The time period 'T' of the artificial satellite of earth depends on the average density `rho` of the earth as

Options

A. `T prop rho`
B. `T prop sqrt(rho)`
C. `T prop (1)/(sqrt(rho))`
D. `T prop (1)/(rho)`

Answer

To solve the problem of how the time period \( T \) of an artificial satellite depends on the average density \( \rho \) of the Earth, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Time Period Formula**:
The time 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. **Expressing Mass in Terms of Density**:
The mass \( M \) of the Earth can be expressed in terms of its average density \( \rho \) and its volume \( V \). The volume \( V \) of the Earth (assuming it is a sphere) is given by:
\[ V = \frac{4}{3} \pi R^3 \]
where \( R \) is the radius of the Earth. Therefore, the mass can be expressed as:
\[ M = \rho V = \rho \left(\frac{4}{3} \pi R^3\right) \]

3. **Substituting Mass into the Time Period Formula**:
Now, substituting the expression for \( M \) into the time period formula:
\[ T = 2\pi \sqrt{\frac{r^3}{G \left(\rho \frac{4}{3} \pi R^3\right)}} \]
Simplifying this, we get:
\[ T = 2\pi \sqrt{\frac{3r^3}{4\pi G \rho R^3}} \]

4. **Analyzing the Dependence on Density**:
From the simplified formula, we can see that \( T \) is proportional to \( \sqrt{\frac{1}{\rho}} \):
\[ T \propto \frac{1}{\sqrt{\rho}} \]
This indicates that as the average density \( \rho \) of the Earth increases, the time period \( T \) decreases.

5. **Conclusion**:
Therefore, the time period \( T \) of the artificial satellite depends inversely on the square root of the average density \( \rho \) of the Earth:
\[ T \propto \frac{1}{\sqrt{\rho}} \]