Question

If a body of mass \(100\) kg is thrown vertically upwards from the surface of the earth with a velocity equal to \[ \frac23 \] times its escape velocity, then the maximum height reached by the body is \[ (\text{Radius of the earth}=6400\text{ km}) \]

• A. \(3200\) km
• B. \(5120\) km
• C. \(8000\) km
• D. \(12800\) km

Hint:

Use conservation of mechanical energy: \[ \boxed{ \frac12mu^2-\frac{GMm}{R} = -\frac{GMm}{R+h}. } \] Also, \[ \boxed{ v_e=\sqrt{\frac{2GM}{R}}. } \]

Answer 1

The Correct Option is B

Step 1: Apply conservation of mechanical energy. Let the radius of the earth be \[ R. \] Escape velocity is \[ v_e=\sqrt{\frac{2GM}{R}}. \] Given, \[ u=\frac23v_e. \] Hence, \[ u^2 = \frac49v_e^2 = \frac49\left(\frac{2GM}{R}\right) = \frac{8GM}{9R}. \] Initially, \[ E_i = \frac12mu^2-\frac{GMm}{R}. \] At the maximum height, \[ E_f = -\frac{GMm}{R+h}. \] Since energy is conserved, \[ \frac12mu^2-\frac{GMm}{R} = -\frac{GMm}{R+h}. \]

Step 2: Substitute the given velocity. Substituting \[ u^2=\frac{8GM}{9R}, \] we get \[ \frac12m\cdot\frac{8GM}{9R} -\frac{GMm}{R} = -\frac{GMm}{R+h}. \] Therefore, \[ \frac{4}{9}-1 = -\frac{R}{R+h}, \] or \[ -\frac59 = -\frac{R}{R+h}. \] Hence, \[ R+h=\frac95R. \] Thus, \[ h=\frac45R. \]

Step 3: Calculate the maximum height. Since \[ R=6400\text{ km}, \] \[ h = \frac45\times6400 = 5120\text{ km}. \] Hence, \[ \boxed{5120\text{ km}}. \] Thus, \[ \boxed{(B)} \] is the correct answer.