A body is projected from the earth's surface with the intention of letting it escape from the earth's gravitational field. What is the minimum escape veloci...

A body is projected from the earth's surface with the intention of letting it escape from the earth's gravitational field. What is the minimum escape velocity of the body? [Earth's radius = 6.4 x 103km, g = 10ms-2]

Answer Details

The minimum escape velocity of a body is the velocity required to escape from the gravitational field of a planet, which means that the body will move away from the planet and never return. The formula for escape velocity is given by: v = √(2GM/r) where G is the universal gravitational constant, M is the mass of the planet, and r is the distance from the center of the planet to the point where the body is located. In this case, the planet is Earth, with a radius of 6.4 x 10^3 km, and g = 10 m/s^2. The mass of Earth is approximately 5.97 x 10^24 kg. The distance from the center of the planet to the surface is equal to the radius of the planet. Substituting these values into the formula, we get: v = √(2 x 6.67 x 10^-11 x 5.97 x 10^24 / 6.4 x 10^3) v ≈ 11.2 km/s Therefore, the minimum escape velocity of the body is approximately 11 km/s, which is closest to option D, 11 kms^-1.