Question

Using the universal gravitational constant G=6.67•10^11 (look up units), derive the Earth’s acceleration of gravity 9.81 m/s2

Answer 1

Solution:

To find the value of the acceleration due to gravity on earth's surface, g :

According to the Newton' s Universal law of Gravitation says that there exists a gravitational attraction between two objects.

It is given by the equation, F = - GMm/ r²

where, M is the mass of the earth

m is the mass of the object on the earth's surface

r is the distance of the object from the center of the earth, it is taken as the radius of the earth since the object is on the surface of the earth

From the Newton's second law of motion, we know that the Force action on a body is given as F= mg

where m is mass of the object and g is the acceleration due to gravity

On equating these two equations we get, mg = -GMm / r²

g = - (GM / r² )

G = 6.67 x 10 ^-11 N/m² /kg² ,

M= 6 x10²⁴ kg

r is the radius of the earth = 6.4 x 10⁶ m

Substituting these values in the equation we get

g = - (9.8) m/s² where the negative sign shows that it acts downwards