Net gravitational force at the center of a square is found to be F1 when four particles having mass M, 2M, 3M and 4M are placed at the four corners of the square as shown in figure and it is F2 when the positions of 3M and 4M are interchanged. The ratio F1F2 is 5. The value of is . [figure]
- 2
- 2 5
- 1
- 3
Answer
(1) 2
Solution
Let a test mass m be at the centre, so each corner is at the same distance r from it, and let g0 = GMmr2. Each mass pulls m towards itself.
Case 1 (4M top-left, 3M top-right, M bottom-left, 2M bottom-right): along diagonal 4M–2M the net pull is (4 - 2)g0 = 2g0 towards 4M, and along diagonal 3M–M it is (3 - 1)g0 = 2g0 towards 3M. The diagonals are perpendicular, so
F1 = (2g0)2 + (2g0)2 = 2 2 g0
Case 2 (3M and 4M interchanged): along diagonal 3M–2M the net pull is (3 - 2)g0 = g0, and along diagonal 4M–M it is (4 - 1)g0 = 3g0, so
F2 = g02 + 9g02 = 10 g0
Hence F1F2 = 2 2 10 = 2 5
= 2
Answer: (1)