Question

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DERIVATION OF GRAVITATIONAL POTENTIAL ENERGY......CLASS 11TH....

Answer 1

hey mate here is your attached.

Answer 2

Stationary roller-coaster
Expression for Gravitational Potential Energy
Case1:- ‘g’ is constant.

Consider an object of mass ‘m’ at point A on the surface of earth.

Work done will be given as :

WBA=F X displacement where F = gravitational force exerted towards the earth)

=mg(h2-h1) (body is brought from position A to B)

=mgh2-mgh1

WAB=VA-VB

where

VA=potential energy at point A

VB= potential energy at point B

From above equation we can say that the work done in moving the particle is just the difference of potential energy between its final and initial positions.

Case2:-‘g’ is not constant.

Calculate Work done in lifting a particle from r = r1 to r = r2 (r2> r1) along a vertical path,

We will get , W=V (r2) – V (r1)

Conclusion: -

In general the gravitational potential energy at a distance ‘r’ is given by :

V(r) = -GMem/r + Vo

where

V(r) = potential energy at distance ‘r’

Vo = At this point gravitational potential energy is zero.

Gravitational potential energy is ∝ to the mass of the particle.

HOPE I HELPED YOU