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.