Question

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)​

Answer 1

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 :

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

Answer 2

#Physics11

#Gravitation