At what height from the surface of earth the gravitation potential and the value of `g` are `- 5.4 xx 10^(7) J kg^(-2)` and `6.0 ms^(-2)` respectively ? Take the radius of earth as `6400 km`:

Options

A. 1600 km
B. 1400 km
C. 2000 km
D. 2600 km

Explanation

To solve the problem, we need to find the height \( h \) from the surface of the Earth where the gravitational potential \( V \) is \( -5.4 \times 10^7 \, \text{J kg}^{-1} \) and the acceleration due to gravity \( g' \) is \( 6.0 \, \text{m s}^{-2} \). We will use the formulas for gravitational potential and acceleration due to gravity at a height \( h \).

### Step-by-step Solution:

1. **Understand the formulas**:
- The gravitational potential \( V \) at a height \( h \) is given by:
\[ V = -\frac{GM}{R + h} \]
- The acceleration due to gravity \( g' \) at height \( h \) is given by:
\[ g' = \frac{GM}{(R + h)^2} \]

2. **Set up the equations**:
- From the first equation, we can express \( GM \) in terms of \( V \):
\[ GM = -V(R + h) \]
- From the second equation, we can express \( GM \) in terms of \( g' \):
\[ GM = g' (R + h)^2 \]

3. **Equate the two expressions for \( GM \)**:
\[ -V(R + h) = g'(R + h)^2 \]

4. **Substitute the known values**:
- Given \( V = -5.4 \times 10^7 \, \text{J kg}^{-1} \), \( g' = 6.0 \, \text{m s}^{-2} \), and \( R = 6400 \, \text{km} = 6400 \times 10^3 \, \text{m} \).
- Substitute these values into the equation:
\[ 5.4 \times 10^7 (R + h) = 6.0 (R + h)^2 \]

5. **Rearranging the equation**:
- Rearranging gives:
\[ 6.0 (R + h)^2 - 5.4 \times 10^7 (R + h) = 0 \]
- This is a quadratic equation in terms of \( R + h \).

6. **Let \( x = R + h \)**:
- The equation becomes:
\[ 6.0 x^2 - 5.4 \times 10^7 x = 0 \]
- Factor out \( x \):
\[ x(6.0 x - 5.4 \times 10^7) = 0 \]

7. **Solve for \( x \)**:
- Ignoring the trivial solution \( x = 0 \):
\[ 6.0 x = 5.4 \times 10^7 \implies x = \frac{5.4 \times 10^7}{6.0} = 9.0 \times 10^6 \, \text{m} \]

8. **Find \( h \)**:
- Recall that \( x = R + h \):
\[ R + h = 9.0 \times 10^6 \implies h = 9.0 \times 10^6 - 6400 \times 10^3 \]
\[ h = 9.0 \times 10^6 - 6.4 \times 10^6 = 2.6 \times 10^6 \, \text{m} = 2600 \, \text{km} \]

### Final Answer:
The height \( h \) from the surface of the Earth is \( 2600 \, \text{km} \).