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

To solve this problem, we need to apply Newton's law of universal gravitation, which states that the force of attraction \( F \) between two masses \( m_1 \) and \( m_2 \) separated by a distance \( r \) is given by:

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

Given that the two objects have equal masses initially, let's assume each mass is \( m \). Therefore, the initial force between the objects is:

F = \frac{G \cdot m \cdot m}{r^2} = \frac{G \cdot m^2}{r^2}

According to the problem, one-third of the mass of one object is transferred to the other. Let's denote the two masses after this transfer as \( m_1 \) and \( m_2 \). So:

• New mass of the first object, m_1 = m - \frac{m}{3} = \frac{2m}{3}
• New mass of the second object, m_2 = m + \frac{m}{3} = \frac{4m}{3}

Now we calculate the new force \( F' \) using the modified masses:

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

We can compare the new force to the original force:

\frac{F'}{F} = \frac{\frac{G \cdot \frac{8m^2}{9}}{r^2}}{\frac{G \cdot m^2}{r^2}} = \frac{8}{9}

Thus, the new force is \(\frac{8}{9}F\).

Therefore, the correct answer is:

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