Step 1: Understanding the Concept:
When an object is thrown upwards, it exchanges its initial kinetic energy for gravitational potential energy as it rises. At its maximum height, its velocity becomes zero.
We use the principle of conservation of mechanical energy to solve this.
Step 2: Key Formula or Approach:
1. Escape Velocity ($v_e$) from Earth's surface: $v_e = \sqrt{\frac{2GM}{R}}$
2. Conservation of Energy: $(K.E. + P.E.)_{\text{surface}} = (K.E. + P.E.)_{\text{max\_height}}$
\[ \frac{1}{2}mv^2 - \frac{GMm}{R} = 0 - \frac{GMm}{r} \]
where $v$ is projection velocity, $R$ is Earth's radius, and $r = R+h$ is the distance from Earth's center at maximum height $h$.
Step 3: Detailed Explanation:
Given projection velocity $v = \frac{1}{3}v_e = \frac{1}{3}\sqrt{\frac{2GM}{R}}$.
Substitute this velocity into the energy conservation equation:
\[ \frac{1}{2}m \left( \frac{1}{3} \sqrt{\frac{2GM}{R}} \right)^2 - \frac{GMm}{R} = - \frac{GMm}{R+h} \]
Square the velocity term:
\[ \frac{1}{2}m \left( \frac{1}{9} \frac{2GM}{R} \right) - \frac{GMm}{R} = - \frac{GMm}{R+h} \]
Simplify the kinetic energy term:
\[ \frac{GMm}{9R} - \frac{GMm}{R} = - \frac{GMm}{R+h} \]
Divide the entire equation by $GMm$ to simplify:
\[ \frac{1}{9R} - \frac{1}{R} = - \frac{1}{R+h} \]
Find a common denominator for the left side:
\[ \frac{1 - 9}{9R} = - \frac{1}{R+h} \]
\[ \frac{-8}{9R} = \frac{-1}{R+h} \]
Multiply by $-1$ and take the reciprocal of both sides:
\[ \frac{9R}{8} = R + h \]
Solve for the height $h$:
\[ h = \frac{9R}{8} - R \]
\[ h = \frac{9R - 8R}{8} = \frac{R}{8} \]
Step 4: Final Answer:
The maximum height reached by the body is $\frac{\text{R}}{8}$.