A ball is thrown vertically upwards with a velocity equal to half the escape velocity from the surface of the earth. The ball rises to a height h above the surface of the earth. If the radius of the earth is R, then the ratio `h/R` is

Options

A. `1/2 `
B. `1/3`
C. 2
D. 3

Explanation

To solve the problem, we need to find the height \( h \) that a ball reaches when thrown vertically upwards with a velocity equal to half the escape velocity from the surface of the Earth. We will also express this height in terms of the Earth's radius \( R \).

### Step-by-Step Solution:

1. **Understand Escape Velocity**:
The escape velocity \( v_e \) from the surface of the Earth is given by the formula:
\[ v_e = \sqrt{\frac{2GM}{R}} \]
where \( G \) is the gravitational constant and \( M \) is the mass of the Earth.

2. **Determine the Initial Velocity**:
The ball is thrown with a velocity \( v \) equal to half the escape velocity:
\[ v = \frac{1}{2} v_e = \frac{1}{2} \sqrt{\frac{2GM}{R}} = \sqrt{\frac{GM}{2R}} \]

3. **Apply Conservation of Energy**:
We will use the principle of conservation of mechanical energy. The total mechanical energy at the surface (point A) and at the maximum height (point B) must be equal.

- At point A (surface of the Earth):
- Kinetic Energy (KE) = \( \frac{1}{2} mv^2 \)
- Potential Energy (PE) = \( -\frac{GMm}{R} \)

- At point B (maximum height \( h \)):
- Kinetic Energy (KE) = 0 (at maximum height, velocity is 0)
- Potential Energy (PE) = \( -\frac{GMm}{R+h} \)

Setting the total energy at point A equal to that at point B:
\[ \frac{1}{2} mv^2 - \frac{GMm}{R} = -\frac{GMm}{R+h} \]

4. **Substituting the Initial Velocity**:
Substitute \( v^2 \) into the energy equation:
\[ \frac{1}{2} m \left(\frac{GM}{2R}\right) - \frac{GMm}{R} = -\frac{GMm}{R+h} \]
Simplifying the left side:
\[ \frac{GMm}{4R} - \frac{GMm}{R} = -\frac{GMm}{R+h} \]
\[ \frac{GMm}{4R} - \frac{4GMm}{4R} = -\frac{GMm}{R+h} \]
\[ -\frac{3GMm}{4R} = -\frac{GMm}{R+h} \]

5. **Canceling Common Terms**:
We can cancel \( GMm \) from both sides (assuming \( m \neq 0 \)):
\[ \frac{3}{4R} = \frac{1}{R+h} \]

6. **Cross-Multiplying**:
Cross-multiply to solve for \( h \):
\[ 3(R+h) = 4R \]
\[ 3R + 3h = 4R \]
\[ 3h = 4R - 3R \]
\[ 3h = R \]
\[ h = \frac{R}{3} \]

7. **Finding the Ratio \( \frac{h}{R} \)**:
Now, we find the ratio \( \frac{h}{R} \):
\[ \frac{h}{R} = \frac{R/3}{R} = \frac{1}{3} \]

### Final Answer:
The ratio \( \frac{h}{R} \) is \( \frac{1}{3} \).

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