At the surface of the moon, the acceleration due to the gravity of the moon is x. At a distance from the center of the moon equal to four times the radius of the moon, the acceleration due to the gravity of the moon is _____.

Respuesta :

Answer:

Gravity at a distance of 4R will be reduced to 1/16 th.

Explanation:

Given:

At the surface of the moon, the acceleration due to the gravity of the moon is x.

We have to find the gravity a t a distance of 4 times from the center of the moon.

Let the radius of the moon be "R".

And

The value of acceleration due to gravity is [tex]g_m[/tex] .

Formula:

⇒ [tex]g_m=\frac{GM}{R^2}[/tex]   ...where M is the mass of the moon.

Now

Gravity of the moon at the its surface:

⇒ [tex]g_m=\frac{GM}{R^2}[/tex]    ...equation (i)

Gravity of the moon at a distance of  [tex]4R[/tex]:

⇒ [tex]g_m_1=\frac{GM}{(4R)^2}[/tex]

⇒ [tex]g_m_1=\frac{GM}{16R^2}[/tex]   ...equation (ii)

Dividing equation (i) with (ii) to find the relationship between the two.

⇒ [tex]\frac{g_m_1}{g_m} =\frac{GM}{16R^2}\times \frac{R^2}{GM}[/tex]

⇒ [tex]\frac{g_m_1}{g_m} =\frac{1}{16}[/tex]

⇒ [tex]g_m_1 =g_m(\frac{1}{16})[/tex]

⇒ [tex]g_m_1 =x(\frac{1}{16})[/tex]    ...as gm=x at the surface.

So,

We can say that the gravity at a distance of 4R will be reduced to 1/16 th.

Answer: The acceleration due to gravity on the surface of the moon is 1.620 m/s2. 2) The radius of the Earth is 6.38 x 106 m.

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