Question

Three identical particle $A, B$ and$ C$ of mass $100 \,kg$ each are placed in a straight line with $AB = BC =13 \,m$ The gravitational force on a fourth particle $P$ of the same mass is $F$, when placed at a distance $13\, m$ from the particle B on the perpendicular bisector of the line $AC$ The value of $F$ will be approximately :

• A. 21 G
• B. 100 G
• C. 59 G
• D. 42 G

Answer 1

The Correct Option is B

To find the gravitational force on a fourth particle \( P \) when placed at a distance of \( 13 \, \text{m} \) from the particle \( B \) on the perpendicular bisector of the line \( AC \), we'll follow these steps using Newton's law of universal gravitation.

1. According to Newton's law of gravitation, the gravitational force between two masses \( m_1 \) and \( m_2 \) is given by: \(F = \frac{G \cdot m_1 \cdot m_2}{r^2}\) where \( G \) is the gravitational constant and \( r \) is the distance between the masses.
2. Consider particles \( A, B, \) and \( C \) aligned in a straight line with \( AB = BC = 13 \, \text{m} \). The fourth particle \( P \) is placed at a distance of \( 13 \, \text{m} \) from particle \( B \) on the perpendicular bisector of the line \( AC \). Thus, \( P \) is also equidistant from \( A \) and \( C \).
3. Assuming \( M = 100 \, \text{kg} \) for each particle \( A, B, C, \) and \( P \), we find \( P \) is positioned on the perpendicular bisector forming a right triangle with \( AB \) and \( BC \). The distances \( PA = PC = \sqrt{13^2 + 6.5^2} = \sqrt{169 + 42.25} = \sqrt{211.25} \approx 14.53 \, \text{m}\).
4. The force due to \( B \): \(F_B = \frac{G \cdot M \cdot M}{13^2} = \frac{G \cdot 100 \cdot 100}{169} = \frac{10000G}{169} \approx 59.17G\)
5. The force on \( P \) due to \( A \) and \( C \) (which are equidistant) add up vectorially. Since they are symmetrically placed and \( PA = PC \): \(F_{A} = F_{C} = \frac{G \cdot 100 \cdot 100}{(14.53)^2} \approx \frac{10000G}{211.25} \approx 47.34G\) The horizontal components cancel, and the vertical components add up: \(F_{\text{net}, AC} = 2 \cdot F_A \cdot \frac{6.5}{14.53} \approx 2 \times 47.34G \times 0.447 \approx 42.34G\)
6. Total gravitational force \( F \): \(F_{\text{total}} = F_B + F_{\text{net}, AC} \approx 59.17G + 42.34G = 101.51G \approx 100G\)

Thus, the gravitational force on the particle \( P \) is approximately 100 G.