Question

If an object of mass$m$ is taken from the surface of earth (radius $R$) to a height $2R,$ then the work done is

• A. $2mgR$
• B. $mgR$
• C. $\left(\frac{2}{3}\right)mgR$
• D. $\frac{3}{2}mgR$
• E. $\frac{1}{3}mgR$

Answer 1

The correct option is C

$\left(\frac{2}{3}\right)mgR$

Step 1: Given information :

An object of mass$m$ is taken from the surface of the earth (radius $R$) to a height $h=2R$

Step 2: Formula used and apply the condition:

The formula of the potential energy$PE=m\left(mass\right)\times g\left(gravity\right)\times h\left(height\right)..............................\left(1\right)$

The value of$g=\frac{GM}{r^2}$, here $G$acceleration constant, $M$ the mass of the earth and $r$radius of the earth.

Step 3: Now substituting the value of $g$in equation (1).

Therefore,

$P{E}_{grav}=-\frac{GMm}{r}$

Here at the surface of the earth $r=R$.

Now the initial energy,

$P{E}_{initial}=-\frac{GMm}{R}$

Therefore at height (h) from the surface,

$$\begin{gathered} r=R+h \\ r=R+2R(giventhath=2R) \\ r=3R \end{gathered}$$

Therefore,

$P{E}_{final}=-\frac{GMm}{3R}$

Now change in the potential energy is,

$$\begin{gathered} ∆P.E=P.E_{final}-P.E_{initail} \\ ∆P.E=-\frac{GMm}{3R}-\left(-\frac{GMm}{R}\right) \\ ∆P.E=\frac{2GMm}{3R} \\ ∆P.E=\frac{2}{3}mgR \end{gathered}$$

Therefore the work done is $\left(\frac{2}{3}\right)mgR$.

Hence, the correct option is (C).