The force of attraction between two bodies separated by a distance of d is F. The distance for which the force of attraction between them is 64 F is ________.
d/8
Understanding Gravitational Force and Distance
The force of attraction between two bodies is described by Newton's Law of Universal Gravitation. This law states that the force is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers.
Mathematically, the gravitational force F between two bodies with masses $m_1$ and $m_2$ separated by a distance d is given by:
\( F = G \frac{m_1 m_2}{d^2} \)
where G is the gravitational constant.
Analyzing the Problem: Initial and Final Conditions
We are given an initial situation where the force of attraction between two bodies is F when the distance between them is d.
- Initial Force: \( F_1 = F \)
- Initial Distance: \( d_1 = d \)
- Using the formula: \( F = G \frac{m_1 m_2}{d^2} \)
We need to find the new distance, let's call it \( d' \), for which the force of attraction between the same two bodies becomes \( 64F \).
- Final Force: \( F_2 = 64F \)
- Final Distance: \( d_2 = d' \)
- Using the formula for the new distance: \( 64F = G \frac{m_1 m_2}{(d')^2} \)
Calculating the New Distance for 64F Force
To find the relationship between the initial distance d and the new distance \( d' \), we can compare the two force equations:
Initial equation: \( F = G \frac{m_1 m_2}{d^2} \quad (1) \)
Final equation: \( 64F = G \frac{m_1 m_2}{(d')^2} \quad (2) \)
Divide equation (2) by equation (1):
\( \frac{64F}{F} = \frac{G \frac{m_1 m_2}{(d')^2}}{G \frac{m_1 m_2}{d^2}} \)
Simplify the equation:
\( 64 = \frac{(d^2)}{(d')^2} \)
\( 64 = \left(\frac{d}{d'}\right)^2 \)
To solve for \( d' \), take the square root of both sides:
\( \sqrt{64} = \sqrt{\left(\frac{d}{d'}\right)^2} \)
\( 8 = \frac{d}{d'} \)
Rearrange the equation to find \( d' \):
\( d' = \frac{d}{8} \)
This calculation shows that to increase the gravitational force to 64 times the original force, the distance between the bodies must be reduced to one-eighth of the original distance.
Conclusion
The distance for which the force of attraction between the two bodies is 64 F is \( \frac{d}{8} \). This corresponds to option 4.
Revision Table: Key Concepts in Gravitational Force
| Concept | Description | Formula (Example) |
|---|---|---|
| Gravitational Force (F) | Force of attraction between any two objects with mass. | \( F = G \frac{m_1 m_2}{d^2} \) |
| Mass (\( m_1, m_2 \)) | Measure of the amount of matter in a body. Force is directly proportional to mass. | \( F \propto m_1 m_2 \) |
| Distance (d) | Distance between the centers of the two bodies. Force is inversely proportional to the square of the distance. | \( F \propto \frac{1}{d^2} \) (Inverse Square Law) |
| Gravitational Constant (G) | A universal constant relating gravitational force to mass and distance. | \( G \approx 6.674 \times 10^{-11} \text{ N m}^2/\text{kg}^2 \) |
Additional Information: Inverse Square Law of Gravitation
The relationship \( F \propto \frac{1}{d^2} \) is known as the inverse square law. This principle is fundamental not only to gravitation but also appears in other areas of physics, such as the intensity of light or sound from a point source, and the electric force between charged particles (Coulomb's Law).
Understanding the inverse square relationship is crucial. It means that if you double the distance, the force becomes \( (1/2)^2 = 1/4 \) of the original force. If you halve the distance, the force becomes \( (1/(1/2))^2 = 2^2 = 4 \) times the original force. In this problem, the force increased by a factor of 64. Since \( 64 = 8^2 \), the distance must have decreased by a factor of 8 (\( 1/8 \)).
Important Questions from Gravitation
The known forces of nature can be divided into four classes, viz., gravity, electromagnetism, weak nuclear force and strong nuclear force. With reference to them, which one of the following statements is not correct?
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