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Physics Question 31 – JEE-MAIN 2026

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/s2)

When a body falls from infinity under the influence of gravity, its total mechanical energy is conserved.

Step 1: Apply Conservation of Mechanical Energy✦ Active

The initial total mechanical energy (Ei) at infinity is zero, as the potential energy is zero and the initial velocity is assumed to be zero. The final total mechanical energy (Ef) at the Earth's surface is the sum of kinetic energy (Kf=12mv2) and gravitational potential energy (Uf=mgR). By conservation of energy:

Ei=Ef0=12mv2mgR

This simplifies to:

12mv2=mgR
Step 2: Calculate Velocity (v)○ Expand

From the energy conservation equation, we can solve for v:

v2=2gRv=2gR

Given R=6400 km=6.4×106 m and g=9.8 m/s2. Substitute these values:

v=2×9.8 m/s2×6.4×106 m=125.44×106 m/s11.2×103 m/s=11.2 km/s
💡 Teacher's Secret Hint

Ensure all units are consistent (SI units) before calculation.

Step 3: Calculate Kinetic Energy (k)○ Expand

The kinetic energy k at the surface is k=12mv2. From Step 1, we established that 12mv2=mgR. Given m=1 kg:

k=mgR=1 kg×9.8 m/s2×6.4×106 m=62.72×106 J=6.27×107 J

Comparing the calculated values (v=11.2 km/s and k=6.27×107 J) with the given options, Option 1 matches.

💡 Teacher's Secret Hint

Remember that the kinetic energy gained is equal to the loss in potential energy when falling from infinity.

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