A satellite is orbiting the earth in an orbit with a velocity `4 km//s`, then the acceleration due to gravity at that height is (in `ms^(-2)`)

Options

A. `0.4`
B. `0.62`
C. `0.87`
D. `1.21`

Explanation

To find the acceleration due to gravity at the height where a satellite is orbiting the Earth with a velocity of 4 km/s, we can use the relationship between gravitational force and centripetal acceleration.

### Step-by-step Solution:

1. **Convert the velocity to meters per second**:
The velocity of the satellite is given as 4 km/s.
\[ v = 4 \text{ km/s} = 4 \times 10^3 \text{ m/s} \]

2. **Use the formula for centripetal acceleration**:
The centripetal acceleration \( a_c \) of the satellite can be expressed as:
\[ a_c = \frac{v^2}{r} \]
where \( r \) is the distance from the center of the Earth to the satellite.

3. **Relate centripetal acceleration to gravitational acceleration**:
The gravitational force provides the necessary centripetal force for the satellite's orbit. Thus, we have:
\[ a_c = g' = \frac{GM}{r^2} \]
where \( g' \) is the acceleration due to gravity at the height of the satellite, \( G \) is the universal gravitational constant, and \( M \) is the mass of the Earth.

4. **Set the equations equal**:
Since both expressions represent the acceleration at the height of the satellite, we can set them equal to each other:
\[ \frac{v^2}{r} = \frac{GM}{r^2} \]

5. **Rearranging the equation**:
Multiply both sides by \( r^2 \):
\[ v^2 \cdot r = GM \]
Now, we can express \( g' \) in terms of \( v \):
\[ g' = \frac{v^2}{r} \]

6. **Substituting the values**:
We need to find \( r \). We can rearrange the equation to find \( r \):
\[ r = \frac{GM}{v^2} \]
However, we can also express \( g' \) in terms of known values. The radius of the Earth \( R \) is approximately \( 6.37 \times 10^6 \) m, and the gravitational acceleration at the surface \( g \) is approximately \( 9.81 \, \text{m/s}^2 \).

7. **Calculate \( g' \)**:
Using the formula:
\[ g' = \frac{v^2}{R} \]
We can substitute \( v = 4 \times 10^3 \text{ m/s} \):
\[ g' = \frac{(4 \times 10^3)^2}{6.37 \times 10^6} \]
\[ g' = \frac{16 \times 10^6}{6.37 \times 10^6} \approx 2.51 \, \text{m/s}^2 \]

8. **Final Calculation**:
After calculating, we find that the acceleration due to gravity at that height is approximately \( 2.51 \, \text{m/s}^2 \).

### Final Answer:
The acceleration due to gravity at that height is approximately \( 2.51 \, \text{m/s}^2 \).