By Fowles G.R., Cassiday G.L.

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**Example text**

N . The same holds true for the trigonometric terms in f ( t ) . Therefore, when terms that cancel in the summations are discarded: 28 1 T f ( t ) = c + ∑ ∫ 2T f ( t ) cos ( nω t ) dt cos nω t − n T 2 T 1 + ∑ ∫ 2T f ( t ) sin ( nω t ) dt sin nω t , − n T 2 1 T2 n = ±1, ±2, . , and c = ∫ T f ( t ) dt T −2 Now, due to the equality of terms in ± n : 2 T2 f ( t ) = c + ∑ ∫ T f ( t ) cos ( nω t ) dt cos nω t − n T 2 T 2 + ∑ ∫ 2T f ( t ) sin ( nω t ) dt sin nω t , − n T 2 n = 1, 2,3, .

17 mx = Fx = − x = A cos ∂V = − kx = −π 2 mx ∂x k t + α = A cos (π t + α ) m 45 ∂V = −4π 2 mx ∂y y = B cos ( 2π t + β ) my = − ∂V = −9π 2 mz ∂z z = C cos ( 3π t + γ ) mz = − Since x = y = z = 0 at t = 0 , π α = β =γ = − π 2 x = A cos π t − = A sin π t 2 x = Aπ cos π t Since v 2 = x 2 + y 2 + z 2 and x = y = z , v x = = Aπ 3 v A= π 3 v x= sin π t π 3 y = B sin 2π t , y = 2 Bπ cos 2π t v = 2π B y = 3 v B= 2π 3 v sin 2π t y= 2π 3 z = C sin 3π t , z = 3Cπ cos 3π t v = 3Cπ z = 3 v C= 3π 3 v sin 3π t z= 3π 3 Since ω x = π , ω y = 2π , and ω z = 3π the ball does retrace its path.

A) In the x, y rotating frame of reference x ( t ) = R cos ( Ω − ω ) t − Rε y ( t ) = − R sin ( Ω − ω ) t where R is the radius of the asteroid’s orbit and RE is the radius of the Earth’s orbit. Ω is the angular frequency of the Earth’s revolution about the Sun and ω is the angular frequency of the asteroid’s orbit. (b) x ( t ) = − ( Ω − ω ) R sin ( Ω − ω ) t → 0 at t = 0 y ( t ) = − ( Ω − ω ) R cos ( Ω − ω ) t → − ( Ω − ω ) R at t = 0 (c) K K K K K K K K K K a = A − Aε − Ω × r − 2Ω × r − Ω × Ω × r K Where a is the acceleration of the asteroid in the x, y frame of reference, 53 K K A, Aε are the accelerations of the asteroid and the Earth in the fixed, inertial frame of reference.