The orbital speed of a satellite revolving around a planet in a circular orbit is `v_(0)` . If its speed is increased by 10 % ,then

Options

A. It will escape from its orbit
B. It will start rotating in an elliptical
C. It will continue to move in the same orbit
D. It will move in a circular orbit of radius 20 % more than radius of intial orbit

Answer

To solve the problem, we need to analyze the effect of increasing the orbital speed of a satellite by 10%. Let's break it down step by step.

### Step 1: Understand the initial conditions
The initial orbital speed of the satellite is given as \( v_0 \). The orbital speed for a satellite in a circular orbit around a planet is determined by the formula:
\[ v_0 = \sqrt{\frac{GM}{R}} \]
where \( G \) is the gravitational constant, \( M \) is the mass of the planet, and \( R \) is the radius of the orbit.

### Step 2: Calculate the new speed
If the speed is increased by 10%, the new speed \( v \) can be calculated as:
\[ v = v_0 + 0.1 v_0 = 1.1 v_0 \]

### Step 3: Determine the relationship between the new speed and escape velocity
The escape velocity \( v_e \) from the planet is given by:
\[ v_e = \sqrt{2} v_0 \]
To find out if the new speed \( v \) is less than, equal to, or greater than the escape velocity, we need to compare \( 1.1 v_0 \) with \( v_e \):
\[ v_e = \sqrt{2} v_0 \approx 1.414 v_0 \]
Since \( 1.1 v_0 < 1.414 v_0 \), we conclude that:
\[ v < v_e \]

### Step 4: Determine the type of orbit
Since the new speed \( v \) is greater than the original orbital speed \( v_0 \) but less than the escape velocity \( v_e \), the satellite will no longer be in a circular orbit. Instead, it will enter an elliptical orbit.

### Conclusion
Thus, the satellite will start rotating in an elliptical orbit.

### Final Answer
The correct conclusion is: **It will start rotating in an elliptical orbit.**

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