Question

Question: The weight of the Empire State Building is . Show that the complete conversion of of mass would provide sufficient energy to putli.is rather a large object in a low Earth orbit or LEO for short. (Orbit radius Earth's radius).

Short answer

Answer:

LEO is the orbit very close to the earth’s surface, mostly under , but can be as low as . The conversion energy of 1kg mass is quite sufficient to get such a massive object that is The Empire State Building into Low Earth Orbit.

Step-by-step solution

A concept:

The conversion energy can be easily calculated using Einstein’s famous mass-energy relation, that is given by,

$\mathit{E}\mathbf{}\mathbf{=}\mathit{m}{\mathit{c}}^{\mathbf{2}}$ ….. (1)

Here, E is the rest mass energy, is the mass, and is the speed of light.

Calculate the conversion energy of mass of 1kg :

In this problem check if conversion energy of 1 kg of mass is sufficient to get the Empire State Building of 365 kilotons into a low earth orbit.

Consider the known data as below.

The mass, m = 1 kg

The speed of light, $c=3\times {10}^{8}m/s$

Substitute these values into equation (1).

$$\begin{gathered} E=1kg\times {\left(3\times {10}^{8}m/s\right)}^2 \\ =9\times {10}^{16}J \end{gathered}$$

Determine the energy required to get the building to low earth orbit:

To get an object in LEO, the first object needs to be moved from the surface to an altitude of or more and it needs a minimum velocity to stay in orbit to avoid crashing. This minimum velocity required to stay in orbit is called orbital velocity, that is given by

$$\begin{gathered} V_o=\sqrt{g{R}_E} \\ =\sqrt{\frac{G{M}_E}{R_E}} \end{gathered}$$

Here, g is the acceleration due to gravity.

For simplicity we will assume that the earth is not rotating that is the building is initially at rest. Now, we will apply the Work-Energy relation for the Earth-Building system,

$W=\Delta K+\Delta U$

$W=\frac{1}{2}{M}_B{\left(V_o\right)}^2+G{M}_E{M}_B\left(\frac{1}{R_E}-\frac{1}{R_o}\right)$ ….. (2)

Here, $V_o$ is the orbital velocity and $R_o$ is the orbital radius.

Consider the known numerical data as beow.

The mass of the building,$M_B=3.65\times {10}^{6}kg$,

The mass of the earth, $M_E=5.97\times {10}^{24}kg$

The radius of the the earth, $R_E=6.38\times {10}^{6}m$

Univeralsal constant, $G=6.67\times {10}^{-11}\raisebox{1ex}{$N{m}^2$}\!\left/ \!\raisebox{-1ex}{$K{g}^2$}\right.$

Substituting the expression of orbital velocity into expression (2) and solving further as follow.

$$\begin{gathered} W=\frac{1}{2}{M}_B{\left(V_o\right)}^2+G{M}_E{M}_B\left(\frac{1}{R_E}-\frac{1}{R_o}\right) \\ =\frac{1}{2}{M}_B{\left(\sqrt{\frac{G{M}_E}{R_E}}\right)}^2+G{M}_E{M}_B\left(\frac{1}{R_E}-\frac{1}{R_o}\right) \\ W=\frac{1}{2}\frac{G{M}_B{M}_E}{R_E}+\frac{G{M}_B{M}_E}{R_E}-\frac{G{M}_B{M}_E}{R_o} \\ =G{M}_B{M}_E\left(\frac{3}{2R_E}-\frac{1}{R_o}\right) \end{gathered}$$

Here, the orbit radius is the same as the earth’s radius as given in the problem statement.

$$\begin{gathered} W=G{M}_B{M}_E\left(\frac{3}{2R_E}-\frac{1}{R_E}\right) \\ =\frac{G{M}_B{M}_E}{2R_E} \end{gathered}$$

Substituting known numerical data in the above equation, and you have,

$$\begin{gathered} W=\frac{\left({\displaystyle 6.67\times {10}^{-11}\raisebox{1ex}{$N{m}^2$}\!\left/ \!\raisebox{-1ex}{$k{g}^2$}\right.}\right)\left(3.65\times {10}^{8}kg\right)\left(5.97\times {10}^{24}kg\right)}{2\left(6.38\times {10}^{6}m\right)} \\ =\frac{\left(1.453\times {10}^{23}N{m}^2\right)}{\left(12.76\times {10}^{6}m\right)} \\ =1.14\times {10}^{15}J \end{gathered}$$

Hence, the value of energy is$1.14\times {10}^{15}J$which is far less compared to conversion energy of of mass.

Therefore, the conversion energy of 1kg mass is quite sufficient to get such a massive object that is The Empire State Building into Low Earth Orbit.