Question

At a certain height above the Earth's surface, the acceleration due to gravity is 1% of its value at the surface of the earth. Determine the height.

Answer 1

Answer:

• The height (h) above the earth's surface is 9 R.

Given:

1. Let the radius of earth be ' R '
2. Acceleration due to gravity is 1 % from the earth's surface.

Explanation:

$\rule{300}{1.5}$

From the formula we know,

$\large \bigstar\; \boxed{\tt g_h = \dfrac{g}{\Bigg(1 + \dfrac{h}{R}\Bigg)^2}}$

$\mathfrak{Here}\begin{cases}\text{g Denotes Acceleration due to gravity at surface} \\ \sf{g_h}\text{ Denotes Acceleration due to gravity at height h} \\ \text{R Denotes Radius of Earth}\end{cases}$

Now,

$\large \boxed{\tt g_h = \dfrac{g}{\Bigg(1 + \dfrac{h}{R}\Bigg)^2}}$

Substituting the values,

We know, the acceleration due to gravity at height  is 1% of its value at the surface of the earth.

Therefore,

$\dashrightarrow \tt g_h = 1 \; \% \; of \; g \\\\ \dashrightarrow\tt g_h = \dfrac{1}{100} \times g \\\\\dashrightarrow\tt g_h = 0.01 \times g \\\\\dashrightarrow\tt \large \underline{\blue{g_h = 0.01 \; g}}$

Substituting the value of Acceleration due to gravity at height 'h' in the above equation.

$\dashrightarrow\tt 0.01\;g = \dfrac{g}{\Bigg(1 + \dfrac{h}{R}\Bigg)^2}\\\\\\\dashrightarrow\tt 0.01\;\cancel{g} = \dfrac{\cancel{g}}{\Bigg(1 + \dfrac{h}{R}\Bigg)^2}\\\\\\\dashrightarrow\tt 0.01 = \dfrac{1}{\Bigg(1 + \dfrac{h}{R}\Bigg)^2}\\\\\\\dashrightarrow \tt\Bigg(1 + \dfrac{h}{R}\Bigg)^2 = \dfrac{1}{0.01}\\\\\\\dashrightarrow \tt\Bigg(1 + \dfrac{h}{R}\Bigg)^2 = \dfrac{100}{1}\\\\\\\dashrightarrow\tt\Bigg(1 + \dfrac{h}{R}\Bigg) = \sqrt{100}\\\\\\\dashrightarrow\tt\Bigg(1 + \dfrac{h}{R}\Bigg) = 10$

$\dashrightarrow\tt \dfrac{h}{R} = 10 - 1\\\\\\\dashrightarrow\tt \dfrac{h}{R} = 9 \\\\\\\dashrightarrow\tt \large \underline{\boxed{\red{\tt h = 9 \; R}}}$

∴ The height (h) above the earth's surface is 9 R.

$\rule{300}{1.5}$