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ivann1987 [24]
3 years ago
14

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]3 years ago
8 0
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]3 years ago
5 0

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 brainly.com/question/2887352

2. A rocket being thrust upward as the force of the fuel being burned pushes downward brainly.com/question/11411375

3. A 50-kg meteorite moving at 1000 m/s strikes earth. Assume the velocity is along the line brainly.com/question/6536722

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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An air hockey game has a puck of mass 30 grams and a diameter of 100 mm. The air film under the puck is 0.1 mm thick. Calculate
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time required after impact for a puck is 2.18 seconds

Explanation:

given data

mass = 30 g = 0.03 kg

diameter = 100 mm = 0.1 m

thick = 0.1 mm = 1 ×10^{-4} m

dynamic viscosity = 1.75 ×10^{-5} Ns/m²

air temperature = 15°C

to find out

time required after impact for a puck to lose 10%

solution

we know velocity varies here 0 to v

we consider here initial velocity = v

so final velocity = 0.9v

so change in velocity is du = v

and clearance dy = h

and shear stress acting on surface is here express as

= µ \frac{du}{dy}

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= µ  \frac{v}{h}   ............1

put here value

= 1.75×10^{-5} × \frac{v}{10^{-4}}

= 0.175 v

and

area between air and puck is given by

Area = \frac{\pi }{4} d^{2}

area  =  \frac{\pi }{4} 0.1^{2}

area = 7.85 × \frac{v}{10^{-3}} m²

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force on puck is express as

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force = 0.175 v × 7.85 × 10^{-3}

force = 1.374 × 10^{-3} v    

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force = mass × acceleration

- force = mass \frac{dv}{dt}

- 1.374 × 10^{-3} v = 0.03 \frac{0.9v - v }{t}

t =  \frac{0.1 v * 0.03}{1.37*10^{-3} v}

time = 2.18

so time required after impact for a puck is 2.18 seconds

3 0
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