Question

Calculate the change in the value of acceleration due to gravity when a body is taken from surface to height h
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Answer 1

Answer:

At a height h above the surface of earth, the change in value of g is same as that at a depth x below the surface on the earth. Both x and h are very small in comparison to the radius of the earth

x=2h

Step-by-step explanation:

I think this is right answer

Answer 2

Answer:

The change in the value of acceleration due to gravity when a body is taken from the surface of the Earth to a height h above the surface is given by $\delta g = G * M * (1/(r1+h)^2 - 1/r1^2)$

Step-by-step explanation:

The acceleration due to gravity, denoted by "g", is the force per unit mass that a body experiences when in a gravitational field. It is given by the equation:

$g = G * M / r^2$

where G is the gravitational constant, M is the mass of the object producing the gravitational field, and r is the distance between the center of the object and the point where the acceleration is being measured.

As a body is taken from the surface of the Earth to a height h above the surface, the distance between the body and the center of the Earth increases from r1 (the radius of the Earth) to r2 (the radius of the Earth plus h). Using the equation above, we can calculate the change in g as follows:

$g2 = G * M / r2^2$

$g2 = G * M / r2^2$

The change in g, denoted by Δg, is:

$\delta g = g2 - g1\\= (G * M / r2^2) - (G * M / r1^2)\\= G * M * (1/r2^2 - 1/r1^2)$

Substituting $r2 = r1 + h$, we get:

$\delta g = G * M * (1/(r1+h)^2 - 1/r1^2)$

Therefore, the change in the value of acceleration due to gravity when a body is taken from the surface of the Earth to a height h above the surface is given by $\delta g = G * M * (1/(r1+h)^2 - 1/r1^2)$.

To learn more about gravity: https://brainly.in/question/46930634

To learn more about acceleration: https://brainly.in/question/16067815

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