Question

Find the value of accceleration due to gravity at height of 6400km from the
surface of the earth. Radius of the earth is 6400km.​

Answer 1

Answer:

The value of accceleration due to gravity at height of 6400km from the surface of the earth will be 2.45 m/s².

Explanation:

We have to find the value of accceleration due to gravity at height of 6400km from the surface of the earth.

Radius of the earth is given as 6400 km.

Let the acceleration due to gravity at surface of the earth be 'g', mass of earth be 'M', radius of the earth be 'R'.

Now, we know that,

$\displaystyle\pink{\sf\:g\:=\:\dfrac{G\:M}{R^2}}\\\\\\\implies\sf\:G\:M\:=\:g\:R^2\:\:\:-\:-\:(\:1\:)$

Now,

Let the acceleration due to gravity at height 'h' from the surface of the earth be g', mass of the earth be M and radius of the earth be ( R + h ).

Now, we know that,

$\displaystyle\pink{\sf\:g'\:=\:\dfrac{G\:M}{(\:R\:+\:h\:)^2}}\\\\\\\implies\sf\:G\:M\:=\:g'\:(\:R\:+\:h\:)^2\:\:\:-\:-\:(\:2\:)$

From ( 1 ) & ( 2 ),

$\displaystyle\sf\:g\:R^2\:=\:g'\:(\:R\:+\:h\:)^2\\\\\\\implies\sf\:g'\:=\:\dfrac{g\:R^2}{(\:R\:+\:h\:)^2}\\\\\\\implies\sf\:g'\:=\:\dfrac{9.8\:\times\:(\:6400\:)^2}{(\:6400\:+\:6400\:)^2}\:\:\:\:-\:-\:[\:Substituting\:the\:values\:]\\\\\\\implies\sf\:g'\:=\:\dfrac{9.8\:\times\:(\:6400\:)^2}{(\:2\:\times\:6400\:)^2}\\\\\\\implies\sf\:g'\:=\:\dfrac{9.8\:\times\:\cancel{6400}\:\times\:\cancel{6400}}{2\:\times\:\cancel{6400}\:\times\:2\:\times\:\cancel{6400}}\\\\\\\implies\sf\:g'\:=\:\dfrac{\cancel{9.8}}{\cancel{2}\:\times\:2}\\\\\\\implies\sf\:g'\:=\:\cancel{\dfrac{4.9}{2}}\\\\\\\implies\boxed{\red{\sf\:g'\:=\:2.45\:m\:/\:s^2}}$

The value of accceleration due to gravity at height of 6400km from the surface of the earth will be 2.45 m/s².