# During a solar eclipse , the moon , earth , and and sun all lie on the same line,with the moon between earth and sun. what force is exerted by the sun on the moon? what force is exerted by earth on the moon? what force is exerted by the sun on earth

Jul 5, 2015

During a solar eclipse, the forces between the sun, earth and moon are:

sun-earth: $3.56 \cdot {10}^{22} N$
earth-moon: $1.97 \cdot {10}^{20} N$
sun-moon: $4.33 \cdot {10}^{20} N$

#### Explanation:

Newton's law of gravitation states that the force exerted by a body on another body is calculated as below:

$F = \frac{G \cdot {m}_{1} \cdot {m}_{2}}{d} ^ 2$

with

$G = 6.67 \cdot {10}^{- 11} {m}^{3} {s}^{- 2} k {g}^{- 1}$ the universal gravitation constant
${m}_{1}$ the first body's mass (expressed in kg)
${m}_{2}$ the second body's mass (expressed in kg)
$d$ the distance between the two bodies' mass centers (expressed in m)

Knowing that:

${m}_{S} = 2.0 \cdot {10}^{30} k g$ is the sun's mass
${m}_{E} = 6.0 \cdot {10}^{24} k g$ is the earth's mass
${m}_{M} = 7.3 \cdot {10}^{22} k g$ is the moon's mass

${d}_{S E} = 1.5 \cdot {10}^{11} m$ is the distance between the sun and the earth
${d}_{E M} = 3.85 \cdot {10}^{8} m$ is the distance between the earth and the moon

We can calculate:

$\textcolor{red}{{F}_{S E}} = \frac{G \cdot {m}_{S} \cdot {m}_{E}}{{d}_{S E}} ^ 2$

$= \frac{6.67 \cdot {10}^{- 11} \cdot 2.0 \cdot {10}^{30} \cdot 6.0 \cdot {10}^{24}}{1.5 \cdot {10}^{11}} ^ 2$

$= \frac{6.67 \cdot 2.0 \cdot 6.0 \cdot {10}^{30 + 24 - 11}}{{1.5}^{2} \cdot {10}^{22}}$

$= \frac{6.67 \cdot 12 \cdot {10}^{43 - 22}}{2.25} = \frac{80.04}{2.25} \cdot {10}^{21} \approx \textcolor{red}{3.56 \cdot {10}^{22} N}$

$\textcolor{b l u e}{{F}_{E M}} = \frac{G \cdot {m}_{E} \cdot {m}_{M}}{{d}_{E M}} ^ 2$

$= \frac{6.67 \cdot {10}^{- 11} \cdot 6.0 \cdot {10}^{24} \cdot 7.3 \cdot {10}^{22}}{3.85 \cdot {10}^{8}} ^ 2$

$= \frac{6.67 \cdot 6.0 \cdot 7.3 \cdot {10}^{24 + 22 - 11}}{{3.85}^{2} \cdot {10}^{16}}$

$= \frac{6.67 \cdot 43.8 \cdot {10}^{35 - 16}}{14.8225} = \frac{292.146}{14.8225} \cdot {10}^{19} \approx \textcolor{b l u e}{1.97 \cdot {10}^{20} N}$

$\textcolor{p u r p \le}{{F}_{S M}} = \frac{G \cdot {m}_{S} \cdot {m}_{M}}{{d}_{S E} - {d}_{E M}} ^ 2$

$= \frac{6.67 \cdot {10}^{- 11} \cdot 2.0 \cdot {10}^{30} \cdot 7.3 \cdot {10}^{22}}{1.5 \cdot {10}^{11} - 3.85 \cdot {10}^{8}} ^ 2$

$= \frac{6.67 \cdot 2.0 \cdot 7.3 \cdot {10}^{30 + 22 - 11}}{{\left(1.5 - 0.00385\right)}^{2} \cdot {10}^{22}}$

$\approx \frac{6.67 \cdot 14.6 \cdot {10}^{41}}{{1.5}^{2} \cdot {10}^{22}}$

$= \frac{6.67 \cdot 14.6 \cdot {10}^{41 - 22}}{2.25} = \frac{97.382}{2.25} \cdot {10}^{19} \approx \textcolor{p u r p \le}{4.33 \cdot {10}^{20} N}$