A satellite whose mass is M , is revolving in circular orbit of radius r around the earth. Time of revolution of satellite is

Options

A. `T prop r^(5)/(GM)`
B. `T prop sqrt(r^(3)/(GM))`
C. `T prop sqrt(r/(GM^(2)//3))`
D. `T prop sqrt(r^(3)/(GM^(1)//4))`

Answer

To find the time of revolution of a satellite of mass \( M \) revolving in a circular orbit of radius \( r \) around the Earth, we can follow these steps:

### Step 1: Understand the relationship between time period, radius, and velocity
The time period \( T \) of a satellite is the time it takes to complete one full revolution around the Earth. It can be expressed as:
\[ T = \frac{\text{Circumference of the orbit}}{\text{Orbital velocity}} \]
The circumference of the orbit is given by \( 2\pi r \).

### Step 2: Express the orbital velocity
The orbital velocity \( v \) of the satellite can be derived from the gravitational force acting on it. The centripetal force required to keep the satellite in circular motion is provided by the gravitational force. Thus, we have:
\[ \frac{Mv^2}{r} = \frac{GMm}{r^2} \]
where \( M \) is the mass of the Earth, \( m \) is the mass of the satellite, and \( G \) is the gravitational constant.

### Step 3: Simplify the equation
From the equation above, we can cancel \( m \) (mass of the satellite) and one \( r \):
\[ Mv^2 = \frac{GMm}{r} \]
This simplifies to:
\[ v^2 = \frac{GM}{r} \]
Taking the square root gives us the orbital velocity:
\[ v = \sqrt{\frac{GM}{r}} \]

### Step 4: Substitute the velocity back into the time period formula
Now we substitute the expression for \( v \) back into the time period formula:
\[ T = \frac{2\pi r}{v} = \frac{2\pi r}{\sqrt{\frac{GM}{r}}} \]

### Step 5: Simplify the time period expression
This can be simplified further:
\[ T = 2\pi r \cdot \sqrt{\frac{r}{GM}} = 2\pi \sqrt{\frac{r^3}{GM}} \]

### Conclusion
Thus, the time of revolution \( T \) of the satellite is given by:
\[ T = 2\pi \sqrt{\frac{r^3}{GM}} \]