If R is the radius of the Earth then the height above the Earth's surface at which the acceleration due to gravity decreases by `20%` is

Options

A. `((sqrt5)/(2)-1)R`
B. `((sqrt5)/(2)+1)R`
C. `(5sqrt2-1)R`
D. `(5sqrt2+1)R`

Answer

To solve the problem of finding the height \( h \) above the Earth's surface at which the acceleration due to gravity decreases by 20%, 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:
\[ g_s = \frac{GM}{R^2} \]
where \( G \) is the universal 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 \( g_h \) is given by:
\[ g_h = \frac{GM}{(R + h)^2} \]

### Step 2: Set up the equation for the decrease in gravity
We know that the acceleration due to gravity decreases by 20%, which means:
\[ g_h = 0.8 g_s \]
Substituting the expressions for \( g_h \) and \( g_s \):
\[ \frac{GM}{(R + h)^2} = 0.8 \left(\frac{GM}{R^2}\right) \]

### Step 3: Simplify the equation
We can cancel \( GM \) from both sides (assuming \( G \) and \( M \) are non-zero):
\[ \frac{1}{(R + h)^2} = 0.8 \cdot \frac{1}{R^2} \]
Rearranging gives:
\[ (R + h)^2 = \frac{R^2}{0.8} \]

### Step 4: Solve for \( R + h \)
Taking the square root of both sides:
\[ R + h = \frac{R}{\sqrt{0.8}} = \frac{R}{\sqrt{\frac{4}{5}}} = R \cdot \frac{\sqrt{5}}{2} \]

### Step 5: Isolate \( h \)
Now, isolate \( h \):
\[ h = R \cdot \frac{\sqrt{5}}{2} - R \]
\[ h = R \left(\frac{\sqrt{5}}{2} - 1\right) \]

### Final Result
Thus, the height \( h \) above the Earth's surface at which the acceleration due to gravity decreases by 20% is:
\[ h = R \left(\frac{\sqrt{5}}{2} - 1\right) \]