At what height from Earth's surface the acceleration due to gravity becomes 20% of its value on the surface of earth. [Radius of earth = 6,400 km.]

Answer

To find the height from the Earth's surface where the acceleration due to gravity becomes 20% of its value on the surface of the Earth, we can follow these steps:

### Step 1: Understand the relationship of gravity at different heights
The acceleration due to gravity at the surface of the Earth is given by:
\[ g = \frac{GM}{R^2} \]
where \( G \) is the gravitational constant, \( M \) is the mass of the Earth, and \( R \) is the radius of the Earth.

At a height \( h \) above the Earth's surface, the acceleration due to gravity \( g' \) is given by:
\[ g' = \frac{GM}{(R + h)^2} \]

### Step 2: Set up the equation for 20% of surface gravity
We want to find the height \( h \) where \( g' \) is 20% of \( g \):
\[ g' = 0.2g \]
Substituting the expressions for \( g \) and \( g' \):
\[ \frac{GM}{(R + h)^2} = 0.2 \left(\frac{GM}{R^2}\right) \]

### Step 3: Cancel out common terms
Since \( GM \) appears in both sides, we can cancel it out:
\[ \frac{1}{(R + h)^2} = 0.2 \cdot \frac{1}{R^2} \]
This simplifies to:
\[ (R + h)^2 = 5R^2 \]

### Step 4: Take the square root of both sides
Taking the square root gives us:
\[ R + h = \sqrt{5}R \]

### Step 5: Solve for height \( h \)
Rearranging the equation to solve for \( h \):
\[ h = \sqrt{5}R - R \]
\[ h = (\sqrt{5} - 1)R \]

### Step 6: Substitute the value of \( R \)
Given that the radius of the Earth \( R = 6400 \) km:
\[ h = (\sqrt{5} - 1) \times 6400 \]

### Step 7: Calculate \( \sqrt{5} \)
Calculating \( \sqrt{5} \):
\[ \sqrt{5} \approx 2.236 \]
Thus,
\[ h = (2.236 - 1) \times 6400 \]
\[ h \approx 1.236 \times 6400 \]

### Step 8: Final calculation
Calculating the height:
\[ h \approx 7910.4 \text{ km} \]

### Conclusion
The height from the Earth's surface where the acceleration due to gravity becomes 20% of its value on the surface of the Earth is approximately:
\[ h \approx 7911 \text{ km} \]

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