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: Understanding the Concept:
When an object falls from infinity to the surface of the Earth, it loses gravitational potential energy and gains kinetic energy.
By the principle of conservation of mechanical energy, the velocity attained is exactly equal to the escape velocity of the Earth.
Step 2: Key Formula or Approach:
The escape velocity from the Earth's surface is given by \(v = \sqrt{2gR}\), where \(g\) is the acceleration due to gravity and \(R\) is the Earth's radius.
The kinetic energy is given by \(k = \frac{1}{2} m v^2\).
Step 3: Detailed Explanation:
Given values:
\(m = 1 \text{ kg}\)
\(g = 9.8 \text{ m/s}^2\)
\(R = 6400 \text{ km} = 6.4 \times 10^6 \text{ m}\)
First, calculate the velocity \(v\):
\[ v = \sqrt{2 \times 9.8 \times 6.4 \times 10^6} \] \[ v = \sqrt{19.6 \times 6.4 \times 10^6} \] \[ v = \sqrt{125.44 \times 10^6} \] \[ v = 11.2 \times 10^3 \text{ m/s} = 11.2 \text{ km/s} \] Next, calculate the kinetic energy \(k\):
\[ k = \frac{1}{2} m v^2 = \frac{1}{2} (1) (11.2 \times 10^3)^2 \] \[ k = \frac{1}{2} (125.44 \times 10^6) \] \[ k = 62.72 \times 10^6 \text{ J} \] Convert this into scientific notation matching the options:
\[ k = 6.272 \times 10^7 \text{ J} \approx 6.27 \times 10^7 \text{ J} \] Step 4: Final Answer:
The values of \(v\) and \(k\) respectively are \(11.2 \text{ km/s}\) and \(6.27 \times 10^7 \text{ J}\).