Question

During a solar eclipse, the Moon, Earth, and Sun all lie on the same line, with the Moon between the Earth and the Sun. Answer all questions to 3 significant figures. (a) What force is exerted by the Sun on the Moon? WebAssign will check your answer for the correct number of significant figures. Incorrect: Your answer is incorrect. Enter a number. (b) What force is exerted by the Earth on the Moon? WebAssign will check your answer for the correct number of significant figures. (c) What force is exerted by the Sun on the Earth? WebAssign will check your answer for the correct number of significant figures.

Answer 1

(a) $4.40\cdot 10^{20}N$

The distance between the Sun and the Earth is

$d_{SE}=1.496 \cdot 10^11 m$

The distance between the Earth and the Moon is

$d_{EM} = 3.84\cdot 10^8 m$

So, the distance between the Sun and the Moon, when the Moon is between the Earth and the Sun, is

$d_SM = 1.496\cdot 10^{11}m -3.84\cdot 10^8 m=1.492\cdot 10^{11} m$

So the gravitational force between the Sun and the Moon is

$F_{SM} = G \frac{M_S M_M}{d_{SM}^2}$

where

G is the gravitational constant

$M_S = 1.988 \cdot 10^{30}kg$ is the mass of the Sun

$M_M = 7.384\cdot 10^{22}kg$ is the mass of the Moon

$d_{SM}=1.492\cdot 10^{11} m$ is their distance

Substituting,

$F_{SM} = (6.67\cdot 10^{-11}) \frac{(1.988\cdot 10^{30} kg)(7.384\cdot 10^{22}kg)}{(1.492\cdot 10^{11} m)^2}=4.40\cdot 10^{20}N$

(b) $2.00\cdot 10^{20}N$

The gravitational force between the Earth and the Moon is

$F_{EM} = G \frac{M_E M_M}{d_{EM}^2}$

where

G is the gravitational constant

$M_E = 5.972 \cdot 10^{24}kg$ is the mass of the Earth

$M_M = 7.384\cdot 10^{22}kg$ is the mass of the Moon

$d_{EM}=3.84\cdot 10^{8} m$ is their distance

Substituting,

$F_{EM} = (6.67\cdot 10^{-11}) \frac{(5.972\cdot 10^{24} kg)(7.384\cdot 10^{22}kg)}{(3.84 \cdot 10^{8} m)^2}=2.00\cdot 10^{20}N$

(c) $3.54\cdot 10^{22}N$

The gravitational force between the Earth and the Sun is

$F_{ES} = G \frac{M_E M_S}{d_{ES}^2}$

where

G is the gravitational constant

$M_E = 5.972 \cdot 10^{24}kg$ is the mass of the Earth

$M_S = 1.988 \cdot 10^{30}kg$ is the mass of the Sun

$d_{SE}=1.496 \cdot 10^{11} m$ is their distance

Substituting,

$F_{ES} = (6.67\cdot 10^{-11}) \frac{(5.972\cdot 10^{24} kg)(1.988\cdot 10^{30}kg)}{(1.496 \cdot 10^{11} m)^2}=3.54\cdot 10^{22}N$