What will be the acceleration due to gravity at height h if h gt gt R . Where R is radius of earth and g is acceleration due to gravity on the surface of earth

Options

A. `(g)/((1+h/R)^(2))`
B. `g(1-(2h)/R)`
C. `(g)/((1-h/R)^(2))`
D. `g(1-h/R)`

Answer

To find the acceleration due to gravity at a height \( h \) where \( h \gg R \) (with \( R \) being the radius of the Earth and \( g \) the acceleration due to gravity at the surface of the Earth), we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Formula for Gravitational Acceleration**:
The acceleration due to gravity \( g \) at a distance \( r \) from the center of the Earth is given by the formula:
\[ g' = \frac{GM}{r^2} \]
where \( G \) is the universal gravitational constant and \( M \) is the mass of the Earth.

2. **Identifying the Distance from the Center of the Earth**:
When we are at a height \( h \) above the Earth's surface, the distance from the center of the Earth becomes:
\[ r = R + h \]
where \( R \) is the radius of the Earth.

3. **Substituting into the Gravitational Formula**:
We can substitute \( r \) into the gravitational acceleration formula:
\[ g' = \frac{GM}{(R + h)^2} \]

4. **Using the Surface Gravity**:
The acceleration due to gravity at the surface of the Earth is:
\[ g = \frac{GM}{R^2} \]
We can express \( GM \) in terms of \( g \):
\[ GM = gR^2 \]

5. **Substituting \( GM \) in the Gravitational Formula**:
Now, substituting \( GM \) into the equation for \( g' \):
\[ g' = \frac{gR^2}{(R + h)^2} \]

6. **Considering the Condition \( h \gg R \)**:
When \( h \) is much greater than \( R \), we can approximate \( R + h \) as \( h \) (since \( h \) dominates \( R \)):
\[ g' \approx \frac{gR^2}{h^2} \]

7. **Final Expression for Acceleration Due to Gravity at Height \( h \)**:
Thus, the acceleration due to gravity at height \( h \) can be approximated as:
\[ g' \approx \frac{gR^2}{h^2} \]

### Final Answer:
The acceleration due to gravity at a height \( h \) where \( h \gg R \) is given by:
\[ g' \approx \frac{gR^2}{h^2} \]