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. `0.7%`
B. `1%`
C. `1.5%`
D. `3%`

Explanation

To solve the problem, we need to determine the difference in the orbital periods of two satellites orbiting the Earth at different radii.

### Step-by-Step Solution:

1. **Understanding the Formula for Orbital Period**:
The orbital 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.

2. **Identifying the Radii**:
Let the radius of the first satellite be \( R \) and the radius of the second satellite be \( r_2 = 1.01R \).

3. **Calculating the Periods**:
- For the first satellite (radius \( R \)):
\[ T_1 = 2\pi \sqrt{\frac{R^3}{GM}} \]
- For the second satellite (radius \( 1.01R \)):
\[ T_2 = 2\pi \sqrt{\frac{(1.01R)^3}{GM}} = 2\pi \sqrt{\frac{1.030301R^3}{GM}} = 2\pi \sqrt{\frac{R^3}{GM}} \cdot \sqrt{1.030301} \]

4. **Expressing \( T_2 \) in Terms of \( T_1 \)**:
We can express \( T_2 \) in terms of \( T_1 \):
\[ T_2 = T_1 \cdot \sqrt{1.030301} \]

5. **Using the Binomial Approximation**:
For small values of \( x \), \( \sqrt{1+x} \approx 1 + \frac{x}{2} \). Here, \( x = 0.030301 \):
\[ \sqrt{1.030301} \approx 1 + \frac{0.030301}{2} \approx 1 + 0.0151505 \]
Therefore,
\[ T_2 \approx T_1 \cdot (1 + 0.0151505) \approx T_1 + 0.0151505 T_1 \]

6. **Calculating the Increase in Period**:
The increase in the period \( \Delta T = T_2 - T_1 \):
\[ \Delta T \approx 0.0151505 T_1 \]
To express this as a percentage increase:
\[ \frac{\Delta T}{T_1} \times 100 \approx 1.51505\% \]

7. **Final Result**:
The period of the second satellite is larger than that of the first satellite by approximately \( 1.5\% \).