A synchronous satellite goes around the earth once 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 = 5.98 xx 10^(24)kg` radius of earth, `r_e=6.37 xx 10^6m` , Universal constant of gravitation, `G=6.67 xx 10^(-11) Nm^2// kg^2` )

Options

A. `2.4 r_e`
B. `3.6 r_e`
C. `4.8 r_e`
D. `6.6 r_e`

Explanation

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

### Step 1: Understand the concept of a synchronous satellite
A synchronous satellite orbits the Earth in such a way that it has an orbital period equal to the Earth's rotation period. For Earth, this period is 24 hours (or 86400 seconds).

### Step 2: Use the formula for gravitational force and centripetal force
The gravitational force acting on the satellite provides the necessary centripetal force for its circular motion. The gravitational force \( F_g \) is given by:
\[ F_g = \frac{G m_e m_s}{r^2} \]
where:
- \( G \) is the universal gravitational constant,
- \( m_e \) is the mass of the Earth,
- \( m_s \) 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 for circular motion is given by:
\[ F_c = \frac{m_s v^2}{r} \]
where \( v \) is the orbital speed of the satellite.

### Step 3: Relate orbital speed to orbital period
The orbital speed \( v \) can be expressed in terms of the orbital radius \( r \) and the period \( T \) (which is 86400 seconds for a synchronous satellite):
\[ v = \frac{2\pi r}{T} \]

### Step 4: Set the gravitational force equal to the centripetal force
Setting \( F_g = F_c \):
\[ \frac{G m_e m_s}{r^2} = \frac{m_s v^2}{r} \]
We can cancel \( m_s \) from both sides (assuming \( m_s \neq 0 \)):
\[ \frac{G m_e}{r^2} = \frac{v^2}{r} \]

### Step 5: Substitute for \( v \)
Substituting \( v = \frac{2\pi r}{T} \) into the equation:
\[ \frac{G m_e}{r^2} = \frac{\left(\frac{2\pi r}{T}\right)^2}{r} \]
This simplifies to:
\[ \frac{G m_e}{r^2} = \frac{4\pi^2 r}{T^2} \]

### Step 6: Rearrange to find \( r \)
Rearranging gives:
\[ G m_e = 4\pi^2 \frac{r^3}{T^2} \]
Thus, we can solve for \( r \):
\[ r^3 = \frac{G m_e T^2}{4\pi^2} \]
\[ r = \left(\frac{G m_e T^2}{4\pi^2}\right)^{1/3} \]

### Step 7: Substitute known values
Substituting the known values:
- \( G = 6.67 \times 10^{-11} \, \text{Nm}^2/\text{kg}^2 \)
- \( m_e = 5.98 \times 10^{24} \, \text{kg} \)
- \( T = 86400 \, \text{s} \)

Calculating \( r \):
\[ r = \left(\frac{(6.67 \times 10^{-11})(5.98 \times 10^{24})(86400^2)}{4\pi^2}\right)^{1/3} \]

### Step 8: Calculate the numerical value
Calculating the above expression will yield the radius \( r \) in meters.

### Step 9: Express \( r \) in terms of Earth's radius \( r_e \)
Finally, express \( r \) in terms of Earth's radius \( r_e = 6.37 \times 10^6 \, \text{m} \):
\[ \text{Radius of synchronous satellite} = \frac{r}{r_e} \]