An object is projected horizontally with speed `1/2 sqrt((GM)/R)` , from a point at height 3 R [ where R is radius and M is mass of earth , then object will ]

Options

A. Fall back on surface of earth by following parabolic path
B. Fall back on surface of earth by following hyperbolic path
C. Start rotating around earth in a circular orbit
D. Escape from gravitational field of earth

Answer

To solve the problem, we need to analyze the motion of an object projected horizontally from a height of \(3R\) with a speed of \(\frac{1}{2} \sqrt{\frac{GM}{R}}\).

### Step-by-Step Solution:

1. **Identify the given parameters**:
- Height \(h = 3R\)
- Speed \(v = \frac{1}{2} \sqrt{\frac{GM}{R}}\)
- Gravitational constant \(G\)
- Mass of the Earth \(M\)
- Radius of the Earth \(R\)

2. **Calculate the escape velocity from the height**:
The escape velocity \(v_e\) from a height \(h\) is given by the formula:
\[ v_e = \sqrt{\frac{2GM}{R + h}} \]
Substituting \(h = 3R\):
\[ v_e = \sqrt{\frac{2GM}{R + 3R}} = \sqrt{\frac{2GM}{4R}} = \sqrt{\frac{GM}{2R}} = \frac{1}{\sqrt{2}} \sqrt{\frac{GM}{R}} \]

3. **Calculate the orbital velocity from the height**:
The orbital velocity \(v_0\) at a height \(h\) is given by:
\[ v_0 = \sqrt{\frac{GM}{R + h}} \]
Again substituting \(h = 3R\):
\[ v_0 = \sqrt{\frac{GM}{R + 3R}} = \sqrt{\frac{GM}{4R}} = \frac{1}{2} \sqrt{\frac{GM}{R}} \]

4. **Compare the given speed with the calculated velocities**:
We have:
- Given speed \(v = \frac{1}{2} \sqrt{\frac{GM}{R}}\)
- Orbital velocity \(v_0 = \frac{1}{2} \sqrt{\frac{GM}{R}}\)

Since the given speed \(v\) is equal to the orbital velocity \(v_0\), the object will enter a circular orbit around the Earth.

5. **Conclusion**:
Therefore, the object will start rotating around the Earth in a circular orbit.

### Final Answer:
The object will start rotating around the Earth in a circular orbit.