Physics, asked by gajanangatade07, 10 months ago

How far from the surface of the earth does the acceleration due to gravity reduces by 51%
of its value at the surface of the Earth? (R=6400 km)​

Answers

Answered by nirman95
6

Answer:

Given:

Radius of Earth = 6400 km

To find:

Height at which gravity reduces by 51%

Concept:

Let gravity at height h be g" , and gravity at Earth surface be g

So , g is reduced by 51%

Hence g" = 49% of (g)

Calculation:

g" =  \dfrac{g}{ {(1 +  \frac{h}{r}) }^{2} }

 =  >  \dfrac{49}{100} g =  \dfrac{g}{ {(1 + \frac{h}{r})  }^{2} }

 =  >  {(1 +  \dfrac{h}{r} )}^{2}  =  \dfrac{100}{49}

 =  > 1 +  \dfrac{h}{r}  =  \dfrac{10}{7}

 =  >  \dfrac{h}{r}  =  \dfrac{3}{7}

 =  > h =  \dfrac{3}{7} r

Putting value of r = 6400 km

h =  \dfrac{3}{7}  \times 6400

 =  > h = 2742.85 \: km

So final answer :

  \boxed{ \red{h = 2742.85 \: km}}

Answered by Anonymous
2

#Physics11

#Gravitation

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