Question

Two objects of equal masses placed at certain distance from each other attracts each other with a force of F. If one-third mass of one object is transferred to the other object, then the new force will be

• A. \(\frac{2}{9}F\)
• B. \(\frac{16}{9}F\)
• C. \(\frac{8}{9}F\)
• D. \(F\)

Answer 1

The Correct Option is C

The problem involves the gravitational force between two objects of equal mass. Let the mass of each object initially be \(m\) and the distance between them be \(d\). According to Newton's law of universal gravitation, the force \(F\) between these two objects can be expressed as:

\(F = \frac{G \cdot m \cdot m}{d^2} = \frac{G \cdot m^2}{d^2}\) 

Where:

• \(G\) is the gravitational constant.
• \(m\) is the mass of each object.
• \(d\) is the distance between the centers of the two objects.

Now, if one-third of the mass of one object is transferred to the other, the situation changes as follows:

• Mass of object 1: \(m - \frac{m}{3} = \frac{2m}{3}\)
• Mass of object 2: \(m + \frac{m}{3} = \frac{4m}{3}\)

The new gravitational force (\(F'\)) between the two objects becomes:

\(F' = \frac{G \cdot (\frac{2m}{3}) \cdot (\frac{4m}{3})}{d^2} = \frac{G \cdot \frac{8m^2}{9}}{d^2}\)

We can compare the new force with the original force:

\(\frac{F'}{F} = \frac{\frac{G \cdot \frac{8m^2}{9}}{d^2}}{\frac{G \cdot m^2}{d^2}} = \frac{8}{9}\)

Thus, the new force \(F'\) is:

\(F' = \frac{8}{9}F\)

Therefore, the correct option is \(\frac{8}{9}F\), which matches the given answer.

Answer 2

The correct answer is (C) : \(\frac{8}{9}F\)

Let the masses are m and distance between them is l, then
\(F=\frac{Gm^2}{I^2}\)
When 1/3^rd mass is transferred to the other then masses will be 4m/3 and 2m/3. So new force will be
\(F^′=\frac{G\frac{4m}{3}×\frac{2m}{3}}{I^2}\)
\(=\frac{8}{9}\frac{Gm^2}{I^2}=\frac{8}{9}F\)