A satellite is launched into a circular orbit of radius R around the earth. A second satellite is launched into an orbit of radius `(1.01)` R. The period of the second satellite is larger than the first one by approximately

Options

A. A. `0.7%`
B. B. `1.0%`
C. C. `1.5%`
D. D. `3.0%`

Answer

To solve the problem, we need to find the difference in the periods of two satellites orbiting the Earth at different radii. Here's the step-by-step solution:

### Step 1: Understand the formula for the period of a satellite
The period \( T \) of a satellite in a circular orbit is given by the formula:
\[ T = 2\pi \sqrt{\frac{R^3}{GM}} \]
where \( R \) is the radius of the orbit, \( G \) is the gravitational constant, and \( M \) is the mass of the Earth.

### Step 2: Calculate the period of the first satellite
Let the radius of the first satellite be \( R \). The period \( T_1 \) of the first satellite is:
\[ T_1 = 2\pi \sqrt{\frac{R^3}{GM}} \]

### Step 3: Calculate the period of the second satellite
The radius of the second satellite is \( 1.01R \). The period \( T_2 \) of the second satellite is:
\[ T_2 = 2\pi \sqrt{\frac{(1.01R)^3}{GM}} = 2\pi \sqrt{\frac{1.01^3 R^3}{GM}} \]

### Step 4: Express \( T_2 \) in terms of \( T_1 \)
We can express \( T_2 \) in terms of \( T_1 \):
\[ T_2 = 2\pi \sqrt{\frac{1.01^3 R^3}{GM}} = 2\pi \sqrt{1.01^3} \cdot \sqrt{\frac{R^3}{GM}} = \sqrt{1.01^3} \cdot T_1 \]

### Step 5: Calculate the difference in periods
Now, we need to find \( T_2 - T_1 \):
\[ T_2 - T_1 = \sqrt{1.01^3} \cdot T_1 - T_1 = ( \sqrt{1.01^3} - 1 ) T_1 \]

### Step 6: Calculate the percentage increase in period
The percentage increase in the period is given by:
\[ \text{Percentage Increase} = \frac{T_2 - T_1}{T_1} \times 100 = ( \sqrt{1.01^3} - 1 ) \times 100 \]

### Step 7: Calculate \( \sqrt{1.01^3} \)
Using the approximation \( \sqrt{1+x} \approx 1 + \frac{x}{2} \) for small \( x \):
\[ \sqrt{1.01^3} = (1.01)^{1.5} \approx 1 + \frac{0.01 \cdot 3}{2} = 1 + 0.015 = 1.015 \]

### Step 8: Substitute back to find the percentage
Now substituting back:
\[ \text{Percentage Increase} \approx (1.015 - 1) \times 100 = 0.015 \times 100 = 1.5\% \]

### Final Answer
Thus, the period of the second satellite is larger than the first one by approximately **1.5%**.

---