Question

An object of mass m is raised from the surface of the earth to a height equal to the radius of the earth that is taken from distance at a distance to hour from the centre of the earth what is the loss in magnitude of acceleration due to gravity

Answer 1

Here is your answer ⤵⤵⤵

given :

mass= m kg

height =h m

acceleration due to gravity=g m/s2

Acceleration due to gravity at height H IS GIVEN BY :

g¹=g(1-2h/R)

Loss of weight =ΔW=mg-mg¹

=m[g-g¹]

=m[g-g(1-2h/R)]

=mg[1-(1-2h/R)]

=mg[1-1+2h/R]

=mg x 2h/R

=2mgh/R

HOPE IT HELPS YOU ☺☺ !!!

Answer 2

Hii Mate.....

Here is your answer ⤵⤵

according to formula,

g' = g/(1 + h/R)²

when h << R

then, g' = g(1 -2h/R)

where body is raised to a

height h from a point on the surface of earth where acceleration due to gravity is g and radius of rath is R .

now, initial mass of body when it is on the surface of earth(W1) = mg

final mass of body when it raised to height h from the Earth surface (W2) = mg(1 - 2h/R)

now, weight loss = initial weight - final weight

= mg - mg(1 - 2h/R)

= mg( 1 - 1 + 2h/R)

= 2mgh/R

hence, loss of weight due to variation of g is 2 mhg/R

HOPE IT HELPS YOU !!