Question

A body of mass m rises to a height $h = \dfrac{R}{5}$ from the earth’s surface where R is the Earth’s radius. If g is acceleration due to gravity at the earth’s surface, the increase in potential energy-
A. $mgh$
B. $\dfrac{4}{5}mgh$
C. $\dfrac{5}{6}mgh$
D. $\dfrac{6}{7}mgh$

Answer 1

Hint: Use the equation of gravitational potential energy to calculate the potential energy at the surface of the earth and at a given height, increase in potential energy is the difference between them.

Complete step-by-step answer:
Step 1: The gravitational potential energy of a body at the earth’s surface is given by,
$P.{E_1} = \dfrac{{ - GMm}}{R}$ ………(1)
Where G=gravitational constant, M=mass of earth, m = mass of body and R=radius of earth
Now at a height of $h = \dfrac{R}{5}$ the gravitational potential energy is given by-
$P.{E_2} = \dfrac{{ - GMm}}{{R + h}} \\\ P.{E_2} = \dfrac{{ - GMm}}{{R + \dfrac{R}{5}}} \\\$
$P.{E_2} = \dfrac{{ - GMm}}{{\dfrac{{6R}}{5}}} \\\ P.{E_2} = \dfrac{{ - 5GMm}}{{6R}} \\\$
…………………(2)
Step 2: Now calculating the increase in P.E of the body which is equal to the difference in the two potential energies.
Therefore, Increase in potential energy=$P.E = P.{E_2} - P.{E_1}$
$P.E = \dfrac{{ - 5GMm}}{{6R}} + \dfrac{{GMm}}{R} \\\ P.E = \left( {1 - \dfrac{5}{6}} \right)\dfrac{{GMm}}{R} \\\ P.E = \dfrac{{GMm}}{{6R}} \\\$
……………….(3)
Step 3: Now convert this P.E in the form gravity
As, $g = \dfrac{{GM}}{{{R^2}}}$ and
$h = \dfrac{R}{5} \\\ \Rightarrow R = 5h \\\$
From equation (3)
$P.E = \dfrac{{GMm}}{{6R}} = \dfrac{{GM}}{{{R^2}}} \times \dfrac{1}{6} \times R \times m \\\ \Rightarrow P.E = g \times \dfrac{1}{6} \times 5h \times m \\\ P.E = \dfrac{5}{6}mgh \\\$
Hence option (C) is the correct option.

Note: Remember that there is always a negative sign in gravitational potential energy which shows the opposition in direction. Always use the equation with sign.