Question

Calculate the work done in moving a body of mass $1000\;kg$ from height 2R to a height 3R above the surface of the earth.
($G = 6.67 \times {10^{11}}\;N{m^2}/k{g^2}$, Radius of the earth$= 6400\;km$)

Answer 1

To calculate the work done in moving a body from height 2R to 3R, we will first determine the total energy of the body at height 2R from the surface of the earth and total energy of the body height 3R from the surface of the earth.

Complete step by step solution:
We know that the body is moving above the earth surface also poses some amount of energy, and the expression for the energy of the moving body is

$E = \dfrac{{GMm}}{{2r}}$.... (1)

Here, $M$ is the mass of the earth, $m$ is the mass of the body and $r$ is the distance of the body from the centre of the earth.

First, we will calculate the body's energy when it is at height ${r_1} = 2R$ from the earth's surface with the help of equation (1).

Therefore, we get

{E_1} = \dfrac{{GMm}}{{2{r_1}}}\\\ {E_1} = \dfrac{{GMm}}{{2\left( {2R} \right)}}\\\ {E_1} = \dfrac{{GMm}}{{4R}} \end{array}$$ We will calculate the body's energy when it is at height ${r_2} = 3R$ from the earth's surface with the help of equation (1). $$\begin{array}{l} {E_2} = \dfrac{{GMm}}{{2{r_2}}}\\\ {E_2} = \dfrac{{GMm}}{{2\left( {3R} \right)}}\\\ {E_2} = \dfrac{{GMm}}{{6R}} \end{array}$$ To calculate the work done in moving a body from$2R$ to $3R$, subtract the object's energy at $2R$ from the object's energy at $3R$. Therefore, we get $$\begin{array}{l} E = {E_2} - {E_1}\\\ E = \dfrac{{GMm}}{{6R}} - \dfrac{{GMm}}{{4R}}\\\ E = \dfrac{{GMm}}{{2R}}\left( {\dfrac{1}{3} - \dfrac{1}{2}} \right)\\\ E = \dfrac{{GMm}}{{24R}} \end{array}$$ We know that the earth's radius is $R = 6.4 \times {10^6}\;{\rm{m}}$, mass of the earth is $M = 6 \times {10^{24}}\;kg$, value of the gravitational constant is $G = 6.67 \times {10^{ - 11}}\;Nk{g^{ - 2}}{m^2}$ and mass of the object is given, so by substituting these values in the above equation, we can obtain the required work done. $$\begin{array}{l} E = \dfrac{{\left( {6.67 \times {{10}^{ - 11}}\;Nk{g^{ - 2}}{m^2}} \right)\left( {6 \times {{10}^{24}}\;kg} \right)\left( {1000\;kg} \right)}}{{24\left( {6.4 \times {{10}^6}\;{\rm{m}}} \right)}}\\\ E = \dfrac{{4.002 \times {{10}^{17}}\;N{m^2}}}{{1.536 \times 8\;m}}\\\ E = 2.605 \times {10^9}\;Nm\\\ E = 2.605 \times {10^9}\;J \end{array}$$ **Therefore, the magnitude of the required work done is $$2.605 \times {10^9}\;J$$.** **Note:** Remember the values of the gravitational constant, the radius of the earth, and the mass of the earth for the correct calculation of the required work done. Because these values are in the power of 10, so the incorrect values result in the wrong calculation and incorrect magnitude of work done.