Find the net gravitational force fnet acting on the earth in the sun-earth-moon system during the new moon (when the moon is loc

Question

ivann1987 [24]

3 years ago

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Find the net gravitational force fnet acting on the earth in the sun-earth-moon system during the new moon (when the moon is loc

ated directly between the earth and the sun. express your answer in newtons to three significant figures.

Physics

2 answers:

Lina20 [59] · Answer 1 · 2022-01-03T18:22:14+03:00

After thorough researching, the net gravitational force Fnet acting on the earth in the sun-earth-moon system during the new moon and when the moon is located directly between the earth and the sun is 3.50x10^22 N. The correct answer to the following given statement above is 3.50x10^22 N.

slega [8] · Answer 2 · 2022-01-03T20:31:13+03:00

The gravitational force of attraction on Earth due to the Sun and the Moon is $\boxed{3.544 \times {{10}^{22}}\,{\text{N}}}$ .

Further Explanation:

The gravitational force of attraction experienced by one body due to the other body is given by the Newton’s law of Gravitation. The newton’s law of Gravitation states that the force experienced by the bodies is directly proportional to the product of the masses of the two bodies and inversely proportional to the square of distance between the bodies.

The force experienced by the Earth due to the moon is;

${F_{EM}} = \dfrac{{G{M_E}{M_M}}}{{r_{EM}^2}}$

Here, ${F_{EM}}$ is the force experienced by Earth due to moon, ${r_{EM}}$ is the distance between Earth and Moon.

The mass of the Earth is $5.987 \times {10^{24}}\,{\text{kg}}$ .

The mass of the Moon is $7.35 \times {10^{22}}\,{\text{kg}}$ .

The distance between the Earth and Moon is $3.84 \times {10^8}\,{\text{m}}$ .

Substitute the values in the above expression.

$\begin{aligned}{F_{EM}}&=\frac{{\left( {6.674 \times {{10}^{ - 11}}} \right)\times\left( {5.987 \times {{10}^{24}}} \right) \times \left( {7.35 \times {{10}^{22}}} \right)}}{{{{\left( {3.84 \times {{10}^8}} \right)}^2}}}\\&= \frac{{2.937 \times {{10}^{37}}}}{{1.47 \times {{10}^{17}}}}\\&= 1.986 \times {10^{20}}\,{\text{N}}\\\end{aligned}$

The force experienced by the Earth due to the Sun is;

${F_{ES}} =\dfrac{{G{M_E}{M_S}}}{{r_{ES}^2}}$

Here, ${F_{ES}}$ is the force experienced by Earth due to Sun, is the distance between Earth and Sun.

The mass of the Sun is $1.99 \times {10^{30}}\,{\text{kg}}$ .

The distance between the Earth and Sun is $1.5 \times {10^{11}}\,{\text{m}}$ .

Substitute the values in the above expression.

$\begin{aligned}{F_{EM}}&= \frac{{\left( {6.674 \times {{10}^{ - 11}}} \right) \times \left( {5.987 \times {{10}^{24}}} \right) \times \left( {1.99 \times {{10}^{30}}\,{\text{kg}}} \right)}}{{{{\left( {1.5 \times {{10}^{11}}{\text{m}}}\right)}^2}}}\\&=\frac{{7.929 \times {{10}^{44}}}}{{2.25 \times {{10}^{22}}}}\\&= 3.524 \times {10^{22}}\,{\text{N}}\\\end{aligned}$

The net force acting on the Earth is given as:

${F_{{\text{net}}}} = {F_{EM}} + {F_{ES}}$

Substitute the values of forces.

$\begin{aligned}{F_{{\text{net}}}}&= 1.986 \times {10^{20}}\,{\text{N}} + 3.524\times {10^{22}}\,{\text{N}}\\&= 3.544 \times {10^{22}}\,{\text{N}} \\\end{aligned}$