A-Level Newton's gravitation, gravitational field strength, potential and energy, escape velocity and orbits: worked solution

Question

A lone uniform spherical planet has surface gravitational potential φ. A particle of mass m is fired straight up from the surface with exactly enough kinetic energy to just reach infinity, with no extra work supplied. Show that the launch speed satisfies v = √(−2φ).

Principle and definitions first, because this is a "show that" and the marks are for the reasoning, not the algebra. Gravitational potential φ at a point is defined as the work done per unit mass in bringing a small test mass from infinity to that point. Infinity is taken as the zero of potential, so for an attractive field φ is always negative. It follows that the gravitational potential energy of a mass m at a place where the potential is φ is E_p = m φ and at infinity, where φ = 0, the potential energy is zero. Step 1 - interpret "just reaches infinity with no extra work supplied". The particle is given all of its energy at launch; nothing pushes it afterwards, and there is no atmosphere, so no energy is lost. "Just reaches infinity" means it arrives there with its speed reduced to zero, so its kinetic energy at infinity is zero. Step 2 - write down the total energy at each of the two places. At the surface, where the potential is φ and the speed is v: kinetic energy = ½ m v² potential energy = m φ At infinity: kinetic energy = 0 (it only just gets there) potential energy = 0 (infinity is the zero of potential) Step 3 - apply conservation of energy: with no resistive forces and no propulsion, the total energy is the same at both places. ½ m v² + m φ = 0 + 0 Step 4 - rearrange. Subtract m φ from both sides: ½ m v² = − m φ The mass m appears in every term, so divide throughout by m (the result does not depend on the mass of the particle): ½ v² = − φ Multiply both sides by 2: v² = − 2 φ Take the positive square root, since v is a speed: v = √(− 2 φ) Note this is a real number: φ is negative, so − 2 φ is positive, as required. Shown as needed.