267×10 (b) The ratio of the magnitude of this acceleration and g is 81=272x109 (c)Since the acceleration is a=or=/2T 4r2 The orbital period is 4zr_4z23_|4×3142×(1×105)2 GMV667×10-×4×10x≈3.84×10-(s) 4. The magnitude of the total gravitational field at the point 1.60×10°m Moon P in Figure 2 is_ 2.37x10- m/s-_, the magnitude of the P9597.36×102kg acceleration experienced by a 4.00 kg salt lick at point P 2.37x10 m/s-, the magnitude of the total gravitational force on the salt lick if it is placed at P 9.48 N 4.16×108m Solution (a)The total gravitational field at the point P is gota =8M+ge=2Mi+2E(cos0i+sin]) ③o Fig 2 667×101×736×102:667×10-11×598×102416×108:3.84×103 (1.6×10 (416×103 4.16×1084.16×103 192×10-i+23×10(0.38+0.92j) (1.07×10-)+(2.12×10-)j The magnitude of the total gravitational field at the point P is gmu=2.37×103m/s (b) The acceleration has no relation with anything at some point. So the magnitude of the acceleration experienced by a 4.00 kg salt lick at point P is also equal to(a) a=gmu=237×10-3m2 (c)The magnitude of the total gravitational force on the salt lick if it is placed at P is F=ma=4×2.37=948N I. Give the solutions of the following problems(b) The ratio of the magnitude of this acceleration and g is 9 10 2.72 10 9.81 2.67 10 = × × = g a (c) Since the acceleration is 2 2 2 2 2 4 T r r T a r π π ω ⎟ = ⎠ ⎞ ⎜ ⎝ ⎛ = = The orbital period is 3.84 10 ( ) 6.67 10 4 10 4 4 4 3.14 (1 10 ) 5 11 30 2 2 3 2 5 2 s GM r a r T − − = × × × × × × × = = = π π 4. The magnitude of the total gravitational field at the point P in Figure 2 is 2.37×10-3 m/s2 ,the magnitude of the acceleration experienced by a 4.00 kg salt lick at point P is 2.37×10-3 m/s2 , the magnitude of the total gravitational force on the salt lick if it is placed at P is 9.48 N . Solution: (a) The total gravitational field at the point P is )ˆ ˆ (cos ˆ sin 2 2 i j r GM i r GM g g g PE E PM M total = M + E = + θ + θ r r r i j i i j i i j ˆ (2.12 10 ) ˆ (1.07 10 ) ) 92 ˆ 0. 38ˆ 2.3 10 (0. 92 10 ˆ 1. )ˆ 4.16 10 3.84 10 ˆ 4.16 10 1.6 10 ( (4.16 10 ) 6.67 10 5.98 10 ˆ (1.6 10 ) 6.67 10 7.36 10 3 3 4 3 8 8 8 8 8 2 11 24 8 2 11 22 − − − − − − = × + × = × + × + × × + × × × × × × + × × × × = The magnitude of the total gravitational field at the point P is 3 2 2.37 10 m/s − gtotal = × (b) The acceleration has no relation with anything at some point. So the magnitude of the acceleration experienced by a 4.00 kg salt lick at point P is also equal to (a). 3 2 2.37 10 m/s − a = gtotal = × (c)The magnitude of the total gravitational force on the salt lick if it is placed at P is F = ma = 4× 2.37 = 9.48 N III. Give the Solutions of the Following Problems Moon Earth 7.36×1022kg 5.98×1024kg 1.60×108 m 4.16×108 m 90° P Fig.2 i ˆ j ˆ