Question

A 1 kgblock rests on the Earth's surface. How much energy is required to move the block very far from the Earth, ending up at rest again?

Short answer

Energy required to move the block very far from the Earth is $6.25-{10}^7\mathrm{J}$

Step-by-step solution

 Step 1: Identification of given data

The mass of the block is, m=1 kg

Significance of the escape energy

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.

Determine the energy required to move the block far from the Earth

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

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.

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

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