Question

the height at which acceleration due to gravity decrease by 36 % of its value on the surface of earth is (assume the radius of earth is R)

Answer 1

acceleration due to gravity = g
here if it is decreased by 36% = 16/25g

g proposnal to 1/R^2
g = k /R^2 ( k = constant of proportionality = G M×m)

when g became 16/ 25 then k remain constant and only R changes so we have to find R

16/ 25 g = 16/25 k / R2

= 16k / 25 R2
= k × 16 / 25 R2
=k (4/5R)^2


so when when R became 5/4 time larger then g will decrease by 36%.

Answer 2

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" $\begin{lgathered}g" = \frac{64}{100} g \\\end{lgathered}$

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

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

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

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

$\begin{lgathered}= > \frac{h}{R} = \frac{2}{8} \\\end{lgathered}$

$\begin{lgathered}= > h = \frac{R}{4} \\\end{lgathered}$

【 So, height is ¼ of radius of Earth.】