if two bodies of masses 1 kg and 4kg are realesed from the heights where gravitational force on them is equal, then find the height of the heavier body if the lighter body is dropped from a height of 1 km. Take radius of earth as 6400 km.

To solve this problem, we need to understand how gravitational force varies with height above the Earth's surface. The gravitational force acting on an object decreases as the distance from the center of the Earth increases. The formula for gravitational force is given by Newton's law of universal gravitation:

Understanding Gravitational Force

The gravitational force \( F \) between two masses \( m_1 \) and \( m_2 \) is described by the equation:

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

Where:

• G is the gravitational constant, approximately \( 6.674 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \).
• r is the distance from the center of the Earth to the object.

Setting Up the Problem

In this scenario, we have two bodies: one with a mass of 1 kg and the other with a mass of 4 kg. The problem states that they are released from heights where the gravitational force on them is equal. The lighter body (1 kg) is dropped from a height of 1 km (or 1000 m) above the Earth's surface.

The radius of the Earth is given as 6400 km (or 6400000 m). Therefore, the distance from the center of the Earth to the lighter body when it is at a height of 1 km is:

r_1 = 6400000 \, \text{m} + 1000 \, \text{m} = 6401000 \, \text{m}

Calculating Gravitational Force on the Lighter Body

The gravitational force acting on the 1 kg body at this height can be calculated as follows:

F_1 = G \frac{m_1 \cdot M}{r_1^2}

Where \( M \) is the mass of the Earth (approximately \( 5.972 \times 10^{24} \, \text{kg} \)).

Finding the Height of the Heavier Body

Now, we need to find the height \( h \) from which the 4 kg body is released, such that the gravitational force acting on it is equal to that acting on the 1 kg body. The distance from the center of the Earth for the heavier body will be:

r_2 = 6400000 \, \text{m} + h

Setting the forces equal gives us:

G \frac{1 \cdot M}{6401000^2} = G \frac{4 \cdot M}{(6400000 + h)^2}

We can simplify this equation by canceling \( G \) and \( M \) from both sides:

\frac{1}{6401000^2} = \frac{4}{(6400000 + h)^2}

Cross-Multiplying to Solve for h

Cross-multiplying gives us:

(6400000 + h)^2 = 4 \cdot 6401000^2

Taking the square root of both sides:

6400000 + h = 2 \cdot 6401000

Now, solving for \( h \):

h = 2 \cdot 6401000 - 6400000

h = 12802000 - 6400000

h = 6402000 \, \text{m} = 6402 \, \text{km}

Final Result

The height from which the heavier body (4 kg) must be released for the gravitational force to be equal to that on the lighter body (1 kg) dropped from 1 km is approximately 6402 km above the Earth's surface.