Question

Drive a formula to calculate the acceleration due to gravity of earth

Answer 1

Answer:

$9.8 \: m/ {s}^{2}$

Explanation:

To calculate the value of g, we should put the values of G, M and R 

Where G=universal gravitational constant, 

$G = 6.7 \times {10}^{ - 11}N {m}^{2} {kg}^{ - 2}$

mass of the earth, 

$M = 6 \times {10}^{24} kg$

and radius of the earth, 

$R = 6.4 \times {10}^{6} m$

$g = \frac{GM}{ {R}^{2} }$

$g = \frac{6.67 \times {10}^{ - 11} \times 6 \times {10}^{24} }{( {6 \times {10}^{6} })^{2} } = 9.8 \:m / {s}^{2}$

 Thus, the value of acceleration due to gravity of the earth, 

$g = 9.8 \: m/ {s}^{2}$

Also we know Weight in nothing but the force exerted by the earth with a value of acceleration 

$= 9.8 \: m/ {s}^{2}$

Now at the poles the value of acceleration due to gravity is higher as compared to the equator as the constant value of 

g depends on the radius of the earth which decreases when we come down from poles to the equator. 

Thank you....