A satellite is revolving round the earth in an orbit of radius r with time period T. If the satellite is revolving round the earth in an orbit of radius `r + Deltar(Deltar lt lt r)` with time period `T + DeltaT(DeltaT lt lt T)` then.

Options

A. `(Delta T)/(T) = 3/2 (Delta r)/(r)`
B. `(Delta T)/T = 2/3 (Delta r)/(r)`
C. `(Delta T)/T = (Delta r)/(r)`
D. `(Delta T)/T = -(Delta r)/(r)`

Answer

To solve the problem, we will use Kepler's Third Law of planetary motion, which relates the time period of a satellite to the radius of its orbit.

### Step-by-Step Solution:

1. **Understanding Kepler's Third Law**:
According to Kepler's Third Law, the square of the time period (T) of a satellite is proportional to the cube of the radius (r) of its orbit around the Earth. Mathematically, this can be expressed as:
\[ T^2 \propto r^3 \]
or
\[ T^2 = k r^3 \]
where \( k \) is a constant.

2. **Differentiating the Equation**:
We will differentiate the equation \( T^2 = k r^3 \) with respect to time. Applying the product rule, we get:
\[ 2T \Delta T = 3k r^2 \Delta r \]

3. **Rearranging the Equation**:
We can rearrange the differentiated equation to express \( \Delta T \) in terms of \( \Delta r \):
\[ \Delta T = \frac{3k r^2}{2T} \Delta r \]

4. **Substituting for k**:
Since \( k = \frac{T^2}{r^3} \), we can substitute this back into the equation:
\[ \Delta T = \frac{3 \cdot \frac{T^2}{r^3} \cdot r^2}{2T} \Delta r \]
Simplifying this gives:
\[ \Delta T = \frac{3T}{2r} \Delta r \]

5. **Expressing the Final Relationship**:
Now, we can express the relationship between \( \Delta T \) and \( \Delta r \):
\[ \frac{\Delta T}{T} = \frac{3}{2} \frac{\Delta r}{r} \]

### Conclusion:
From the derived equation, we can conclude that the change in time period (\( \Delta T \)) is proportional to the change in radius (\( \Delta r \)) of the orbit, with a proportionality constant of \( \frac{3}{2} \).

### Correct Option:
Comparing this result with the options given in the question, we find that the correct option is **Option A**.