If R is radius of the earth , the height above the surface of the earth where the weight of a body is 36% less than its weight on the surface of the earth is

Options

A. 4R/5
B. R/5
C. R/6
D. R/4

Answer

To solve the problem of finding the height above the surface of the Earth where the weight of a body is 36% less than its weight on the surface, we can follow these steps:

### Step 1: Understand the Weight Reduction
The weight of a body on the surface of the Earth is given by \( W = mg \), where \( m \) is the mass of the body and \( g \) is the acceleration due to gravity at the surface. If the weight is 36% less at height \( h \), then the weight at height \( h \) is:
\[ W_h = W - 0.36W = 0.64W \]
This means:
\[ W_h = 0.64mg \]

### Step 2: Relate Weight at Height to Gravity
The weight at height \( h \) can also be expressed in terms of the acceleration due to gravity at that height, \( g_h \):
\[ W_h = mg_h \]
Setting the two expressions for \( W_h \) equal gives:
\[ mg_h = 0.64mg \]
Dividing both sides by \( m \) (assuming \( m \neq 0 \)):
\[ g_h = 0.64g \]

### Step 3: Use the Formula for Gravity at Height
The acceleration due to gravity at a height \( h \) above the surface of the Earth is given by:
\[ g_h = \frac{gR^2}{(R + h)^2} \]
Substituting \( g_h = 0.64g \) into this equation gives:
\[ 0.64g = \frac{gR^2}{(R + h)^2} \]
We can cancel \( g \) from both sides (assuming \( g \neq 0 \)):
\[ 0.64 = \frac{R^2}{(R + h)^2} \]

### Step 4: Rearranging the Equation
Cross-multiplying gives:
\[ 0.64(R + h)^2 = R^2 \]
Expanding the left side:
\[ 0.64(R^2 + 2Rh + h^2) = R^2 \]
This simplifies to:
\[ 0.64R^2 + 1.28Rh + 0.64h^2 = R^2 \]
Rearranging gives:
\[ R^2 - 0.64R^2 - 1.28Rh - 0.64h^2 = 0 \]
\[ 0.36R^2 - 1.28Rh - 0.64h^2 = 0 \]

### Step 5: Solving the Quadratic Equation
This is a quadratic equation in terms of \( h \):
\[ 0.64h^2 + 1.28Rh - 0.36R^2 = 0 \]
Using the quadratic formula \( h = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 0.64 \), \( b = 1.28R \), and \( c = -0.36R^2 \):
\[ h = \frac{-1.28R \pm \sqrt{(1.28R)^2 - 4 \cdot 0.64 \cdot (-0.36R^2)}}{2 \cdot 0.64} \]
Calculating the discriminant:
\[ (1.28R)^2 + 4 \cdot 0.64 \cdot 0.36R^2 = 1.6384R^2 + 0.9216R^2 = 2.56R^2 \]
Thus:
\[ h = \frac{-1.28R \pm \sqrt{2.56}R}{1.28} \]
Since \( \sqrt{2.56} = 1.6 \):
\[ h = \frac{-1.28R \pm 1.6R}{1.28} \]
Taking the positive root:
\[ h = \frac{0.32R}{1.28} = \frac{R}{4} \]

### Final Answer
The height \( h \) above the surface of the Earth where the weight of a body is 36% less than its weight on the surface is:
\[ h = \frac{R}{4} \]