Assuming density d of a planet to be uniform, we can say that the time period of its artificial satellite is proportional to

Options

A. d
B. `sqrtd`
C. `(1)/(sqrtd)`
D. `1/d`

Explanation

To find the relationship between the time period of an artificial satellite and the uniform density of a planet, we can follow these steps:

### Step 1: Understand the Formula for Time Period
The time period \( T \) of a satellite orbiting a planet can be expressed using the formula:
\[ T = 2\pi \sqrt{\frac{r^3}{GM}} \]
where:
- \( T \) is the time period of the satellite,
- \( r \) is the radius of the orbit (which is approximately equal to the radius of the planet if the satellite is close to the surface),
- \( G \) is the gravitational constant,
- \( M \) is the mass of the planet.

### Step 2: Calculate the Mass of the Planet
Since the planet has a uniform density \( d \), we can express the mass \( M \) of the planet in terms of its volume and density. The volume \( V \) of a sphere (the planet) is given by:
\[ V = \frac{4}{3} \pi r^3 \]
Thus, the mass \( M \) can be expressed as:
\[ M = V \cdot d = \frac{4}{3} \pi r^3 d \]

### Step 3: Substitute Mass into the Time Period Formula
Now, substituting the expression for mass \( M \) into the time period formula:
\[ T = 2\pi \sqrt{\frac{r^3}{G \left(\frac{4}{3} \pi r^3 d\right)}} \]

### Step 4: Simplify the Expression
This simplifies to:
\[ T = 2\pi \sqrt{\frac{r^3}{\frac{4}{3} \pi G r^3 d}} \]
The \( r^3 \) terms cancel out:
\[ T = 2\pi \sqrt{\frac{1}{\frac{4}{3} \pi G d}} \]
This can be rewritten as:
\[ T = 2\pi \sqrt{\frac{3}{4\pi G d}} \]

### Step 5: Identify the Relationship
From the final expression, we can see that:
\[ T \propto \frac{1}{\sqrt{d}} \]
This indicates that the time period \( T \) of the satellite is inversely proportional to the square root of the density \( d \) of the planet.

### Conclusion
Thus, the time period of the artificial satellite is proportional to \( \frac{1}{\sqrt{d}} \).