Time period of a satellite to very close to earth`s surface, around the earth is approximately

Options

A. 1.42 h
B. 2.42 h
C. 24 h
D. 0.72 h

Answer

To find the time period of a satellite very close to the Earth's surface, we can follow these steps:

### Step 1: Understand the Forces Involved
The gravitational force acting on the satellite provides the necessary centripetal force for it to maintain its circular orbit around the Earth.

### Step 2: Write the Equations for Gravitational and Centripetal Forces
The gravitational force \( F_g \) is given by:
\[ F_g = \frac{GMm}{r^2} \]
where:
- \( G \) is the gravitational constant (\( 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \)),
- \( M \) is the mass of the Earth (\( 5.972 \times 10^{24} \, \text{kg} \)),
- \( m \) is the mass of the satellite,
- \( r \) is the distance from the center of the Earth to the satellite (approximately equal to the radius of the Earth when the satellite is very close to the surface).

The centripetal force \( F_c \) required for circular motion is given by:
\[ F_c = \frac{mv^2}{r} \]
where \( v \) is the orbital velocity of the satellite.

### Step 3: Set the Gravitational Force Equal to the Centripetal Force
Since the gravitational force provides the centripetal force, we can set them equal to each other:
\[ \frac{GMm}{r^2} = \frac{mv^2}{r} \]
We can cancel \( m \) from both sides (assuming \( m \neq 0 \)):
\[ \frac{GM}{r^2} = \frac{v^2}{r} \]

### Step 4: Solve for Orbital Velocity \( v \)
Rearranging the equation gives:
\[ v^2 = \frac{GM}{r} \]
Taking the square root, we find:
\[ v = \sqrt{\frac{GM}{r}} \]

### Step 5: Calculate the Time Period \( T \)
The time period \( T \) of the satellite is the time it takes to complete one full orbit. The circumference of the orbit is \( 2\pi r \), so:
\[ T = \frac{\text{Circumference}}{\text{Velocity}} = \frac{2\pi r}{v} \]
Substituting the expression for \( v \):
\[ T = \frac{2\pi r}{\sqrt{\frac{GM}{r}}} \]
This simplifies to:
\[ T = 2\pi \sqrt{\frac{r^3}{GM}} \]

### Step 6: Substitute Values
For a satellite very close to the Earth's surface, we take \( r \) as the radius of the Earth, approximately \( 6.4 \times 10^6 \, \text{m} \). Substituting the values:
- \( G = 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \)
- \( M = 5.972 \times 10^{24} \, \text{kg} \)

Calculating:
\[ T = 2\pi \sqrt{\frac{(6.4 \times 10^6)^3}{6.67 \times 10^{-11} \times 5.972 \times 10^{24}}} \]

### Step 7: Calculate the Result
After performing the calculations, we find:
\[ T \approx 84.6 \, \text{minutes} \quad \text{or} \quad 1.42 \, \text{hours} \]

### Conclusion
Thus, the time period of a satellite very close to the Earth's surface is approximately \( 1.42 \) hours.

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