At what height above the surface of the Earth will the acceleration due to gravity be reduced to g/16 ?

At what height above the surface of the Earth will the acceleration due to gravity be reduced to \( g/16 \)?

A) \( 3R \)
B) \( 4R \)
C) \( 5R \)
D) \( 2R \)

... Answer is A)

To find the height where the acceleration due to gravity is reduced to \( g/16 \), we use the formula:

\( g_h = g \left( \frac{R}{R + h} \right)^2 \)

Here, \( g_h = \frac{g}{16} \). Substituting into the equation:

\( \frac{g}{16} = g \left( \frac{R}{R + h} \right)^2 \)

Cancel \( g \) on both sides:

\( \frac{1}{16} = \left( \frac{R}{R + h} \right)^2 \)

Take the square root of both sides:

\( \frac{1}{4} = \frac{R}{R + h} \)

Rearranging for \( R + h \):

\( R + h = 4R \)

Solve for \( h \):

\( h = 4R - R = 3R \)

Thus, the height above the Earth's surface where gravity is reduced to \( g/16 \) is:

\( h = 3R \)

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