Question

Find the
Magnitude of the
Acceleration
due to gravity at the
Surface of earth
​

Answer 1

Answer:

hey mate here is ur ans

Explanation:

g = Gm/r² = 9.8 where m = mass of earth G = 6.75×10-¹¹Nm ²/kg²and r = radius of earth

hope it help u

Answer 2

Answer:

The magnitude of the acceleration due to gravity at the surface of the earth is 9.77 m/s² (approx.).

Explanation:

Given data:

We know that,

$\bullet\sf\:Mass\:of\:earth\:(\:M\:)\:=\:6\:\times\:10^{24}\:kg\\\\\\\bullet\sf\:Radius\:of\:earth\:(\:R\:)\:=\:6400\:km\:=\:6.4\:\times\:10^6\:m\\\\\\\bullet\sf\:Universal\:gravitational\:constant\:(\:G\:)\:=\:6.67\:\times\:10^{\:-\:11}\:N\:.\:m^2\:/\:kg^2$

To find:

Magnitude of the acceleration due to gravity ( g )

Solution:

We know that, the value of acceleration due to gravity is given by,

$\pink{\sf\:g\:=\:\dfrac{GM}{R^2}}\\\\\\\implies\sf\:g\:=\:\dfrac{6.67\:\times\:10^{\:-\:11}\:\times\:6\:\times\:10^{24}}{(\:6.4\:\times\:10^6\:)^2}\\\\\\\implies\sf\:g\:=\:\dfrac{6.67\:\times\:10\:\times\:6\:\times\:10^{\:-\:12}\:\times\:10^{24}}{6.4\:\times\:6.4\:\times\:10^6\:\times\:10^6}\\\\\\\implies\sf\:g\:=\:\dfrac{6.67\:\times\:10\:\times\:6\:\times\:10^{\:24\:-\:12}}{6.4\:\times\:10^{6\:+\:6}}\:\:\:-\:-\:[\:\because\:a^m\:\times\:a^n\:=\:a^{m\:+\:n}\:]\\\\\\\implies\sf\:g\:=\:\dfrac{6.67\:\times\:10\:\times\:6\:\times\:\cancel{10^{12}}}{6.4\:\times\:6.4\:\times\:\cancel{10^{12}}}\\\\\\\implies\sf\:g\:=\:\dfrac{66.7\:\times\:6}{6.4\:\times\:6.4}\\\\\\\implies\sf\:g\:=\:\cancel{\dfrac{400.2}{40.96}}\\\\\\\implies\sf\:g\:=\:9.770\\\\\\\implies\boxed{\red{\sf\:g\:\approx\:9.77\:m\:/\:s^2}}$

The magnitude of the acceleration due to gravity at the surface of the earth is 9.77 m/s² (approx.).