A satellite is moving around the earth's with speed `v` in a circular orbit of radius `r`. If the orbit radius is decreases by `1%` , its speed will

Options

A. Increases by `1%`
B. Increases by `0.5%`
C. decreases by `1%`
D. decreases by `0.5%`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the relationship between orbital speed and radius
The orbital speed \( v \) of a satellite in a circular orbit is given by the formula:
\[ v = \sqrt{\frac{GM}{r}} \]
where \( G \) is the gravitational constant, \( M \) is the mass of the Earth, and \( r \) is the radius of the orbit.

### Step 2: Express the speed in terms of radius
From the formula, we can see that the speed \( v \) is inversely proportional to the square root of the radius \( r \):
\[ v \propto \frac{1}{\sqrt{r}} \]

### Step 3: Determine the effect of a 1% decrease in radius
If the radius \( r \) is decreased by 1%, we can express this mathematically as:
\[ r' = r - 0.01r = 0.99r \]

### Step 4: Calculate the new speed
Substituting \( r' \) into the speed formula gives:
\[ v' = \sqrt{\frac{GM}{r'}} = \sqrt{\frac{GM}{0.99r}} = \sqrt{\frac{GM}{r}} \cdot \sqrt{\frac{1}{0.99}} = v \cdot \sqrt{\frac{1}{0.99}} \]

### Step 5: Simplify the expression for new speed
Now, we can simplify \( \sqrt{\frac{1}{0.99}} \):
\[ \sqrt{\frac{1}{0.99}} \approx 1 + \frac{1}{2}(0.01) = 1 + 0.005 = 1.005 \]
This approximation comes from the binomial expansion for small values.

### Step 6: Calculate the percentage change in speed
The new speed \( v' \) can be expressed as:
\[ v' \approx v \cdot 1.005 \]
This indicates that the speed increases by approximately 0.5%:
\[ \text{Percentage increase in speed} = (v' - v) / v \times 100\% \approx 0.5\% \]

### Conclusion
Thus, if the radius of the orbit decreases by 1%, the speed of the satellite will increase by approximately 0.5%.

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