Question

Variation of accleration due to gravity above the earth surface​

Answer 1

Answer:

let we consider,

radius of earth R and mass is M

and we know that acceleration due to gravity at surface

g=

R

2

GM

---- (i)

G is universal gravitational constant.

consider, a mass m is at above the earth at height h and g

′

is acceleration due to gravity at height h.

Therefore, g′=

(R+h)

2

GM

----- (ii)

divide, (ii) by (i)

g

g

′

=

(R+h)

2

R

2

g

′

=

(R+h)

2

gR

2

For h<<<R,

g

′

=g(1−

R

2h

)

Answer 2

Answer:

g is the invesly proportional to the height of the  above the earth surface so that decreases with increase in height,

Explanation:

A driving force hold  everything together. acceleration of gravity earth sueface  is 9.8 m/s2 as a result for every second free fall body its speed rises 9.8m/s2 which means matter is the proportional to the gravity.

the value of the g maximum atb the earth surface nad decrease with the deptht d the value of the affects other factors. so that as soon as body goes the upwards variation of the accelerationn due to garvity.