Q.The escape speed of a projectile on the earth's surface is $11.2\text{ km s}^{-1}$. A body is projected out with thrice this speed. What is the speed of the body far away from the earth? Ignore the presence of the sun and other planets.

The key idea here is the Conservation of Mechanical Energy. Since only the conservative gravitational force acts on the body, its total mechanical energy (kinetic + potential) remains constant throughout its motion.

1. The total mechanical energy $E$ of a body of mass $m$ at a distance $r$ from the center of Earth (mass $M$, radius $R$) with speed $v$ is $E = \frac{1}{2}mv^2 - \frac{GMm}{r}$.
2. At the Earth's surface ($r=R$), the initial speed is $v_i = 3v_e$, where $v_e$ is the escape speed. The initial energy is $E_i = \frac{1}{2}m(3v_e)^2 - \frac{GMm}{R}$.
3. Far away from Earth (at infinity, $r \to \infty$), the gravitational potential energy approaches zero. Let the final speed be $v_f$. The final energy is $E_f = \frac{1}{2}mv_f^2 + 0$.
4. By conservation of energy, $E_i = E_f$: $\frac{1}{2}m(3v_e)^2 - \frac{GMm}{R} = \frac{1}{2}mv_f^2$ We know that for escape velocity, $\frac{1}{2}mv_e^2 = \frac{GMm}{R}$. Substituting this into the equation: …

When a body moves under the influence of only conservative forces, such as gravity, its total mechanical energy remains constant. This is the principle of conservation of mechanical energy. In this problem, we are considering a body projected from Earth's surface, and the only significant force acting on it is Earth's gravity. We are told to ignore the Sun and other planets, simplifying the system to just the Earth and the projectile.

"Far away from the earth" implies a distance so large that the gravitational potential energy due to Earth becomes negligible, effectively zero. This is often referred to as "at infinity."

Let's break down the solution:

1. Understanding Escape Speed

   The escape speed ($v_e$) is the minimum initial speed required for a projectile to completely escape Earth's gravitational pull and never return. This means that when the body reaches an infinite distance from Earth, its kinetic energy will be zero, and its gravitational potential energy will also be zero.

   Let $M$ be the mass of Earth, $R$ be the radius of Earth, and $m$ be the mass of the projectile.

   At Earth's surface, the initial kinetic energy is $K_i = \frac{1}{2}mv_e^2$, and the initial gravitational potential energy is $U_i = -\frac{GMm}{R}$.

   At infinity, the final kinetic energy is $K_f = 0$, and the final gravitational potential energy is $U_f = 0$.

   The principle of conservation of mechanical energy states:

   $K_i + U_i = K_f + U_f$

   Applying this for escape speed:

   $\frac{1}{2}mv_e^2 - \frac{GMm}{R} = 0 + 0$

   $\frac{1}{2}mv_e^2 = \frac{GMm}{R}$

   $v_e^2 = \frac{2GM}{R}$

   $v_e = \sqrt{\frac{2GM}{R}}$

   From this equation, we can see that $\frac{GM}{R} = \frac{1}{2}v_e^2$. This relationship will be very useful.

2. Initial Conditions for the Projected Body

   The problem states that the body is projected with thrice the escape speed.

   So, the initial speed of the body is $v_{initial} = 3v_e$.

   The initial kinetic energy is $K_{initial} = \frac{1}{2}m(3v_e)^2 = \frac{1}{2}m(9v_e^2) = \frac{9}{2}mv_e^2$.

   The initial gravitational potential energy (at Earth's surface) is $U_{initial} = -\frac{GMm}{R}$.

3. Final Conditions for the Projected Body

   We want to find the speed of the body "far away from the earth." As discussed, this means at an infinite distance.

   At infinity, the final gravitational potential energy is $U_{final} = 0$.

   Let the speed of the body far away from Earth be $v_f$.

   The final kinetic energy is $K_{final} = \frac{1}{2}mv_f^2$.

4. Applying Conservation of Mechanical Energy for the Projected Body

   Using the conservation of mechanical energy principle:

   $K_{initial} + U_{initial} = K_{final} + U_{final}$

   $\frac{9}{2}mv_e^2 - \frac{GMm}{R} = \frac{1}{2}mv_f^2 + 0$

5. Solving for the Final Speed ($v_f$) …

Step 1: Escape speed satisfies $\dfrac12mv_e^2 = \dfrac{GMm}{R}$, i.e. $GM/R = v_e^2/2$ — the energy per unit mass needed to just reach infinity with zero speed.

Step 2: For launch speed $v_0 = 3v_e$, energy conservation from the surface to "far away" ($r\to\infty$, $U\to0$) gives $\dfrac12v_0^2 - \dfrac{GM}{R} = \dfrac12v_f^2$. …