Physics, asked by alishakadian, 5 months ago

Derive expression for variation of acceleration due to gravity (g)
with height and depth. Also plot a graph for variation of g with distance.​

Answers

Answered by malathydamotharan
2

Answer:

Explanation:Variation of g with Height

Acceleration due to Gravity at a height (h) from the surface of the earth

Consider a test mass (m) at a height (h) from the surface of the earth. Now, the force acting on the test mass due to gravity is;

F = GMm/(R+h)2

Where M is the mass of earth and R is the radius of the earth. The acceleration due to gravity at a certain height is ‘h’ then,

mgh= GMm/(R+h)2

⇒ gh= GM/[R2(1+ h/R)2 ] . . . . . . (2)

The acceleration due to gravity on the surface of the earth is given by;

g = GM/R2 . . . . . . . . . (3)

On dividing equation (3) and (2) we get,

gh = g (1+h/R)-2. . . . . . (4)

This is the acceleration due to gravity at a height above the surface of the earth. Observing the above formula we can say that the value of g decreases with increase in height of an object and the value of g becomes zero at infinite distance from the earth.

⇒ Check: Kepler’s Laws of Planetary Motion

Approximation Formula:

From Equation (4)

when h << R, the value of g at height ‘h’ is given by gh = g/(1 – 2h/R)

Variation of g with Depth

Consider a test mass (m) taken to a distance (d) below the earth’s surface, the acceleration due to gravity that point (gd) is obtained by taking the value of g in terms of density.

On the surface of the earth, the value of g is given by;

g = 4/3 × πρRG

At a distance (d) below the earth’s surface, the acceleration due to gravity is given by;

gd = 4/3 × πρ × (R – d) G

On dividing the above equations we get,

gd = g (R – d)/R

When the depth d = 0, the value of g on the surface of the earth gd = g.

When the depth d = R, the value of g at the centre of the earth gd = 0.

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