Question

Take the ratio of the Moon’s acceleration to the Sun’s and comment on why the tides are predominantly due to the Moon in spite of this number.Calculate the magnitude of the acceleration due to gravity on the surface of Earth due to the Moon. (b) Calculate the magnitude of the acceleration due to gravity at Earth due to the Sun. (c) Take the ratio of the Moon's acceleration to the Sun's and comment on why the tides are predominantly due to the Moon in spite of this number.

Answer 1

Answer:

$3.31777\times 10^{-5}\ m/s^2$

$0.00612273\ m/s^2$

$\dfrac{g_{me}}{g_{es}}=0.00541$

Explanation:

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

$M_m$ = Mass of Moon = $7.35\times 10^{22}\ kg$

$M_s$ = Mass of Sun = $1.989\times 10^{30}\ kg$

$r_{me}$ = Distance between Moon and Earth = $384.4\times 10^6\ m$

$r_{es}$ = Distance between Sun and Earth = $147.2\times 10^9\ m$

Acceleration due to gravity is given by

$g_{me}=\dfrac{GM_m}{r_{me}^2}\\\Rightarrow g_{me}=\dfrac{6.67\times 10^{-11}\times 7.35\times 10^{22}}{(384.4\times 10^6)^2}\\\Rightarrow g_{me}=3.31777\times 10^{-5}\ m/s^2$

The magnitude of the acceleration due to gravity on the surface of Earth due to the Moon is $3.31777\times 10^{-5}\ m/s^2$

$g_{es}=\dfrac{GM_s}{r_{es}^2}\\\Rightarrow g_{es}=\dfrac{6.67\times 10^{-11}\times 1.989\times 10^{30}}{(147.2\times 10^{9})^2}\\\Rightarrow g_{es}=0.00612273\ m\s^2$

The magnitude of the acceleration due to gravity at Earth due to the Sun is $0.00612273\ m/s^2$

Dividing the equations we get

$\dfrac{g_{me}}{g_{es}}=\dfrac{\dfrac{6.67\times 10^{-11}\times 7.35\times 10^{22}}{(384.4\times 10^6)^2}}{\dfrac{6.67\times 10^{-11}\times 1.989\times 10^{30}}{(147.2\times 10^{9})^2}}\\\Rightarrow \dfrac{g_{me}}{g_{es}}=0.00541\\\Rightarrow g_{es}=184.54g_{me}$

The acceleration due to gravity by the Sun is 184.54 times the acceleration of the Moon.

The Moon is still responsible for the tides because of there being a difference in the gravity exerted on Earth's near and far side.