A planet has twice the mass of earth and of identical size. What will be the height above the surface of the planet where its acceleration due to gravity reduces by 36% of its value on its surface ?

Explanation

To solve the problem step by step, we need to find the height above the surface of a planet where the acceleration due to gravity is reduced by 36% of its value on the surface. Here’s how we can approach this:

### Step 1: Determine the acceleration due to gravity on the planet's surface.

The formula for acceleration due to gravity \( g \) on the surface of a planet is given by:
\[ g = \frac{GM}{R^2} \]
where:
- \( G \) is the gravitational constant,
- \( M \) is the mass of the planet,
- \( R \) is the radius of the planet.

Given that the planet has twice the mass of Earth (\( M = 2M_e \)) and identical size (\( R = R_e \)), we can express the acceleration due to gravity on this planet as:
\[ g_p = \frac{G(2M_e)}{R_e^2} = 2 \cdot \frac{GM_e}{R_e^2} = 2g_e \]
where \( g_e \) is the acceleration due to gravity on Earth (approximately \( 9.8 \, \text{m/s}^2 \)). Thus:
\[ g_p = 2 \cdot 9.8 = 19.6 \, \text{m/s}^2 \]

### Step 2: Calculate the reduced acceleration due to gravity.

The problem states that we need to find the height where the acceleration due to gravity is reduced by 36%. This means:
\[ g' = g_p - 0.36g_p = 0.64g_p \]
Substituting \( g_p \):
\[ g' = 0.64 \cdot 19.6 = 12.544 \, \text{m/s}^2 \]

### Step 3: Use the formula for gravity at height \( h \).

The formula for the acceleration due to gravity at a height \( h \) above the surface of the planet is:
\[ g' = \frac{g_p R^2}{(R + h)^2} \]
Setting \( g' = 12.544 \, \text{m/s}^2 \) and \( g_p = 19.6 \, \text{m/s}^2 \):
\[ 12.544 = \frac{19.6 R^2}{(R + h)^2} \]

### Step 4: Rearranging the equation.

Cross-multiplying gives:
\[ 12.544(R + h)^2 = 19.6 R^2 \]
Expanding the left side:
\[ 12.544(R^2 + 2Rh + h^2) = 19.6 R^2 \]
This simplifies to:
\[ 12.544R^2 + 25.088Rh + 12.544h^2 = 19.6R^2 \]
Rearranging gives:
\[ 25.088Rh + 12.544h^2 = 19.6R^2 - 12.544R^2 \]
\[ 25.088Rh + 12.544h^2 = 7.056R^2 \]

### Step 5: Solve for \( h \).

To simplify, we can assume \( h \) is much smaller than \( R \) (which is valid for small heights compared to the radius of the planet), thus we can neglect \( h^2 \):
\[ 25.088Rh \approx 7.056R^2 \]
\[ h \approx \frac{7.056R^2}{25.088R} = \frac{7.056R}{25.088} \]
Calculating \( h \):
\[ h \approx 0.28R \]

### Step 6: Substitute the radius of the planet.

Given that the radius of Earth \( R_e \) is approximately \( 6400 \, \text{km} \), the radius of the planet is also \( 6400 \, \text{km} \):
\[ h \approx 0.28 \times 6400 \, \text{km} \approx 1792 \, \text{km} \]

### Final Step: Conclusion.

Thus, the height above the surface of the planet where the acceleration due to gravity reduces by 36% of its value on its surface is approximately:
\[ h \approx 1792 \, \text{km} \]