18. A body is at a height equal to the radius of Earth from the surface of Earth. With what velocity it is to be thrown so that it goes out of the gravitational field of the Earth. Use the mass and the radius of Earth as given in the questions above.

We start by applying energy conservation. The body is initially at a height equal to Earth’s radius (R) above the surface, so its distance from Earth’s center is r = R + R = 2R. The gravitational potential energy at a distance r from Earth’s center is given by   U = –GMm/r where   G is the gravitational constant,   M is the mass of Earth, and   m is the mass of the body. At the starting point (r = 2R), the potential energy is   U₁ = –GMm/(2R). If we throw the body upward with speed v, its kinetic energy is   K₁ = (1/2) m v². For the body to just escape Earth’s gravitational field, its total energy at infinity should be zero. At infinity, the potential energy is zero and we want the kinetic energy to be zero as well (i.e., it barely reaches infinity). Thus, the energy conservation equation is   K₁ + U₁ = 0,   (1/2) m v² – GMm/(2R) = 0. Cancel m (assuming m ≠ 0) and multiply both sides by 2:   v² = GM/R. Taking the square root gives   v = √(GM/R). We can also express GM/R in terms of the gravitational acceleration g at Earth’s surface. Since   g = GM/R²  ⇒  GM = gR², it follows that   v = √(gR²/R) = √(gR). If you use the typical values g ≈ 9.8 m/s² and R ≈ 6.37 × 10⁶ m, then   v ≈ √(9.8 × 6.37 × 10⁶) ≈ √(6.24 × 10⁷) ≈ 7.9 × 10³ m/s. Thus, the required initial velocity is   v = √(GM/R)  or  v = √(gR), which is approximately 7.9 km/s.