Radius of earth is around 6000 km . The weight of body at height of 6000 km from earth surface becomes

Options

A. Half
B. One-fourth
C. One third
D. No change

Answer

To find the weight of a body at a height of 6000 km from the Earth's surface, we can use the formula for gravitational force, which is given by:

\[ F = \frac{G \cdot m_1 \cdot m_2}{r^2} \]

where:
- \( F \) is the gravitational force (weight of the body),
- \( G \) is the universal gravitational constant (\( 6.674 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \)),
- \( m_1 \) is the mass of the Earth,
- \( m_2 \) is the mass of the body,
- \( r \) is the distance from the center of the Earth to the body.

### Step 1: Determine the total distance from the center of the Earth
The radius of the Earth is approximately 6000 km. If the body is at a height of 6000 km above the Earth's surface, the total distance \( r \) from the center of the Earth to the body is:

\[ r = \text{Radius of Earth} + \text{Height above surface} = 6000 \, \text{km} + 6000 \, \text{km} = 12000 \, \text{km} \]

Convert this distance into meters (since 1 km = 1000 m):

\[ r = 12000 \, \text{km} = 12000 \times 1000 \, \text{m} = 12000000 \, \text{m} \]

### Step 2: Calculate the weight at the new distance
The weight of the body at this height can be calculated using the formula for gravitational force:

\[ F = \frac{G \cdot m_1 \cdot m_2}{r^2} \]

However, we can simplify the calculation by using the ratio of the weights at the surface and at height \( h \):

\[ \frac{F_h}{F_0} = \left(\frac{R}{R + h}\right)^2 \]

where:
- \( F_h \) is the weight at height \( h \),
- \( F_0 \) is the weight at the surface,
- \( R \) is the radius of the Earth,
- \( h \) is the height above the surface.

In our case, \( R = 6000 \, \text{km} \) and \( h = 6000 \, \text{km} \):

\[ \frac{F_h}{F_0} = \left(\frac{6000}{6000 + 6000}\right)^2 = \left(\frac{6000}{12000}\right)^2 = \left(\frac{1}{2}\right)^2 = \frac{1}{4} \]

### Step 3: Conclusion
This means that the weight of the body at a height of 6000 km is one-fourth of its weight at the surface of the Earth.

If we denote the weight of the body at the surface as \( W_0 \), then:

\[ F_h = \frac{1}{4} W_0 \]

### Final Answer
The weight of the body at a height of 6000 km from the Earth's surface becomes one-fourth of its weight at the surface.