An object is projected from the earth's surface with escape velocity at `30^(@)` with horizontal. What is the angle made by the velocity with horizontal when the object reaches a height 2R from the earth's surface ? R is the radius of the earth. Horizontal can be considered as a line parallel to the tangent at the earth's surface just below the object .

Options

A. `30^(@)`
B. `45^(@)`
C. `60^(@)`
D. `15^(@)`

Answer

To solve the problem, let's break it down step by step:

### Step 1: Understand the initial conditions
The object is projected from the surface of the Earth with an escape velocity at an angle of \(30^\circ\) with the horizontal. The escape velocity \(v_e\) from the surface of the Earth is given by:

\[ v_e = \sqrt{2gR} \]

where \(g\) is the acceleration due to gravity and \(R\) is the radius of the Earth.

### Step 2: Resolve the initial velocity into components
The initial velocity \(v_e\) can be resolved into horizontal and vertical components:

- Horizontal component \(v_{e_x} = v_e \cos(30^\circ) = v_e \cdot \frac{\sqrt{3}}{2}\)
- Vertical component \(v_{e_y} = v_e \sin(30^\circ) = v_e \cdot \frac{1}{2}\)

### Step 3: Determine the height reached
The object reaches a height of \(2R\) above the Earth's surface, which means the total height from the center of the Earth is:

\[ h = R + 2R = 3R \]

### Step 4: Use conservation of energy
At the maximum height, the total mechanical energy (kinetic + potential) remains constant. The potential energy at height \(h\) is given by:

\[ PE = -\frac{GMm}{h} = -\frac{GMm}{3R} \]

At the surface, the potential energy is:

\[ PE_{initial} = -\frac{GMm}{R} \]

The initial kinetic energy is:

\[ KE_{initial} = \frac{1}{2}mv_e^2 = \frac{1}{2}m(2gR) = mgR \]

At height \(h\), the kinetic energy is:

\[ KE_{final} = \frac{1}{2}mv^2 \]

Using conservation of energy:

\[ KE_{initial} + PE_{initial} = KE_{final} + PE_{final} \]

Substituting the values:

\[ mgR - \frac{GMm}{R} = \frac{1}{2}mv^2 - \frac{GMm}{3R} \]

### Step 5: Solve for the final velocity \(v\)
Rearranging gives:

\[ mgR - \frac{GMm}{R} + \frac{GMm}{3R} = \frac{1}{2}mv^2 \]

This simplifies to:

\[ mgR - \frac{2GMm}{3R} = \frac{1}{2}mv^2 \]

Now, substituting \(g = \frac{GM}{R^2}\):

\[ m \left( \frac{GM}{R^2} \right) R - \frac{2GMm}{3R} = \frac{1}{2}mv^2 \]

### Step 6: Find the angle of velocity with horizontal at height \(2R\)
At height \(2R\), the vertical component of velocity will change due to gravity. The vertical component of velocity can be calculated using the kinematic equations, taking into account that the object is under the influence of gravity.

The time taken to reach height \(2R\) can be calculated, and then we can find the vertical component of velocity at that height. The angle \( \theta \) with the horizontal can then be found using:

\[ \tan(\theta) = \frac{v_{y}}{v_{x}} \]

### Final Step: Calculate the angle
Using the components of velocity at height \(2R\), we can find the angle \( \theta \) made by the velocity with the horizontal.