The relative uncertainty in the period of a satellite orbiting around the earth is `10^(-2)` . If the relative uncertainty in the radius of the orbit is negligible, the relative uncertainty in the mass of the earth is :

Options

A. `3xx10^(-2)`
B. `6xx10^(-2)`
C. `10^(-2)`
D. `2xx10^(-2)`

Answer

To solve the problem, we need to find the relative uncertainty in the mass of the Earth based on the given relative uncertainty in the period of a satellite orbiting the Earth.

### Step-by-step Solution:

1. **Understanding the Formula for the Period of a Satellite**:
The period \( T \) of a satellite in 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. **Squaring the Period Equation**:
To simplify our calculations, we can square both sides of the equation:
\[ T^2 = \frac{4\pi^2 R^3}{GM} \]

3. **Taking the Logarithm**:
We take the natural logarithm of both sides to facilitate the calculation of uncertainties:
\[ \ln(T^2) = \ln\left(\frac{4\pi^2 R^3}{GM}\right) \]
This can be rewritten using logarithmic properties:
\[ 2\ln(T) = \ln(4\pi^2) + 3\ln(R) - \ln(G) - \ln(M) \]

4. **Applying Uncertainty Relations**:
The relative uncertainty in a product or quotient can be expressed as the sum of the relative uncertainties:
\[ \frac{\Delta T}{T} = \frac{1}{2}\frac{\Delta M}{M} + 3\frac{\Delta R}{R} + \frac{\Delta G}{G} \]
Given that the relative uncertainty in the radius \( R \) is negligible (\(\Delta R/R \approx 0\)) and \( G \) is a constant with no uncertainty, we can simplify this to:
\[ \frac{\Delta T}{T} = \frac{1}{2}\frac{\Delta M}{M} \]

5. **Substituting the Given Values**:
We know from the problem statement that the relative uncertainty in the period \( \frac{\Delta T}{T} = 10^{-2} \):
\[ 10^{-2} = \frac{1}{2}\frac{\Delta M}{M} \]

6. **Solving for the Relative Uncertainty in Mass**:
Rearranging the equation gives:
\[ \frac{\Delta M}{M} = 2 \times 10^{-2} = 2 \times 10^{-2} \]

Thus, the relative uncertainty in the mass of the Earth is:
\[ \frac{\Delta M}{M} = 2 \times 10^{-2} \]

### Final Answer:
The relative uncertainty in the mass of the Earth is \( 2 \times 10^{-2} \).