Step 1: Understanding the Concept:
The initial velocity (\( v = 8000 \) m/s) is comparable to the escape velocity, so we cannot assume \( g \) is constant. We must use the conservation of mechanical energy (kinetic + gravitational potential energy).
Step 2: Key Formula or Approach:
Energy Conservation:
\[ \frac{1}{2} m v^2 - \frac{GMm}{R} = 0 - \frac{GMm}{R+h} \]
where \( R \) is Earth's radius, \( M \) is Earth's mass, \( h \) is maximum height.
Also, use \( GM = gR^2 \) to eliminate \( GM \).
Step 3: Detailed Explanation:
Given:
\( v = 8000 \, \text{m/s} = 8 \times 10^3 \, \text{m/s} \)
\( R = 6400 \, \text{km} = 6.4 \times 10^6 \, \text{m} \)
\( g = 10 \, \text{m/s}^2 \)
From energy conservation:
\[ \frac{v^2}{2} = \frac{GM}{R} - \frac{GM}{R+h} = GM \left( \frac{1}{R} - \frac{1}{R+h} \right) \]
Substitute \( GM = gR^2 \):
\[ \frac{v^2}{2} = gR^2 \left( \frac{R+h - R}{R(R+h)} \right) = gR^2 \left( \frac{h}{R(R+h)} \right) = \frac{gRh}{R+h} \]
Rearrange to solve for \( h \):
\[ \frac{v^2}{2} (R+h) = gRh \]
\[ v^2 R + v^2 h = 2gRh \]
\[ v^2 R = h(2gR - v^2) \]
\[ h = \frac{v^2 R}{2gR - v^2} \]
Or divide numerator and denominator by \( R \):
\[ h = \frac{R}{\frac{2gR}{v^2} - 1} \]
Now substitute numerical values:
\( v^2 = (8000)^2 = 64 \times 10^6 \).
\( 2gR = 2 \times 10 \times 6.4 \times 10^6 = 128 \times 10^6 \).
\[ h = \frac{64 \times 10^6 \times R}{128 \times 10^6 - 64 \times 10^6} \]
\[ h = \frac{64 \times 10^6 \times R}{64 \times 10^6} \]
\[ h = R \]
Since \( R = 6400 \) km, \( h = 6400 \) km.
Step 4: Final Answer:
The maximum height reached is 6400 km.