Answer:
[tex]d = 384,500\,km + (21,500\,km)\cdot \cos \left[\frac{2\pi}{27.3}\cdot (t-1) \right][/tex]
Step-by-step explanation:
The trigonometric model of the distance between Earth and the Moon is:
[tex]d = A + \Delta A \cdot \cos (\omega\cdot t + \phi)[/tex]
Where:
[tex]A[/tex] - Apogee, measured in kilometers.
[tex]\Delta A[/tex] - Amplitude, measured in kilometers.
[tex]\omega[/tex] - Angular frequency, measured in radians.
[tex]\phi[/tex] - Phase angle, measured in radians.
[tex]t[/tex] - Time, measured in days.
The required information are derived below:
[tex]A = \frac{406,000\,km+363,000\,km}{2}[/tex]
[tex]A = 384,500\,km[/tex]
[tex]\Delta A = \frac{406,000\,km-363,000\,km}{2}[/tex]
[tex]\Delta A = 21,500\,km[/tex]
[tex]\omega = \frac{2\pi}{27.3}[/tex]
[tex]\phi = -\frac{2\pi}{27.3}[/tex]
The expression is:
[tex]d = 384,500\,km + (21,500\,km)\cdot \cos \left[\frac{2\pi}{27.3}\cdot (t-1) \right][/tex]
Answer:
D(t) = 21500 cos (2pi/27.3(t-1)) + 384500
Step-by-step explanation:
