Q56 / 75 · JEE Main 2026 (02 April Shift 2) · Physics
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 earth. The values of $v$ and $k$ respectively are _______. (Take radius of earth to be $6400$ km and $g = 9.8$ m/s$^2$)
- Correct answerA$11.2$ 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
Solution
By conservation of mechanical energy, the total energy of the body at infinity is zero.
When the body reaches the surface of the earth, its potential energy is $-\dfrac{GMm}{R}$ and kinetic energy is $k = \dfrac{1}{2}mv^2$.
$0 = k - \dfrac{GMm}{R}$
$k = \dfrac{GMm}{R} = mgR$
Substituting $m = 1$ kg, $g = 9.8$ m/s$^2$, and $R = 6400 \times 10^3$ m:
$k = 1 \times 9.8 \times 6400 \times 10^3 = 6.272 \times 10^7$ J
The velocity $v$ is given by $\dfrac{1}{2}mv^2 = k$.
$v = \sqrt{2gR} = \sqrt{2 \times 9.8 \times 6400 \times 10^3}$
$v = \sqrt{125.44 \times 10^6} = 11.2 \times 10^3$ m/s $= 11.2$ km/s
Answer: $11.2$ km/s; $6.27 \times 10^7$ J
When the body reaches the surface of the earth, its potential energy is $-\dfrac{GMm}{R}$ and kinetic energy is $k = \dfrac{1}{2}mv^2$.
$0 = k - \dfrac{GMm}{R}$
$k = \dfrac{GMm}{R} = mgR$
Substituting $m = 1$ kg, $g = 9.8$ m/s$^2$, and $R = 6400 \times 10^3$ m:
$k = 1 \times 9.8 \times 6400 \times 10^3 = 6.272 \times 10^7$ J
The velocity $v$ is given by $\dfrac{1}{2}mv^2 = k$.
$v = \sqrt{2gR} = \sqrt{2 \times 9.8 \times 6400 \times 10^3}$
$v = \sqrt{125.44 \times 10^6} = 11.2 \times 10^3$ m/s $= 11.2$ km/s
Answer: $11.2$ km/s; $6.27 \times 10^7$ J
Hence, the correct answer is option: 'A'.