Question

Define Acceleration due to gravity Derive an mathematical expression of g = GM/R?
where G and Mare Gravitation constant and Mass of an object.

Answer 1

Answer:

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

Where G=universal gravitational constant, G=6.7×10

–11

Nm

2

kg

−2

mass of the earth, M=6×10

24

kg,

and radius of the earth, R=6.4×10

6

m.

g=

R

2

GM

g=

(6×10

6

)

2

6.67×10

−11

×6×10

24

=9.8m/s

2

Thus, the value of acceleration due to gravity of the earth, g=9.8m/s

2

Also we know Weight in nothing but the force exerted by the earth with a value of acceleration =9.8m/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.

Answer 2

Answer:

Acceleration due to gravity is the acceleration gained by an object due to the gravitational force. Its SI unit is m/s2. It has both magnitude and direction, hence, it's a vector quantity. Acceleration due to gravity is represented by g. The standard value of g on the surface of the earth at sea level is 9.8 m/s2.

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

Where G=universal gravitational constant, G=6.7×10–11Nm2kg−2

mass of the earth, M=6×1024kg,

and radius of the earth, R=6.4×106m.

g=R2GM​

g=(6×106)26.67×10−11×6×1024​=9.8m/s2

Thus, the value of acceleration due to gravity of the earth, g=9.8m/s2

Also we know Weight in nothing but the force exerted by the earth with a value of acceleration =9.8m/s2

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.

your welcome hope it helps you