A satellite is moving around the earth with speed v in a circular orbit of radius r. IF the orbit radius is decreased by 1%, its speed will

Options

A. increase by 1%
B. increase by 0.5 %
C. decrease by 1%
D. decrease by 0.5 %

Explanation

To solve the problem step by step, we will use the relationship between the speed of a satellite in a circular orbit and the radius of that orbit.

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

### Step 2: Differentiate the equation
To find how the speed changes when the radius changes, we can take the logarithm of both sides:
\[ \log v = \frac{1}{2} \log G + \frac{1}{2} \log M - \frac{1}{2} \log r \]
Differentiating both sides with respect to \( r \):
\[ \frac{d(\log v)}{dt} = 0 + 0 - \frac{1}{2} \frac{d(\log r)}{dt} \]
This simplifies to:
\[ \frac{dv}{v} = -\frac{1}{2} \frac{dr}{r} \]

### Step 3: Express the percentage change
Now, we can express the percentage change in speed \( \delta v \) in terms of the percentage change in radius \( \delta r \):
\[ \frac{\delta v}{v} = -\frac{1}{2} \frac{\delta r}{r} \]
Multiplying both sides by 100 gives us:
\[ \delta v \% = -\frac{1}{2} \delta r \% \]

### Step 4: Substitute the given change in radius
We are given that the orbit radius is decreased by 1%. Therefore:
\[ \delta r \% = -1\% \]
Substituting this into our equation:
\[ \delta v \% = -\frac{1}{2} \times (-1\%) = 0.5\% \]

### Step 5: Conclusion
Thus, the speed of the satellite will increase by 0.5% when the orbit radius is decreased by 1%.

### Final Answer
The speed of the satellite increases by 0.5%.

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