If the satellite is stopped suddenly in its orbit which is at a distnace = radius of earth from earth’s surface and allowed to fall freely into the earth, the speed with which it hits the surface of earth will be -

Options

A. `7.919 m//s`
B. `7.919 km//s`
C. `11.2 m//s`
D. `11.2 km//s`

Answer

To find the speed with which a satellite, initially in orbit at a distance equal to the radius of the Earth from the Earth's surface, hits the surface of the Earth after being stopped suddenly and allowed to fall freely, we can follow these steps:

### Step 1: Understand the Initial and Final Positions
The satellite is at a distance equal to the radius of the Earth (r) from the Earth's surface. Therefore, the total distance from the center of the Earth to the satellite is:
\[ d = r + r = 2r \]
where \( r \) is the radius of the Earth.

### Step 2: Calculate Initial Potential Energy
The gravitational potential energy (U) of the satellite when it is at a distance of \( 2r \) from the center of the Earth is given by:
\[ U_i = -\frac{G M m}{2r} \]
where:
- \( G \) is the universal gravitational constant,
- \( M \) is the mass of the Earth,
- \( m \) is the mass of the satellite.

Since the satellite is stopped suddenly, its initial kinetic energy (K) is:
\[ K_i = 0 \]

Thus, the total initial mechanical energy (E_i) is:
\[ E_i = U_i + K_i = -\frac{G M m}{2r} + 0 = -\frac{G M m}{2r} \]

### Step 3: Calculate Final Potential and Kinetic Energy
When the satellite falls to the surface of the Earth, its distance from the center of the Earth is \( r \). The potential energy at this position is:
\[ U_f = -\frac{G M m}{r} \]
Let \( v \) be the final speed of the satellite just before it hits the surface. The kinetic energy at this point is:
\[ K_f = \frac{1}{2} m v^2 \]

Thus, the total final mechanical energy (E_f) is:
\[ E_f = U_f + K_f = -\frac{G M m}{r} + \frac{1}{2} m v^2 \]

### Step 4: Apply Conservation of Energy
According to the conservation of mechanical energy, the initial energy is equal to the final energy:
\[ E_i = E_f \]
Substituting the expressions we derived:
\[ -\frac{G M m}{2r} = -\frac{G M m}{r} + \frac{1}{2} m v^2 \]

### Step 5: Solve for Final Speed \( v \)
Rearranging the equation:
\[ \frac{1}{2} m v^2 = -\frac{G M m}{2r} + \frac{G M m}{r} \]
\[ \frac{1}{2} m v^2 = \frac{G M m}{2r} \]
Cancelling \( m \) from both sides (assuming \( m \neq 0 \)):
\[ \frac{1}{2} v^2 = \frac{G M}{2r} \]
Multiplying both sides by 2:
\[ v^2 = \frac{G M}{r} \]
Taking the square root:
\[ v = \sqrt{\frac{G M}{r}} \]

### Step 6: Relate to Gravitational Acceleration
We know that the gravitational acceleration \( g \) at the surface of the Earth is given by:
\[ g = \frac{G M}{r^2} \]
Thus, we can express \( \frac{G M}{r} \) as:
\[ \frac{G M}{r} = g r \]
Substituting this back into our expression for \( v \):
\[ v = \sqrt{g r} \]

### Step 7: Calculate the Final Speed
Using the approximate values:
- \( g \approx 9.8 \, \text{m/s}^2 \)
- \( r \approx 6400 \, \text{km} = 6400 \times 10^3 \, \text{m} \)

Calculating:
\[ v = \sqrt{9.8 \times 6400 \times 10^3} \]
\[ v \approx \sqrt{62720000} \approx 7919 \, \text{m/s} \]

### Conclusion
The speed with which the satellite hits the surface of the Earth is approximately:
\[ v \approx 7.919 \, \text{km/s} \]