1. Escape velocity from Earth’s surface

For a body projected from Earth’s surface,

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

Given that the projection speed is one-third of escape velocity,

$u = \frac{v_e}{3}$

So,

$u^2 = \frac{v_e^2}{9} = \frac{1}{9}\cdot \frac{2GM}{R} = \frac{2GM}{9R}$

2. Use conservation of mechanical energy

Let the maximum height attained be $h$ above Earth’s surface. Then the distance from Earth’s center at the highest point is $R+h$.

At projection point:

$E_i = \frac{1}{2}mu^2 - \frac{GMm}{R}$

At maximum height, velocity becomes zero:

$E_f = -\frac{GMm}{R+h}$

By conservation of energy,

$\frac{1}{2}mu^2 - \frac{GMm}{R} = -\frac{GMm}{R+h}$

Substitute $u^2 = \dfrac{2GM}{9R}$:

$\frac{1}{2}m\cdot \frac{2GM}{9R} - \frac{GMm}{R} = -\frac{GMm}{R+h}$

$\frac{GMm}{9R} - \frac{GMm}{R} = -\frac{GMm}{R+h}$

Factor out $\frac{GMm}{R}$:

$\left(\frac{1}{9} - 1\right)\frac{GMm}{R} = -\frac{GMm}{R+h}$

$-\frac{8}{9}\frac{GMm}{R} = -\frac{GMm}{R+h}$

Cancel common factors:

$\frac{8}{9R} = \frac{1}{R+h}$

Thus,

$R+h = \frac{9R}{8}$

$h = \frac{9R}{8} - R = \frac{R}{8}$

3. Substitute Earth’s radius

Given:

$R = 6400\text{ km}$

So,

$h = \frac{6400}{8} = 800\text{ km}$

4. Option check

• A: $800$ km ✅
• B: $1600$ km ❌
• C: $2133$ km ❌
• D: $4800$ km ❌

Therefore, the correct option is A.