The gravitational force of attraction between two bodies is ‘F’ N. If each mass and distance between them is doubled, then the force of attraction between them is:
remains same
Understanding Gravitational Force and Its Dependence
The question asks about the change in gravitational force between two bodies when both their masses and the distance separating them are doubled. To solve this, we need to use Newton's Law of Universal Gravitation.
Newton's Law of Universal Gravitation
According to Newton's Law of Universal Gravitation, the force of attraction between two bodies is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers.
The formula for gravitational force (\(F\)) is given by:
\(F = G \frac{m_1 m_2}{r^2}\)
Where:
- \(G\) is the gravitational constant
- \(m_1\) is the mass of the first body
- \(m_2\) is the mass of the second body
- \(r\) is the distance between the centers of the two bodies
Analyzing the Changes
Initially, let the masses be \(m_1\) and \(m_2\), and the distance be \(r\). The initial force is given as \(F\).
The changes described in the question are:
- Each mass is doubled: New mass \(m_1' = 2m_1\) and \(m_2' = 2m_2\).
- The distance between them is doubled: New distance \(r' = 2r\).
Calculating the New Force
Let the new force be \(F'\). We can calculate \(F'\) by substituting the new values of masses and distance into the gravitational force formula:
\(F' = G \frac{m_1' m_2'}{r'^2}\)
Substitute \(m_1' = 2m_1\), \(m_2' = 2m_2\), and \(r' = 2r\):
\(F' = G \frac{(2m_1)(2m_2)}{(2r)^2}\)
Simplify the expression:
\(F' = G \frac{4m_1 m_2}{4r^2}\)
We can cancel out the '4' from the numerator and the denominator:
\(F' = G \frac{m_1 m_2}{r^2}\)
Comparing Original and New Force
By comparing the expression for \(F'\) with the original formula for \(F\), we see that:
\(F' = F\)
This means the new force of attraction is equal to the original force of attraction.
Therefore, when each mass and the distance between them are doubled, the gravitational force of attraction between them remains the same.
Summary of Calculation Steps
| Parameter | Original Value | New Value | Change Factor |
|---|---|---|---|
| Mass 1 | \(m_1\) | \(m_1' = 2m_1\) | 2 |
| Mass 2 | \(m_2\) | \(m_2' = 2m_2\) | 2 |
| Distance | \(r\) | \(r' = 2r\) | 2 |
| Force | \(F = G \frac{m_1 m_2}{r^2}\) | \(F' = G \frac{(2m_1)(2m_2)}{(2r)^2} = G \frac{4m_1 m_2}{4r^2} = G \frac{m_1 m_2}{r^2}\) | 1 |
The calculation shows that the new force \(F'\) is exactly the same as the original force \(F\).
Revision Table: Gravitational Force Factors
| Factor | Relationship with Force \(F\) | Impact of Doubling |
|---|---|---|
| Mass 1 (\(m_1\)) | Directly Proportional (\(F \propto m_1\)) | Force would double (if only \(m_1\) changes) |
| Mass 2 (\(m_2\)) | Directly Proportional (\(F \propto m_2\)) | Force would double (if only \(m_2\) changes) |
| Distance Squared (\(r^2\)) | Inversely Proportional (\(F \propto \frac{1}{r^2}\)) | Force would be quartered (if only \(r\) changes) |
Additional Information: Gravitational Force Concepts
- Inverse Square Law: Newton's law follows an inverse square law with respect to distance. This means if the distance is doubled, the force becomes \(1/(2^2) = 1/4\) times the original force. If the distance is tripled, the force becomes \(1/(3^2) = 1/9\) times, and so on.
- Dependence on Mass: The force is directly proportional to the product of the masses. If one mass is doubled, the force doubles. If both masses are doubled, the product of masses becomes \((2m_1)(2m_2) = 4m_1 m_2\), making the force four times stronger (if distance remains constant).
- Combined Effect: In this specific problem, both the numerator (due to masses doubling) and the denominator (due to distance doubling and squaring) change by a factor of 4. The \(4\) in the numerator from \((2m_1)(2m_2)\) is cancelled by the \(4\) in the denominator from \((2r)^2\), resulting in no net change in the force.
Important Questions from Gravitation
Which of the following forces is responsible for the tides, due to the Moon and the Sun?
The weight of an object was 60 N when measured on the surface of the earth. What would be its weight when measured on the surface of the moon?
Seven people, A, B, C, L, X, Y, and Z are sitting in a row, facing north. No one sits to the right of Y. Only three people sit between Y and C. Only two people sit between C and Z. B sits third to the left of X. L sits to the immediate right of X.
How many people sit between A and Z?
The universal constant of gravitation G has the unit
- An object's apparent weight is slightly less at the Earth's equator compared to its poles. This difference is primarily attributed to:
