Calculating Earth's Escape Velocity

As we know the gravitation formula is:

$F = \frac{G \times M \times m}{R^2}$

Where:

• $F$ is the gravitational force between two objects,
• $G$ is the gravitational constant,
• $M$ is the mass of the first object,
• $m$ is the mass of the second object, and
• $R$ is the distance between the centers of the two objects.

We can calculate the potential energy formula integrating the force formula:

$U = -\int F \cdot dr = -\int \frac{G \times M \times m}{r^2} \cdot dr = \frac{G \times M \times m}{r}$

The potential energy at the surface of the Earth is:

$U = \frac{G \times M \times m}{R}$

We have to have at least a kinetic energy equal to the potential energy to escape the Earth’s gravitational field. The kinetic energy is:

$K = \frac{1}{2} m v^2$

The escape velocity is the velocity at which the kinetic energy is equal to the potential energy:

$\frac{1}{2} m v_e^2 = \frac{G \times M \times m}{R}$

$v_e = \sqrt{\frac{2 \times G \times M}{R}}$

Substitute the given values (showing units):

$v_e = \sqrt{\frac{2 \times (6.67 \times 10^{-11} \, \text{Nm²/kg²}) \times (5.97 \times 10^{24} \, \text{kg})}{6.37 \times 10^6 \, \text{m}}}$

$v_e = \sqrt{\frac{2 \times 3.99 \times 10^{14} \, \text{Nm²/kg}}{6.37 \times 10^6 \, \text{m}}}$

$v_e = \sqrt{\frac{7.98 \times 10^{14} \, \text{Nm/kg}}{6.37 \times 10^6 \, \text{m}}}$

$v_e = \sqrt{1.25 \times 10^8 \, \text{m²/s²}}$

$v_e = 1.12 \times 10^4 \, \text{m/s}$

Therefore, the escape velocity required for an object to leave Earth’s gravitational field and escape into space is approximately $11,200 \, \text{m/s}$ or $11.2 \, \text{km/s}$.