Physics, 19.09.2019 20:00 annarain2004

Find the net gravitational force fnet acting on the earth in the sun-earth-moon system during the new moon (when the moon is located directly between the earth and the sun. express your answer in newtons to three significant figures.

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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.

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

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

Learn More:

1. Calculate the total force on the earth due to Venus, Jupiter and Saturn

2. A rocket being thrust upward as the force of the fuel being burned pushes downward

3. A 50-kg meteorite moving at 1000 m/s strikes earth. Assume the velocity is along the line

Answer Details:

Grade: Senior School

Subject: Physics

Chapter: Newton’s Law of Gravitation

Keywords: Gravitation, sun-moon-earth, gravitational force, new moon, directly between the earth and Sun, newton’s law, 3.544x10^22, net force.

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