The orbiting time .T. of a surface satellite is given by the relation

Options

A. `2pisqrt(R/g)`
B. `2pisqrt(g/R)`
C. `2pisqrt(2gR)`
D. `2pisqrt((2R)/g)`

Answer

To solve the problem of finding the orbiting time \( T \) of a surface satellite, we can follow these steps:

### Step 1: Understand the Concept
A satellite in orbit around the Earth will have a certain time period \( T \) for completing one full revolution. For a satellite that is very close to the surface of the Earth, we can assume its orbital radius \( r \) is approximately equal to the Earth's radius \( R \).

### Step 2: Establish the Relationship for Orbital Velocity
The orbital velocity \( v \) of a satellite can be expressed as the distance traveled in one complete revolution divided by the time taken for that revolution. The distance traveled in one complete orbit is the circumference of the circle, which is given by:
\[ v = \frac{2 \pi r}{T} \]
where \( r \) is the radius of the orbit.

### Step 3: Use the Gravitational Force
The gravitational force provides the necessary centripetal force for the satellite to stay in orbit. The gravitational force acting on the satellite can be expressed as:
\[ F = \frac{G M m}{r^2} \]
where \( G \) is the gravitational constant, \( M \) is the mass of the Earth, and \( m \) is the mass of the satellite.

### Step 4: Set Up the Centripetal Force Equation
The centripetal force required to keep the satellite in circular motion is given by:
\[ F_c = \frac{m v^2}{r} \]
Setting the gravitational force equal to the centripetal force gives us:
\[ \frac{G M m}{r^2} = \frac{m v^2}{r} \]
We can cancel \( m \) from both sides (assuming \( m \neq 0 \)):
\[ \frac{G M}{r^2} = \frac{v^2}{r} \]
Multiplying both sides by \( r \) gives:
\[ \frac{G M}{r} = v^2 \]

### Step 5: Substitute for Velocity
Now, substituting \( v^2 \) back into the equation for \( v \):
\[ v = \sqrt{\frac{G M}{r}} \]

### Step 6: Substitute for Time Period
Now, we can substitute this expression for \( v \) back into the equation for \( T \):
\[ T = \frac{2 \pi r}{v} = \frac{2 \pi r}{\sqrt{\frac{G M}{r}}} \]
This simplifies to:
\[ T = 2 \pi \sqrt{\frac{r^3}{G M}} \]

### Step 7: Special Case for Surface Satellite
For a satellite that is at the surface of the Earth, we can replace \( r \) with \( R \) (the radius of the Earth):
\[ T = 2 \pi \sqrt{\frac{R^3}{G M}} \]

### Conclusion
Thus, the time period \( T \) of a surface satellite is given by:
\[ T = 2 \pi \sqrt{\frac{R}{g}} \]
where \( g \) is the acceleration due to gravity at the surface of the Earth.

### Final Answer
The time period \( T \) of a surface satellite is:
\[ T = 2 \pi \sqrt{\frac{R}{g}} \]