Question

the height at which the acceleration due to gravity decreases by 36% of its value on surface of the Earth is assume radius of earth is R​

Answer 1

Answer:

To find:

Height at which acceleration of gravity decreases by 36% , assuming radius of Earth as R

Concept:

As per the question , the gravity decreases by 36% , so the new acceleration due to gravity will be 64% as that of gravity at surface.

Calculation:

Let new gravity be g"

$g" = \frac{64}{100} g$

$= > \frac{g}{ {(1 + \frac{h}{R}) }^{2} } = \frac{64}{100} g$

$= > {(1 + \frac{h}{R}) }^{2} = \frac{100}{64}$

$= > 1 + \frac{h}{R} = \frac{10}{8}$

$= > \frac{h}{R} = \frac{10}{8} - 1$

$= > \frac{h}{R} = \frac{2}{8}$

$= > h = \frac{R}{4}$

So, height is ¼ of radius of Earth.

Answer 2

Answer:

${\boxed{\bold{Answer=\frac{1}{4}\:of\:earth}}}$

Explanation:

Solution:-

As the acceleration decreases by 36% ,so the new acceleration will be (100%-36%) or 64% of gravity of earth.

Now,

$\tt g" = 64\% \times g \\ \rightarrow\tt g"= \frac{64}{100} \times g \\ \rightarrow\tt g" = \frac{16}{25} \times g \\ \rightarrow\tt \frac{g}{(1 + \frac{h}{R}) {}^{2} } = \frac{16g}{25} \\ \rightarrow\tt (1 + \frac{h}{R} ) {}^{2} = \frac{25}{16} \\ \rightarrow\tt 1 + \frac{h}{R} = \sqrt{ \frac{25}{16} } \\ \rightarrow\tt 1 + \frac{h}{R} = \frac{5}{4} \\ \rightarrow\tt \frac{h}{R} = \frac{5}{4} - 1 \\ \rightarrow\tt \frac{h}{R} = \frac{5 - 4}{4} \\ \rightarrow\tt \frac{h}{R} = \frac{1}{4} \\ \rightarrow\tt 4h = R \\ \rightarrow\tt h =\frac{R}{4}$

$\therefore\bold{\underline{Height=\frac{1}{4}\:of\:earth's\:radius}}$