If the value of `g` at the surface of the earth is `9.8 m//sec^(2)`, then the value of `g` at a place `480` km above the surface of the earth will be (Radius of the earth is `6400` km)

Options

A. `8.4 m//sec^(2)`
B. `9.8 m//sec^(2)`
C. `7.2 m//sec^(2)`
D. `4.2 m//sec^(2)`

Answer

To find the value of acceleration due to gravity \( g' \) at a height \( h \) above the Earth's surface, we can use the formula:

\[ g' = \frac{g}{(1 + \frac{h}{R})^2} \]

where:
- \( g \) is the acceleration due to gravity at the surface of the Earth (given as \( 9.8 \, \text{m/s}^2 \)),
- \( h \) is the height above the Earth's surface (given as \( 480 \, \text{km} \)),
- \( R \) is the radius of the Earth (given as \( 6400 \, \text{km} \)).

### Step 1: Convert heights to the same unit
First, we need to ensure that both \( h \) and \( R \) are in the same unit. We can convert \( h \) from kilometers to meters:

\[ h = 480 \, \text{km} = 480,000 \, \text{m} \]
\[ R = 6400 \, \text{km} = 6,400,000 \, \text{m} \]

### Step 2: Substitute values into the formula
Now we can substitute the values into the formula:

\[ g' = \frac{9.8}{\left(1 + \frac{480,000}{6,400,000}\right)^2} \]

### Step 3: Simplify the fraction
Calculate the fraction inside the parentheses:

\[ \frac{480,000}{6,400,000} = \frac{480}{6400} = \frac{3}{40} \]

Now, substituting this back into the equation:

\[ g' = \frac{9.8}{\left(1 + \frac{3}{40}\right)^2} \]

### Step 4: Calculate the expression in the parentheses
Now calculate \( 1 + \frac{3}{40} \):

\[ 1 + \frac{3}{40} = \frac{40}{40} + \frac{3}{40} = \frac{43}{40} \]

### Step 5: Substitute back into the equation
Now substitute this back into the equation for \( g' \):

\[ g' = \frac{9.8}{\left(\frac{43}{40}\right)^2} \]

### Step 6: Calculate \( \left(\frac{43}{40}\right)^2 \)
Calculate \( \left(\frac{43}{40}\right)^2 \):

\[ \left(\frac{43}{40}\right)^2 = \frac{1849}{1600} \]

### Step 7: Final calculation for \( g' \)
Now substitute this into the equation for \( g' \):

\[ g' = 9.8 \times \frac{1600}{1849} \]

Calculating this gives:

\[ g' \approx 8.4 \, \text{m/s}^2 \]

### Final Answer
Thus, the value of \( g \) at a place 480 km above the surface of the Earth is approximately:

\[ g' \approx 8.4 \, \text{m/s}^2 \]