The height above the surface of earth at which the gravitational filed intensity is reduced to `1%` of its value on the surface of earth is

Options

A. `(a)100 R_(e)`
B. `(b)10 R_(e)`
C. `(c)99 R_(e)`
D. `(d)9 R_(e)`

Explanation

To find the height above the surface of the Earth at which the gravitational field intensity is reduced to 1% of its value on the surface of the Earth, we can follow these steps:

### Step 1: Understand the gravitational field intensity at the surface of the Earth
The gravitational field intensity \( E \) at the surface of the Earth is given by the formula:
\[ E = \frac{GM}{R_e^2} \]
where:
- \( G \) is the universal gravitational constant,
- \( M \) is the mass of the Earth,
- \( R_e \) is the radius of the Earth.

### Step 2: Gravitational field intensity at height \( H \)
At a height \( H \) above the surface of the Earth, the gravitational field intensity \( E_H \) is given by:
\[ E_H = \frac{GM}{(R_e + H)^2} \]

### Step 3: Set up the equation for 1% of the surface intensity
We need to find the height \( H \) where the gravitational field intensity is 1% of the surface value:
\[ E_H = 0.01 E \]
Substituting the expressions for \( E_H \) and \( E \):
\[ \frac{GM}{(R_e + H)^2} = 0.01 \cdot \frac{GM}{R_e^2} \]

### Step 4: Simplify the equation
We can cancel \( GM \) from both sides (assuming \( G \) and \( M \) are non-zero):
\[ \frac{1}{(R_e + H)^2} = \frac{0.01}{R_e^2} \]
Cross-multiplying gives:
\[ R_e^2 = 0.01 (R_e + H)^2 \]

### Step 5: Expand and rearrange the equation
Expanding the right-hand side:
\[ R_e^2 = 0.01 (R_e^2 + 2R_eH + H^2) \]
Rearranging gives:
\[ R_e^2 = 0.01 R_e^2 + 0.02 R_e H + 0.01 H^2 \]
Subtracting \( 0.01 R_e^2 \) from both sides:
\[ 0.99 R_e^2 = 0.02 R_e H + 0.01 H^2 \]

### Step 6: Rearranging to form a quadratic equation
Rearranging the equation:
\[ 0.01 H^2 + 0.02 R_e H - 0.99 R_e^2 = 0 \]

### Step 7: Solve the quadratic equation
Using the quadratic formula \( H = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
- Here, \( a = 0.01 \), \( b = 0.02 R_e \), and \( c = -0.99 R_e^2 \).

Calculating the discriminant:
\[ b^2 - 4ac = (0.02 R_e)^2 - 4 \cdot 0.01 \cdot (-0.99 R_e^2) \]
\[ = 0.0004 R_e^2 + 0.0396 R_e^2 = 0.04 R_e^2 \]

Now substituting back into the quadratic formula:
\[ H = \frac{-0.02 R_e \pm \sqrt{0.04 R_e^2}}{2 \cdot 0.01} \]
\[ = \frac{-0.02 R_e \pm 0.2 R_e}{0.02} \]

### Step 8: Calculate the two possible values for \( H \)
Calculating both possible values:
1. \( H = \frac{0.18 R_e}{0.02} = 9 R_e \)
2. \( H = \frac{-0.22 R_e}{0.02} \) (not physically meaningful since height cannot be negative)

Thus, the height \( H \) at which the gravitational field intensity is reduced to 1% of its value on the surface of the Earth is:
\[ H = 9 R_e \]

### Final Answer
The height above the surface of the Earth at which the gravitational field intensity is reduced to 1% of its value on the surface of the Earth is \( 9 R_e \).

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