The value of `g` on the earth's surface is `980 cm//sec^(2)` . Its value at a height of `64` km from the earth's surface is

Options

A. `960.40 cm//sec^(2)`
B. `984.90 cm//sec^(2)`
C. `982.45 cm//sec^(2)`
D. `977.55 cm//sec^(2)`

Answer

To find the value of `g` at a height of `64 km` from the Earth's surface, we can use the formula for gravitational acceleration at a height `h` above the Earth's surface:

\[ g' = g \left(1 - \frac{2h}{R}\right) \]

Where:
- \( g' \) is the gravitational acceleration at height \( h \).
- \( g \) is the gravitational acceleration at the Earth's surface (given as \( 980 \, \text{cm/s}^2 \)).
- \( h \) is the height above the Earth's surface (given as \( 64 \, \text{km} \)).
- \( R \) is the radius of the Earth (approximately \( 6400 \, \text{km} \)).

### Step-by-Step Solution:

1. **Convert the height from kilometers to meters**:
\[ h = 64 \, \text{km} = 64 \times 1000 \, \text{m} = 64000 \, \text{m} \]

2. **Convert the radius of the Earth from kilometers to meters**:
\[ R = 6400 \, \text{km} = 6400 \times 1000 \, \text{m} = 6400000 \, \text{m} \]

3. **Substitute the values into the formula**:
\[ g' = 980 \, \text{cm/s}^2 \left(1 - \frac{2 \times 64000}{6400000}\right) \]

4. **Calculate the fraction**:
\[ \frac{2 \times 64000}{6400000} = \frac{128000}{6400000} = 0.02 \]

5. **Substitute back into the equation**:
\[ g' = 980 \, \text{cm/s}^2 \left(1 - 0.02\right) = 980 \, \text{cm/s}^2 \times 0.98 \]

6. **Calculate \( g' \)**:
\[ g' = 960.4 \, \text{cm/s}^2 \]

### Final Answer:
The value of \( g \) at a height of \( 64 \, \text{km} \) from the Earth's surface is approximately \( 960.4 \, \text{cm/s}^2 \).

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