Question

A particle of mass 'm' is raised to a height h = R from the surface of earth. Find increase in potential energy. R = radius of earth. g = acceleration due to gravity on the surface of earth.

Answer 1

Answer:

the increase in potential is 579

Answer 2

Given:-

Radius of Earth = R.

Acceleration due to gravity = g.

To Find:-

Increase in potential energy.

Solution:-

Potential Energy of a particle of mass m is given by:-

P.E = $\frac{-GMm}{R}$,       where M is the mass of earth and m mass of particle.

Therefore the initial potential energy given by

(P.E)₁ = $\frac{-GMm}{R}$

and the final potential energy is given by

(P.E.)₂ = $\frac{-GMm}{(R+h)}$,    according to question R increase by h.

Therefore the change in Potential Energy is given by

ΔU = Final Potential energy - Initial Potential Energy

ΔU = (P.E)₂ - (P.E)₁

ΔU = $\frac{-GMm}{(R+h)} - ( \frac{-GMm}{R} )$,

taking common, we get

ΔU = $GMm[\frac{1}{R} - \frac{1}{R+h}]$,

Now taking LCM and solve them we get

ΔU = $\frac{GMmh}{R(R+h)}$,

Taking R² comman then, we get

ΔU = $\frac{GMmh}{R^{2}(1+\frac{h}{R}) }$

ΔU = $\frac{GM}{R^{2} }{\frac{(mh)}{(1+ \frac{h}{R}) }$........as we know that g =$\frac{GM}{R^{2} }$, now putting the value of g

then we get change in Potential Energy will be

ΔU = $\frac{mgh}{(1+\frac{h}{R}) }$, This is the increase in Potential Energy/ Change in Potential Energy.