Gravitation

JEE Main level, single correct, medium

104 of 223

Question 104 of 223

If a body of mass 11 kg falls on the earth from infinity, it attains velocity (v)(v) and kinetic energy (k)(k) on reaching the surface of earth. The values of vv and kk respectively are _______. (Take radius of earth to be 64006400 km and g=9.8g = 9.8 m/s2^2)

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Answer: (A) 11.211.2 km/s; 6.27×1076.27 \times 10^7 J

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 −GMmR-\dfrac{GMm}{R} and kinetic energy is k=12mv2k = \dfrac{1}{2}mv^2.

0=k−GMmR0 = k - \dfrac{GMm}{R}

k=GMmR=mgRk = \dfrac{GMm}{R} = mgR

Substituting m=1m = 1 kg, g=9.8g = 9.8 m/s2^2, and R=6400×103R = 6400 \times 10^3 m:

k=1×9.8×6400×103=6.272×107k = 1 \times 9.8 \times 6400 \times 10^3 = 6.272 \times 10^7 J

The velocity vv is given by 12mv2=k\dfrac{1}{2}mv^2 = k.

v=2gR=2×9.8×6400×103v = \sqrt{2gR} = \sqrt{2 \times 9.8 \times 6400 \times 10^3}

v=125.44×106=11.2×103v = \sqrt{125.44 \times 10^6} = 11.2 \times 10^3 m/s =11.2= 11.2 km/s

Answer: 11.211.2 km/s; 6.27×1076.27 \times 10^7 J

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