The value of acceleration due to gravity will be 1% of its value at the surface of earth at a height of `(R_(e )=6400 km)`

Options

A. 6400 km
B. 577600 km
C. 2560 km
D. 6400 km

Answer

To solve the problem, we need to find the height \( h \) at which the acceleration due to gravity \( g_h \) is 1% of the acceleration due to gravity at the surface of the Earth \( g \).

### Step-by-Step Solution:

1. **Understanding the relationship between gravity at height and at the surface:**
The formula for the acceleration due to gravity at a height \( h \) above the Earth's surface is given by:
\[ g_h = \frac{g R_e}{(R_e + h)^2} \]
where \( R_e \) is the radius of the Earth and \( g \) is the acceleration due to gravity at the surface.

2. **Setting up the equation for 1% of surface gravity:**
We want to find \( h \) such that:
\[ g_h = \frac{g}{100} \]
Substituting this into our formula gives:
\[ \frac{g R_e}{(R_e + h)^2} = \frac{g}{100} \]

3. **Canceling \( g \) from both sides:**
Since \( g \) is common on both sides, we can cancel it out:
\[ \frac{R_e}{(R_e + h)^2} = \frac{1}{100} \]

4. **Cross-multiplying to eliminate the fraction:**
Cross-multiplying gives:
\[ 100 R_e = (R_e + h)^2 \]

5. **Expanding the right-hand side:**
Expanding the equation results in:
\[ 100 R_e = R_e^2 + 2R_e h + h^2 \]

6. **Rearranging the equation:**
Rearranging gives us a quadratic equation:
\[ h^2 + 2R_e h + (R_e^2 - 100 R_e) = 0 \]

7. **Using the quadratic formula:**
We can use the quadratic formula \( h = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 1, b = 2R_e, c = (R_e^2 - 100 R_e) \):
\[ h = \frac{-2R_e \pm \sqrt{(2R_e)^2 - 4 \cdot 1 \cdot (R_e^2 - 100 R_e)}}{2 \cdot 1} \]

8. **Calculating the discriminant:**
The discriminant simplifies to:
\[ (2R_e)^2 - 4(R_e^2 - 100R_e) = 4R_e^2 - 4R_e^2 + 400R_e = 400R_e \]

9. **Substituting back into the formula:**
Thus, we have:
\[ h = \frac{-2R_e \pm \sqrt{400R_e}}{2} \]
Simplifying gives:
\[ h = -R_e \pm 10\sqrt{R_e} \]

10. **Choosing the positive root:**
Since height cannot be negative, we take:
\[ h = -R_e + 10\sqrt{R_e} \]

11. **Substituting \( R_e = 6400 \, \text{km} \):**
Now substituting \( R_e \):
\[ h = -6400 + 10 \cdot \sqrt{6400} \]
\[ \sqrt{6400} = 80 \quad \Rightarrow \quad h = -6400 + 800 = 57600 \, \text{km} \]

### Final Answer:
Thus, the height \( h \) at which the acceleration due to gravity is 1% of its value at the surface of the Earth is:
\[ h = 57600 \, \text{km} \]