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 use the formula derived from the balance of gravitational force and centripetal force acting on the satellite. Here’s a step-by-step solution:

### Step 1: Understand the Forces Acting on the Satellite
The gravitational force acting on the satellite is given by:
\[ F_g = \frac{GMm}{r^2} \]
where:
- \( G \) is the universal gravitational constant, \( 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \)
- \( M \) is the mass of the Earth, approximately \( 6 \times 10^{24} \, \text{kg} \)
- \( m \) 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)

### Step 2: Set Up the Equation for Circular Motion
The centripetal force required to keep the satellite in orbit is given by:
\[ F_c = \frac{mv^2}{r} \]
where \( v \) is the orbital velocity of the satellite.

### Step 3: Equate the Gravitational Force and Centripetal Force
For the satellite to remain in orbit, these forces must be equal:
\[ \frac{GMm}{r^2} = \frac{mv^2}{r} \]

### Step 4: Simplify the Equation
We can cancel \( m \) from both sides (assuming \( m \neq 0 \)):
\[ \frac{GM}{r^2} = \frac{v^2}{r} \]
Multiplying both sides by \( r \):
\[ \frac{GM}{r} = v^2 \]
Thus, we have:
\[ v = \sqrt{\frac{GM}{r}} \]

### Step 5: Relate Velocity to Time Period
The orbital velocity \( v \) can also be expressed in terms of the time period \( T \):
\[ v = \frac{2\pi r}{T} \]
Setting the two expressions for \( v \) equal gives:
\[ \sqrt{\frac{GM}{r}} = \frac{2\pi r}{T} \]

### Step 6: Solve for Time Period \( T \)
Squaring both sides:
\[ \frac{GM}{r} = \left(\frac{2\pi r}{T}\right)^2 \]
\[ \frac{GM}{r} = \frac{4\pi^2 r^2}{T^2} \]
Rearranging gives:
\[ T^2 = \frac{4\pi^2 r^3}{GM} \]
Taking the square root:
\[ T = 2\pi \sqrt{\frac{r^3}{GM}} \]

### Step 7: Substitute the Values
For a satellite very close to the Earth's surface, \( r \) is approximately equal to the radius of the Earth, \( R \approx 6400 \times 10^3 \, \text{m} \). Now substituting the values:
- \( G = 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \)
- \( M = 6 \times 10^{24} \, \text{kg} \)

Calculating:
\[ T = 2\pi \sqrt{\frac{(6400 \times 10^3)^3}{(6.67 \times 10^{-11})(6 \times 10^{24})}} \]

### Step 8: Calculate the Final Value
After performing the calculations, you will find:
\[ T \approx 1.41 \, \text{hours} \]

### Final Answer
The time period of a satellite very close to the Earth's surface is approximately **1.41 hours**.

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