Question

Derive the equations for a particle moving near the surface of the earth, given by

?̈= 2? cos ? ?̇

?̈= −2( cos ? ?̇ + ? sin ? ?̇)

?̈= −? + 2? sin ? ?̇

? = v/r; v=velocity; r-radius

where the earth’s rotational speed in the ?̂ direction is ?.

Answer 1

At rest we have:
V0x , V0y and V0z ,
Recall eqs. (2.43),
ẍ= 2 ω cos? ẏ
ÿ= −2 ω ( ẋcos? + ż sin?)
z̈ = - g +2 ω ẏ sin?
By integrating both ẍand z̈ , we get:
ẋ = 2 ω cos? y + V0x
ż = - g t +2 ωy sin?+ V0z
Substitute ẋ , ż in ÿ, we get:
ÿ= −2 ω[ (2 w cos? y + V0x) cos? +(- g t +2 ωy sin?+ V0z) sin?
= −2 ω[ (2 ω cos2? y + V0x cos?) +(- g t sin? +2 ω sin2?+ V0z sin?) ]
= −4 ω2cos2? y − 2 ω V0x cos? +2ω g t sin? −4 ω2 sin2? - 2 ω V0z sin?
= −2 ω V0x cos? + 2ω g t sin? -2 ω V0z sin?
ÿ= −2 ω ( V0x cos? - g t sin? + V0z sin?)
By integrating ÿ we get:

ẏ=−2 ω ( V0x cos? t - 1/2 g t2 sin? + V0z sin? t) + C1
If t = 0, then C1 = V0y
By integrating ẏ we get,
y= −2 ω ( 1/2 g t2 V0x cos? − 1/6 g t3 sin? + 1/2t2 V0z sin? ) + V0y t + C2
If t = 0, then C2 = 0
y = 1/3ω g t3 sin? - ω t2 V0x cos? − ω t2 V0z sin? + V0y t
y = t [ V0y + ω/3 g t2 sin? – t ω ( V0x cos? + V0z sin? ) ]
If ω =0, then: y = V0y t
ż = - g t +2 ω sin? y + V0z
ż = - g t + V0z +2 ω sin? . t [ V0y + ω/3 g t2 sin? – t ω (V0x cos? + V0z sin? ) ]
ż = - g t + V0z +2 ω V0y sin? . t + 2/3ω2 g sin2 ? t3 – 2 ω2 sin? t2 (V0x cos? + V0z sin? )
Put ω2 = 0, we get:
ż = - g t + V0z +2 ω V0y sin? . t
By integrating ż , we get:

Z = V0z t -g/2t2 + ω V0y sin? t2 + C3
If t = 0, then C3 = 0,
z= t [ V0z - g/2t2 + ω V0y sin? .t ]
If ω =0, then: z = V0z -g/2. t2
ẋ = 2 ω cos? y + V0x
Substitute eq. (3.6) in (3.1), we get:
ẋ = 2 ω cos? .t [V0y + ω/3 g t2 sin? – t ω (V0x cos? + V0z sin?) ] + V0x
ẋ = 2 ω V0y cos?.t + 2/3ω2 g t3 sin? cos? –2 ω2 t2 cos? (V0x cos? + V0z sin? ) +V0x

Put ω2 = 0, we get:
ẋ = 2 ω V0y cos? t + V0x
By integrating ẋ , we get:
x =V0x t + ω V0y cos? t2 + C4
İf t = 0, then C4 = 0, then:
x = t ( V0x +ω V0y cos? t )
İf ω = 0, then: x =V0x t