A body hanging from a massless spring stretches it by 3 cm on earth's surface. At a place 800 km above the earth's surface, the same body will stretch the spring by
(Radius of Earth = 6400 km)

Options

A. `((34)/(27))cm`
B. `((64)/(27))cm`
C. `((27)/(64))cm`
D. `((27)/(34))cm`

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the problem
We have a body hanging from a massless spring that stretches it by 3 cm at the Earth's surface. We need to find out how much it will stretch when the same body is at a height of 800 km above the Earth's surface.

### Step 2: Identify the known values
- Stretch of spring on Earth's surface, \( x = 3 \) cm
- Height above Earth's surface, \( h = 800 \) km
- Radius of Earth, \( R = 6400 \) km

### Step 3: Calculate the acceleration due to gravity at height \( h \)
The formula for the acceleration due to gravity at a height \( h \) above the Earth's surface is given by:

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

Where:
- \( g \) is the acceleration due to gravity at the Earth's surface.
- \( R \) is the radius of the Earth.
- \( h \) is the height above the Earth's surface.

### Step 4: Substitute the values
First, we need to convert the height into the same units as the radius (km to m):
- \( h = 800 \) km = \( 800,000 \) m
- \( R = 6400 \) km = \( 6,400,000 \) m

Now, substituting the values into the equation:

\[ g' = \frac{g \cdot (6400 \times 10^3)^2}{(6400 \times 10^3 + 800 \times 10^3)^2} \]

### Step 5: Simplify the equation
Calculating \( R + h \):

\[ R + h = 6400 \times 10^3 + 800 \times 10^3 = 7200 \times 10^3 \text{ m} \]

Now substituting back into the equation for \( g' \):

\[ g' = \frac{g \cdot (6400 \times 10^3)^2}{(7200 \times 10^3)^2} \]

### Step 6: Relate the spring stretch at height \( h \) to the stretch at the surface
Using Hooke's law, we know that the force exerted by the spring is equal to the weight of the body:

\[ Kx = mg \quad \text{(at Earth's surface)} \]
\[ Kx' = mg' \quad \text{(at height \( h \))} \]

Where \( K \) is the spring constant, \( x \) is the stretch at the surface, and \( x' \) is the stretch at height \( h \).

### Step 7: Set up the ratio of stretches
From the above equations, we can set up the ratio:

\[ \frac{x'}{x} = \frac{g'}{g} \]

### Step 8: Calculate \( x' \)
Rearranging gives us:

\[ x' = x \cdot \frac{g'}{g} \]

Substituting \( x = 3 \) cm:

\[ x' = 3 \cdot \frac{g'}{g} \]

### Step 9: Substitute \( g' \) in terms of \( g \)
Using the ratio we derived earlier:

\[ \frac{g'}{g} = \left(\frac{R}{R + h}\right)^2 = \left(\frac{6400}{7200}\right)^2 = \left(\frac{64}{72}\right)^2 = \left(\frac{8}{9}\right)^2 = \frac{64}{81} \]

### Step 10: Final calculation
Now substituting back:

\[ x' = 3 \cdot \frac{64}{81} = \frac{192}{81} \text{ cm} \approx 2.37 \text{ cm} \]

### Conclusion
The stretch of the spring at a height of 800 km above the Earth's surface will be approximately **2.37 cm**.

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