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What is the magnitude of the free-fall acceleration at a point that is a distance 2R above the surface of the Earth, where R is the radius of the Earth

Respuesta :

hamzaahmeds

Answer:

g' = g/9 = 1.09 m/s²

Explanation:

The magnitude of free fall acceleration at the surface of earth is given by the following formula:

g = GM/R²   ----- equation 1

where,

g = free fall acceleration

G = Universal Gravitational Constant

M = Mass of Earth

R = Distance between the center of earth and the object

So, in our case,

R = R + 2 R = 3 R

Therefore,

g' = GM/(3R)²

g' = (1/9) GM/R²

using equation 1:

g' = g/9

g' = (9.8 m/s)/9

g' = 1.09 m/s²

okpalawalter8

Answer:

  • The magnitude of the free-fall acceleration [tex]g_h = 1.09m/s^2[/tex]

Explanation:

Surface of earth,

[tex]g = \frac{GM}{R^2}\\\\g = 9.8m/s^2[/tex]

free fall acceleration at height h,

[tex]g_h = \frac{GM}{(R+h)^2}[/tex]

where

G = gravitational constant

R = Radius of earth

M = mass of earth

therefore,

[tex]\frac{g_h}{g} = \frac{\frac{GM}{(R+h)^2}}{\frac{GM}{R^2}}\\\\ \frac{g_h}{g} = \frac{R^2}{(R+h)^2}\\\\g_h = g\frac{R^2}{(R+h)^2}[/tex]

Where height h = 2R

[tex]g_h = 9.8\frac{R^2}{(R+2R)^2}\\\\g_h = 9.8\frac{R^2}{(3R)^2}\\\\g_h = 9.8\frac{R^2}{(9R^2}\\\\g_h = 1.09m/s^2[/tex]

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https://brainly.com/question/17162343

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