Question

A body of massm is raised to a height h from surface of earth where acceleration due to gravity is g . prove that the loss in weight due to variation in g is approximately (2mgh/r) where r = radius of earth​

Answer 1

Answer:

A body of mass m is raised to a height h from the surface of the earth where the acceleration due to gravity is g. Prove that the loss in weight due to variation in g is approximately 2 mgh/R, where R is the radius of the earth

Answer 2

Given data : A body of mass ‘m’ is raised to a height ‘h’ from surface of earth where acceleration due to gravity is ‘g’. { where, ‘R’ is the radius of the earth}

To prove : The loss in weight due to variation in g is approximately (2*m*g*h/R)

Proof : Let, initial weight of object be W1 = m*g and final weight of object be W2 = m*gh.

⟹ Loss in weight due to variation in g

= initial weight - final weight

⟹ Loss in weight due to variation in g

= m*g - m*gh

Here, when, h << R { height of particle/object is less than radius of the earth } then,

⟹ gh = g*{ 1 - 2h/R }

Where, ‘gh’ is acceleration due to gravity at height ‘h’.

Now,

⟹ gh = g - 2*g*h/R

⟹ gh - g = - 2*g*h/R

{multiply both side by minus ( - ) sign}

⟹ - gh + g = 2*g*h/R i.e.

⟹ g - gh = 2*g*h/R

⟹ m*g - m*gh = 2*m*g*h/R

Answer : Hence, it's prove that, the loss in weight due to variation in g is approximately 2*m*g*h/R.

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