Physics, asked by Maqudom123, 9 months ago

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

Answers

Answered by nirman95
75

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.

Answered by rajsingh24
77

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.

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