Question

The moon's distance from Earth varies in a periodic way that can be modeled by a trigonometric function.

Answer 1

Answer:

$d = 384,500\,km + (21,500\,km)\cdot \cos \left[\frac{2\pi}{27.3}\cdot (t-1) \right]$

Step-by-step explanation:

The trigonometric model of the distance between Earth and the Moon is:

$d = A + \Delta A \cdot \cos (\omega\cdot t + \phi)$

Where:

$A$ - Apogee, measured in kilometers.

$\Delta A$ - Amplitude, measured in kilometers.

$\omega$ - Angular frequency, measured in radians.

$\phi$ - Phase angle, measured in radians.

$t$ - Time, measured in days.

The required information are derived below:

$A = \frac{406,000\,km+363,000\,km}{2}$

$A = 384,500\,km$

$\Delta A = \frac{406,000\,km-363,000\,km}{2}$

$\Delta A = 21,500\,km$

$\omega = \frac{2\pi}{27.3}$

$\phi = -\frac{2\pi}{27.3}$

The expression is:

$d = 384,500\,km + (21,500\,km)\cdot \cos \left[\frac{2\pi}{27.3}\cdot (t-1) \right]$