A synchronous satellite goes around the earth one in every 24 h. What is the radius of orbit of the synchronous satellite in terms of the earth's radius ? (Given: Mass of the earth , `M_(E)=5.98xx10^(24) kg,` radius of the earth, `R_(E)=6.37xx10^(6)m`, universal constant of gravitational , `G=6.67xx10^(-11)Nm^(2)kg^(-2)`)

Options

A. `2.4R_(E)`
B. `3.6R_(E)`
C. `4.8R_(E)`
D. `6.6R_(E)`

Explanation

To find the radius of the orbit of a synchronous satellite in terms of the Earth's radius, we can follow these steps:

### Step 1: Understand the relationship between gravitational force and centripetal force
For a satellite in orbit, the gravitational force provides the necessary centripetal force to keep it in circular motion. The gravitational force \( F_g \) can be expressed as:
\[ F_g = \frac{G M_E m}{r^2} \]
where:
- \( G \) is the universal gravitational constant,
- \( M_E \) is the mass of the Earth,
- \( m \) is the mass of the satellite,
- \( r \) is the distance from the center of the Earth to the satellite.

The centripetal force \( F_c \) required to keep the satellite in circular motion is given by:
\[ F_c = \frac{m v^2}{r} \]
where \( v \) is the orbital speed of the satellite.

### Step 2: Set the gravitational force equal to the centripetal force
Setting \( F_g = F_c \):
\[ \frac{G M_E m}{r^2} = \frac{m v^2}{r} \]
We can cancel \( m \) from both sides (assuming \( m \neq 0 \)):
\[ \frac{G M_E}{r^2} = \frac{v^2}{r} \]
Multiplying both sides by \( r \):
\[ \frac{G M_E}{r} = v^2 \]

### Step 3: Relate orbital speed to the time period
The orbital speed \( v \) can also be expressed in terms of the orbital radius \( r \) and the time period \( T \):
\[ v = \frac{2 \pi r}{T} \]
Substituting this into the previous equation:
\[ \frac{G M_E}{r} = \left(\frac{2 \pi r}{T}\right)^2 \]
Expanding the right side:
\[ \frac{G M_E}{r} = \frac{4 \pi^2 r^2}{T^2} \]

### Step 4: Rearranging the equation
Multiplying both sides by \( r \):
\[ G M_E = \frac{4 \pi^2 r^3}{T^2} \]
Rearranging gives:
\[ r^3 = \frac{G M_E T^2}{4 \pi^2} \]

### Step 5: Substitute the values
Given:
- \( M_E = 5.98 \times 10^{24} \, \text{kg} \)
- \( G = 6.67 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \)
- \( T = 24 \, \text{hours} = 24 \times 60 \times 60 \, \text{seconds} = 86400 \, \text{s} \)

Substituting these values into the equation:
\[ r^3 = \frac{(6.67 \times 10^{-11}) (5.98 \times 10^{24}) (86400)^2}{4 \pi^2} \]

### Step 6: Calculate \( r^3 \)
Calculating the right-hand side:
1. Calculate \( (86400)^2 = 7.46496 \times 10^9 \).
2. Calculate \( G M_E = (6.67 \times 10^{-11}) (5.98 \times 10^{24}) = 3.986 \times 10^{14} \).
3. Now, calculate:
\[ r^3 = \frac{(3.986 \times 10^{14}) (7.46496 \times 10^9)}{4 \pi^2} \]
4. Calculate \( 4 \pi^2 \approx 39.478 \).
5. Finally, calculate \( r^3 \) and then take the cube root to find \( r \).

### Step 7: Convert \( r \) to terms of Earth's radius
Once \( r \) is calculated, convert it into terms of Earth's radius \( R_E = 6.37 \times 10^6 \, \text{m} \):
\[ \text{Radius of orbit in terms of } R_E = \frac{r}{R_E} \]

### Final Answer
After performing the calculations, you will find that the radius of the synchronous satellite is approximately \( 6.6 R_E \).

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