Find the percentage decrease in the acceleration due to gravity when a body is taken from the surface of earth of a height of 64 km from its surface [ Take `R_(e) = 6.4 xx10^(6)` m ]

Explanation

To find the percentage decrease in the acceleration due to gravity when a body is taken from the surface of the Earth to a height of 64 km, we can follow these steps:

### Step 1: Understand the formula for acceleration due to gravity at height
The acceleration due to gravity at a height \( h \) from the surface of the Earth is given by the formula:
\[ g' = g \left( 1 - \frac{2h}{R} \right) \]
where:
- \( g' \) is the acceleration due to gravity at height \( h \),
- \( g \) is the acceleration due to gravity at the surface of the Earth (approximately \( 9.81 \, \text{m/s}^2 \)),
- \( R \) is the radius of the Earth (given as \( 6.4 \times 10^6 \, \text{m} \)),
- \( h \) is the height above the surface of the Earth (given as \( 64 \, \text{km} = 64000 \, \text{m} \)).

### Step 2: Substitute the values into the formula
We can substitute \( h \) and \( R \) into the formula:
\[ g' = g \left( 1 - \frac{2 \times 64000}{6.4 \times 10^6} \right) \]

### Step 3: Calculate \( \frac{2h}{R} \)
First, we need to calculate \( \frac{2h}{R} \):
\[ \frac{2 \times 64000}{6.4 \times 10^6} = \frac{128000}{6400000} = 0.02 \]

### Step 4: Calculate \( g' \)
Now, we can substitute this back into the equation for \( g' \):
\[ g' = g \left( 1 - 0.02 \right) = g \times 0.98 \]

### Step 5: Find the percentage decrease in \( g \)
The percentage decrease in acceleration due to gravity is given by:
\[ \text{Percentage decrease} = \left( \frac{g - g'}{g} \right) \times 100 \]
Substituting \( g' = 0.98g \):
\[ \text{Percentage decrease} = \left( \frac{g - 0.98g}{g} \right) \times 100 = \left( \frac{0.02g}{g} \right) \times 100 = 2\% \]

### Final Answer
The percentage decrease in the acceleration due to gravity when a body is taken to a height of 64 km from the surface of the Earth is **2%**.

---