m \) at a distance of \( 20R \) from the center of a planet of mass \( M \) and radius \( R \) has an initial velocity \( u \). The velocity with which the object hits the surface of the planet is: (G - Universal gravitational constant) \vspace{0.5cm}

• A. \( \left[ u^2 + \frac{19GM}{10R} \right]^{\frac{1}{2}} \)
• B. \( \left[ u^2 + \frac{19Gm}{10R} \right]^{\frac{1}{2}} \)
• C. \( \left[ u^2 - \frac{19GM}{10R} \right]^{\frac{1}{2}} \)
• D. \( \left[ u^2 - \frac{19Gm}{10R} \right]^{\frac{1}{2}} \)

Hint:

For gravitational problems, use conservation of energy: \[ \text{Initial Energy} = \text{Final Energy} \] accounting for kinetic and potential energy at both points.

Answer 1

The Correct Option is A

Step 1: Apply Energy Conservation The total mechanical energy at a distance \( 20R \) from the planet’s center is: \[ E_{\text{initial}} = \frac{1}{2} m u^2 - \frac{GMm}{20R} \] At the surface (\( R \)), the energy is: \[ E_{\text{final}} = \frac{1}{2} m v^2 - \frac{GMm}{R} \] Applying the conservation of energy: \[ \frac{1}{2} m u^2 - \frac{GMm}{20R} = \frac{1}{2} m v^2 - \frac{GMm}{R} \]

Step 2: Solve for \( v \) Rearranging: \[ \frac{1}{2} m v^2 = \frac{1}{2} m u^2 + \frac{GMm}{R} - \frac{GMm}{20R} \] \[ \frac{1}{2} m v^2 = \frac{1}{2} m u^2 + \frac{19GMm}{20R} \] Dividing by \( m \) and multiplying by 2: \[ v^2 = u^2 + \frac{38GM}{20R} \] \[ v^2 = u^2 + \frac{19GM}{10R} \] Taking the square root: \[ v = \left[ u^2 + \frac{19GM}{10R} \right]^{\frac{1}{2}} \] Thus, the velocity with which the object hits the surface of the planet is: \[ \mathbf{\left[ u^2 + \frac{19GM}{10R} \right]^{\frac{1}{2}}} \]