Question

The Moon and Earth rotate about their common center of mass, which is located about RcM 4700 km from the center of Earth. (This is 1690 km below the surface.) Take the distance from the center of the earth to the center of the moon to be 3.84x108 m, and the mass of the moon to be 7.35x1022 kg Part (a) Write an expression for the acceleration of Earth due to the Moon's gravity using M as the mass of the moon and R as the distance between the center of the Earth and the center of the Moon. Expression: ам- Select from the variables below to write your expression. Note that all variables may not be required. a, B, T, 0, d, g. G, h, j, k, M, P, R, RCM t Part (b) Calculate the acceleration of Earth due to the Moon's gravity in ms Numeric :A numeric value is expected and not an expression. ам Part (c) Calculate the centripetal acceleration of the center of Earth as it rotates about their common center of mass once each lunar month (about 27.3 d) in m/s2 Numeric :A numeric value is expected and not an expression.

Answer 1

To solve this problem it is necessary to apply the concepts related to gravity as an expression of a celestial body, as well as the use of concepts such as centripetal acceleration, angular velocity and period.

PART A) The expression to find the acceleration of the earth due to the gravity of another celestial body as the Moon is given by the equation

$g = \frac{GM}{(d-R_{CM})^2}$

Where,

G = Gravitational Universal Constant

d = Distance

M = Mass

$R_{CM} =$ Radius earth center of mass

PART B) Using the same expression previously defined we can find the acceleration of the moon on the earth like this,

$g = \frac{GM}{(d-R_{CM})^2}$

$g = \frac{(6.67*10^{-11})(7.35*10^{22})}{(3.84*10^8-4700*10^3)^2}$

$g = 3.4*10^{-5}m/s^2$

PART C) Centripetal acceleration can be found throughout the period and angular velocity, that is

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

At the same time we have that centripetal acceleration is given as

$a_c = \omega^2 r$

Replacing

$a_c = (\frac{2\pi}{T})^2 r$

$a_c = (\frac{2\pi}{26.3d(\frac{86400s}{1days})})^2 (4700*10^3m)$

$a_c = 3.34*10^{-5}m/s^2$