Two bodies of mass M each are placed R distance apart. In another system, two bodies of mass 2M each are placed R/2 distance apart. If F be the gravitational force between the bodies in the first system, then the gravitational force between the bodies in the second system will be
16 F
Calculating Gravitational Force Between Bodies
The question asks us to compare the gravitational force between two pairs of bodies based on their masses and the distance separating them. We will use Newton's Law of Universal Gravitation to solve this problem.
Understanding Newton's Law of Universal Gravitation
Newton's Law of Universal Gravitation states that the gravitational force (\(F\)) between two point masses (\(m_1\) and \(m_2\)) is directly proportional to the product of their masses and inversely proportional to the square of the distance (\(r\)) between their centers. Mathematically, the formula is:
$$F = G \frac{m_1 m_2}{r^2}$$
Here, \(G\) is the gravitational constant.
Analyzing the First System
In the first system, we have two bodies, each with mass \(M\). The distance between them is \(R\). The gravitational force between them is given as \(F\).
Using the formula, the gravitational force in the first system (\(F_1\)) is:
$$F_1 = G \frac{M \cdot M}{R^2}$$
$$F_1 = G \frac{M^2}{R^2}$$
We are told that \(F_1 = F\). So, we can write:
$$F = G \frac{M^2}{R^2} \quad \text{(Equation 1)}$$
Analyzing the Second System
In the second system, we have two bodies, each with mass \(2M\). The distance between them is \(R/2\). Let the gravitational force in this system be \(F_2\).
Using the gravitational force formula for the second system:
The masses are \(m_1' = 2M\) and \(m_2' = 2M\).
The distance is \(r' = R/2\).
$$F_2 = G \frac{m_1' m_2'}{r'^2}$$
Substitute the values of the masses and distance:
$$F_2 = G \frac{(2M)(2M)}{(R/2)^2}$$
Simplify the expression:
$$F_2 = G \frac{4M^2}{R^2/4}$$
To simplify further, invert the denominator and multiply:
$$F_2 = G \frac{4M^2}{1} \cdot \frac{4}{R^2}$$
$$F_2 = G \frac{16M^2}{R^2}$$
Comparing Forces in Both Systems
Now, we want to express \(F_2\) in terms of \(F\). We have the expression for \(F_2\):
$$F_2 = G \frac{16M^2}{R^2}$$
We can factor out 16 from the expression:
$$F_2 = 16 \left( G \frac{M^2}{R^2} \right)$$
From Equation 1, we know that \(F = G \frac{M^2}{R^2}\). Substitute \(F\) into the expression for \(F_2\):
$$F_2 = 16 F$$
Thus, the gravitational force between the bodies in the second system is 16 times the gravitational force in the first system.
Summary of Calculations
| System | Masses (\(m_1, m_2\)) | Distance (\(r\)) | Gravitational Force Formula | Calculated Force |
|---|---|---|---|---|
| First System | \(M, M\) | \(R\) | \(F_1 = G \frac{m_1 m_2}{r^2}\) | \(F_1 = G \frac{M \cdot M}{R^2} = G \frac{M^2}{R^2} = F\) |
| Second System | \(2M, 2M\) | \(R/2\) | \(F_2 = G \frac{m_1' m_2'}{r'^2}\) | \(F_2 = G \frac{(2M)(2M)}{(R/2)^2} = G \frac{4M^2}{R^2/4} = G \frac{16M^2}{R^2}\) |
Comparing \(F_2\) and \(F\):
$$F_2 = G \frac{16M^2}{R^2} = 16 \left( G \frac{M^2}{R^2} \right) = 16 F$$
Conclusion
The gravitational force between the bodies in the second system is \(16F\).
Revision Table: Gravitational Force Concepts
| Concept | Description | Impact on Force (all else constant) |
|---|---|---|
| Masses (\(m_1, m_2\)) | Product of the two interacting masses. | Directly proportional. If mass increases, force increases. |
| Distance (\(r\)) | Distance between the centers of the two bodies. | Inversely proportional to the square of the distance. If distance increases, force decreases rapidly. |
| Gravitational Constant (\(G\)) | A fundamental constant relating gravitational force to mass and distance. | A fixed value, approximately \(6.674 \times 10^{-11} \, \text{N} \cdot \text{m}^2/\text{kg}^2\). |
Additional Information: Factors Affecting Gravitational Force
The gravitational force is a fundamental force of nature that attracts any two objects with mass towards each other. The strength of this force depends primarily on two factors:
- Mass of the Objects: The greater the mass of the objects, the stronger the gravitational pull between them. The force is directly proportional to the product of their masses. Doubling one mass doubles the force, doubling both masses quadruples the force.
- Distance Between the Objects: The distance between the centers of the objects significantly affects the gravitational force. This relationship follows an inverse square law. If the distance is doubled, the force becomes four times weaker ($1/2^2 = 1/4$). If the distance is halved, the force becomes four times stronger ($1/(1/2)^2 = 1/(1/4) = 4$).
In this problem, both the masses and the distance were changed simultaneously, leading to a combined effect on the gravitational force.
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