Question

How much abive the surface of the earth does the acceleration due to gravity reduce by 36 percent of itsvalue on the earth surface

Answer 1

Answer:

To find :

Height at which the acceleration due to gravity reduces by 36% as that of the Earth surface .

Formulas used:

Let height be h , radius of Earth be r, gravity at Earth surface be g and gravity at that height be g2.

$\boxed{ \red{g_{2} = \dfrac{g}{ {(1 + \frac{h}{r}) }^{2} }}}$

Calculation:

As per the question , the gravity decreases by 36% , so we can say that :

$g_{2} = \{(100 - 36)\% \} \: g \\ = > g_{2} = (64\%)g$

So , continuing with the Calculation :

$\therefore g_{2} = \dfrac{g}{ {(1 + \frac{h}{r}) }^{2} }$

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

$= > \dfrac{100}{64} = ( {1 + \dfrac{h}{r} )}^{2}$

$= > \dfrac{10}{8} = 1 + \dfrac{h}{r}$

$= > \dfrac{h}{r} = \dfrac{2}{8} = \dfrac{1}{4}$

$= > h = \dfrac{r}{4}$

So at a height equal to ¼ of radius of Earth, we will get this type of reduction of gravity.

Answer 2

ANSWER:-

used formula:-

let the height be h and radius of the earth be r or gravity at earth surface be g and gravity at that height be g2.

.°. g2 = (g/1+h/r)²

Solution:-

↦the surface of the earth does the acceleration due to gravity reduce by 36 %.

↦ .°. g2=[(100-36)%]g

↦ .°. g2=(64%)g -------(1)

.°. put all values in a formula,

↦ .°. g2 = (g/1+h/r)²

↦ 64/100g = (g/1+h/r)²------(from eq.1)

↦ 100/64 = 1+h/r

↦ 10/8 = 1+h/r

↦ h/r = 2/8 =1/4

.°. h = r/4

.°. the height equal to 1/4 of radius of earth.