Answer:
Explanation:
Given
x(t) = r cos(ωt)
y(t) = r sin(ωt)
where r = 3.84×10⁸ m and ω = 2.46×10-⁶radians/s
Average velocity = Δx /Δt
At t = 0s
x(0) = 3.84×10⁸×cos(2.46×10-⁶×0)
x(0) = 3.84×10⁸×1 = 3.84×10⁸m
At t = 3.92days = 3.92×86400s = 3.39×10⁵s
x(3.39×10⁵) = 3.84×10⁸×cos(2.46×10-⁶×3.39×10⁵) = –2.58×10⁸m
Δx = x(3.39×10⁵) – x(0) = –2.58×10⁸–3.84×10⁸m = –6.42×10⁸m
Δt = 3.39×10⁵ – 0 = 3.39×10⁵
vx = –6.42×10⁸/3.39×10⁵ = –1894m/s
At t = 0s
y(0) = 3.84×10⁸×sin(2.46×10-⁶×0)
y(0) = 3.84×10⁸×0= 0m
At t = 3.92days = 3.92×86400s = 3.39×10⁵s
y(3.39×10⁵) = 3.84×10⁸×sin(2.46×10-⁶×3.39×10⁵) = 2.84×10⁸m
Δy = y(3.39×10⁵) – y(0) = 2.84×10⁸ – 0 = 2.84×10⁸m
Δt = 3.39×10⁵ – 0 = 3.39×10⁵
vy = 2.84×10⁸/3.39×10⁵= 838m/s
vy = 838m/s
v = √(vx² + vy²) = √((-1894)² + 838²)
v = 2071m/s
θ = tan-¹(vy/vx) = tan-¹(838/1894) = -23.9°