Question

If a body of mass 1 kg falls on the earth from infinity, it attains velocity \( v \) and kinetic energy \( k \) on reaching the surface of the earth. The values of \( v \) and \( k \) respectively are _______.

• A. 11.2 km/s; \( 6.27 \times 10^7 \) J
• B. 11.2 km/s; \( 12.54 \times 10^7 \) J
• C. 8.8 km/s; \( 6.27 \times 10^7 \) J
• D. 8.8 km/s; \( 12.54 \times 10^7 \) J

Answer 1

The Correct Option is B

Step 1: Calculating the velocity.
When a body falls from infinity, its velocity on reaching the surface of the Earth can be calculated using the formula for gravitational potential energy: \[ \frac{GMm}{R} = \frac{1}{2} mv^2 \] where \( G \) is the gravitational constant, \( M \) is the mass of the Earth, \( R \) is the radius of the Earth, \( m \) is the mass of the body, and \( v \) is the velocity. Substituting the known values: \[ G = 6.67 \times 10^{-11} \, \text{Nm}^2/\text{kg}^2, \, M = 5.97 \times 10^{24} \, \text{kg}, \, R = 6400 \times 10^3 \, \text{m} \] We find that the velocity \( v \) is approximately 11.2 km/s.
Step 2: Calculating the kinetic energy.
The kinetic energy of the body can be calculated using the formula: \[ K = \frac{1}{2} mv^2 \] Substituting the value of \( v \) and the mass of the body: \[ K = \frac{1}{2} (1)(11.2 \times 10^3)^2 = 12.54 \times 10^7 \, \text{J} \]
Step 3: Conclusion.
Thus, the velocity \( v = 11.2 \, \text{km/s} \) and the kinetic energy \( k = 12.54 \times 10^7 \, \text{J} \).
Final Answer: (B) 11.2 km/s; \( 12.54 \times 10^7 \) J