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In this problem we are going to compare the strength of the gravitational interaction between the Moon and the Earth and the Sun and the Earth. We will do this by finding the gravitational field g due to the Moon or the Sun, which is the acceleration that the Earth would have if it were interacting with each of them. A. Calculate the magnitude of the gravitational field of the moon at the location of Earth, in meters per square second..B. Calculate the magnitude of the gravitational field of the Sun at the location of Earth, in meters per square second.C. Calculate the ratio of the gravitational field of the Sun to the gravitational field of the Moon, at the location of Earth.

Respuesta :

Answer:

a)   g = moon = - 3,329 10⁻⁵ m / s², b)g_sum = - 5,934 10⁻³ m / s²

c)      g_sum / g_moon = 1.78 10²

Explanation:

a) From the law of universal gravitation, the gravitational field is

        F = - G M m / r²

the gravitational field is

        g = F / m

        g = - G M / r²

let's calculate this field the Earth and the Moon

        g_moon = - G M_moon / r²

   M_moon = 7.36 10²² kg

   r = 3.84  10⁸ m

let's calculate

      g_moon = - 6.67 10⁻¹¹ 7.36 10²² / (3.84 10⁸)²

      g = moon = - 3,329 10⁻⁵ m / s²

B) let's calculate the field of the sun in the position of the Earth

     M = 1,991 10³⁰ kg

     r = 1,496 10¹¹ m

     g_ Sun = - 6.67 10⁻¹¹ 1,991 10³⁰ / (1,496 10¹¹)²

     g_sum = - 5,934 10⁻³ m / s²

c) the relationship between the two gravitational fields is

      g_sun / g_moon = 5,934 10⁻³ / 3,329 10⁻⁵

      g_sum / g_moon = 1.78 10²

shubhamchouhanvrVT

The ratios of gravitational field of the sun and the moon are-

a)    [tex]g(moon)=-3.329\times10^{-5} \dfrac{m}{s^{2} }[/tex]

b)    [tex]g(sun)=-5.934\times10^{-3} \dfrac{m}{s^{2} }[/tex]

c)    [tex]g\dfrac{sun}{moon} =1.78\times10^{2}[/tex]

What will be the ratio of gravitational field of sun to the moon?

a) From Universal law of gravitational force

       [tex]F=-\dfrac{GMm}{r^{2} }[/tex]

the gravitational field is

        [tex]g=\dfrac{F}{m}[/tex]

     [tex]g=-\dfrac{GM}{r^{2} }[/tex]

let's calculate this field for the Earth and the Moon

     [tex]g(moon)=-\dfrac{GM_{m} }{r^{2} }[/tex]

   [tex]M(moon)=7.36\times10^{22}[/tex]kg

                 

                  [tex]r=3.84\times10^{8} m[/tex]

let's calculate

   

[tex]g(moon)=\dfrac{(-6.67\times10^{-11} \times7.36\times10^{22} )}{(3.36\times10^{8} )}[/tex]

   

[tex]g(moon)=-3.329\times10^{-5} \dfrac{m}{s^{2} }[/tex]

B) let's calculate the field of the sun in the position of the Earth

   

[tex]M(sun)=1991\times10^{30}[/tex]

   

[tex]r=1496\times10^{11}[/tex]

   

[tex]g(sun)=\dfrac{(-6.67\times10^{-11}\times1991\times10^{30} )}{1496\times10^{11}\times1496\times10^{11} }[/tex]

     

[tex]g(sun)=-5.934\times10^{-3} \dfrac{m}{s^{2} }[/tex]

c) the relationship between the two gravitational fields is

[tex]g\dfrac{sun}{moon}=\dfrac{5934\times10^{-3} }{3329\times10^{-5} }[/tex]

   

[tex]g\dfrac{sun}{moon} = 1.78\times10^{2}[/tex]

Hence the ratio of the gravitational field of the sun to the moon will be

[tex]g\dfrac{sun}{moon} =1.78\times10^{2}[/tex]

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