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-by-Step Solution:

1. **Identify the Given Values:**
- Height \( h = 64 \) km = \( 64 \times 10^3 \) m = \( 64000 \) m
- Radius of the Earth \( R_e = 6.4 \times 10^6 \) m

2. **Use the Formula for Acceleration due to Gravity at Height \( h \):**
The acceleration due to gravity at height \( h \) is given by:
\[ g' = g \left( \frac{R}{R + h} \right)^2 \]
However, since \( h \) is much smaller than \( R \), we can use the approximation:
\[ g' = g \left( 1 - \frac{2h}{R} \right) \]

3. **Calculate the Original Acceleration due to Gravity \( g \):**
The standard value of \( g \) at the surface of the Earth is approximately \( 9.81 \, \text{m/s}^2 \).

4. **Substitute the Values into the Approximation:**
\[ g' = g \left( 1 - \frac{2 \times 64000}{6400000} \right) \]
Simplifying the fraction:
\[ \frac{2 \times 64000}{6400000} = \frac{128000}{6400000} = \frac{128}{6400} = 0.02 \]
Therefore,
\[ g' = g (1 - 0.02) = g \times 0.98 \]

5. **Calculate the Percentage Decrease in \( g \):**
The percentage decrease in \( g \) can be calculated as:
\[ \text{Percentage Decrease} = \left( \frac{g - g'}{g} \right) \times 100 \]
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 is **2%**.