Question

The weight of the Empire State Building is $\mathbf{365}$kilotons. Show that the complete conversion of $\mathbf{1}\mathbf{}\mathit{k}\mathit{g}$of mass would provide sufficient energy to putli.is rather large object in a low Earth orbit or LEO for short. (Orbit radius $\mathbf{\simeq }$Earth's radius).

Short answer

LEO is the orbit very close to the earth’s surface, mostly under $\mathbf{1000}\mathbf{}\mathit{k}\mathit{m}$, but can be as low as $\mathbf{160}\mathbf{}\mathit{k}\mathit{m}$. Here, in this problem we have to check if conversion energy of $\mathbf{1}\mathbf{}\mathit{k}\mathit{g}$ of mass is sufficient to get the Empire State Building of $\mathbf{365}$ kilotons into a low earth orbit.

Step-by-step solution

Calculate the conversion energy of mass of1 kg

Here, the conversion energy can be easily calculated using Einstein’s famous mass-energy relation, that is $E=m{c}^2$

$=9\times {10}^{16}\mathrm{J}$

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 $160km$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.

Velocity $\left(V_o=\sqrt{g{R}_E}=\sqrt{\frac{G{M}_E}{R_E}}\right)$. For simplicity we will assume that the earth is not rotating that is the building is initially at rest. Now, we will apply Work-Energy relation for Earth-Building system,

$W=\Delta K+\Delta U$

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

Here, the orbit radius is the same as the earth’s radius as given in the problem statement. And by putting values of quantities in the above equation we get the value of energy$1.14\times {10}^{16}J$ as which is far less compared to conversion energy of $1kg$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.