A satellite is launched into a circular orbit of radius R around the earth. A second satellite is launched into an orbit of radius (1.02)R. The period of the second satellite is larger than the first one by approximately

Options

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

Answer

To solve the problem of finding how much larger the period of the second satellite is compared to the first satellite, we will follow these steps:

### 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:
- \( T \) is the period of the satellite,
- \( r \) is the radius of the orbit,
- \( G \) is the gravitational constant,
- \( M \) is the mass of the Earth.

### Step 2: Calculate the period of the first satellite
For the first satellite with radius \( R \):
\[ T_1 = 2\pi \sqrt{\frac{R^3}{GM}} \]

### Step 3: Calculate the period of the second satellite
For the second satellite with radius \( 1.02R \):
\[ T_2 = 2\pi \sqrt{\frac{(1.02R)^3}{GM}} = 2\pi \sqrt{\frac{1.02^3 R^3}{GM}} \]

### Step 4: Simplify the expression for \( T_2 \)
Using the property of exponents:
\[ T_2 = 2\pi \sqrt{1.02^3} \cdot \sqrt{\frac{R^3}{GM}} = \sqrt{1.02^3} \cdot T_1 \]

### Step 5: Calculate \( 1.02^3 \)
Calculating \( 1.02^3 \):
\[ 1.02^3 \approx 1.061208 \]
Thus,
\[ T_2 \approx \sqrt{1.061208} \cdot T_1 \approx 1.0306 \cdot T_1 \]

### Step 6: Find the difference in periods
The difference in periods \( T_2 - T_1 \) can be expressed as:
\[ T_2 - T_1 \approx (1.0306 - 1) \cdot T_1 \approx 0.0306 \cdot T_1 \]

### Step 7: Calculate the percentage increase
To find the percentage increase:
\[ \text{Percentage Increase} = \frac{T_2 - T_1}{T_1} \times 100 = 0.0306 \times 100 \approx 3.06\% \]

### Step 8: Final approximation
Rounding this value gives approximately:
\[ \text{Percentage Increase} \approx 3.0\% \]

### Conclusion
The period of the second satellite is larger than the first one by approximately **3.0%**.