91影视

The mass of the block is, m=1 kg

Escape energy is necessary to move the block away from Earth (until it is at rest). It is the kinetic energy that allows an object to get away from a gravitational field (in this case Earth's), while decelerating and asymptotically approaching zero velocity.

The expression for the gravitational potential energy is as follows,

$E=-\frac{G{M}_{e}m}{R_e}$

Here, G is the gravitational potential energy with value $6.674脳{10}^{-11}{\mathrm{m}}^3/\mathrm{kg}-{\mathrm{s}}^2,M_e$is the mass of the Earth with value $6脳{10}^{24},m$is the mass of the object, and $R_e$is the radius of the Earth with value $6脳{10}^{24},m$.

The escape energy and the potential energy will be equal according to the law of conservation of energy. So, the expression is as follows,

$$\begin{gathered} E_s=E \\ {E}_s=\frac{G{M}_{e}m}{R_e} \end{gathered}$$

Here, $E_s$is the escape energy.

Substitute all the values in the above expression.

$$\begin{gathered} E_s=\frac{\left(6.674卤{10}^{-11}{\mathrm{m}}^3/\mathrm{kg}-{\mathrm{s}}^2\right)\left(6脳{10}^{24}\mathrm{kg}\right)\left(1\mathrm{kg}\right)}{6.4脳{10}^6} \\ =6.25卤{10}^7/\mathrm{kg}-m^2/{\mathrm{s}}^2脳\frac{1\mathrm{J}}{1\mathrm{kg}-{\mathrm{m}}^2/{\mathrm{s}}^2} \\ =6.25卤{10}^7\mathrm{J} \end{gathered}$$

Thus, the energy required to move the block very far from the Earth is $6.25卤{10}^7\mathrm{J}$.