Escape Velocity

Introduction

Escape velocities are some velocities which have special importance in astronomy and can be determined from the physical conditions of the earth and celestial mechanics.

1. Cosmic velocity

$$ v_1 = \sqrt{G \, \dfrac{m_\rm{E}}{r_\rm{E}}} = 7,9 \, \rm{\dfrac{km}{s}} $$ A missile theoretically requires at least the first cosmic velocity in order to remain in a circular orbit slightly above the Earth's surface without falling to the ground.

An example would be a horizontally thrown stone with the first cosmic velocity which would not fall back to the ground, but flies in an orbit around the Earth. However, this is practically not possible due to the high air resistance on the Earth's surface.

Deriving the first cosmic velocity using calculus

The gravitational force of the Earth acts in this case as a centripetal force which forces the missile onto a circular path.

\begin{aligned} F_\rm{Z} &= F_\rm{G} \\ \\ \dfrac{\cancel m \cdot (v_1)^2}{\cancel r_\rm{E}} &= G \, \dfrac{\cancel m \cdot m_\rm{E}}{(r_\rm{E})^{\cancel 2}} \\ \\ (v_1)^2 &= G \, \dfrac{m_\rm{E}}{r_\rm{E}} \\ \\ v_1 &= \sqrt{G \, \dfrac{m_\rm{E}}{r_\rm{E}}} \\ \\ \end{aligned}

2. Cosmic velocity

$$ v_2 = \sqrt{2 \, G \, \dfrac{m_\rm{E}}{r_\rm{E}}} = 11,2 \, \rm{\dfrac{km}{s}} $$ A missile theoretically requires at least the second cosmic velocity,
to escape the gravitational field of the earth.

An example would be a launching a missile of the Earth with the second cosmic velocity so that it flies further and further away from the Earth without further acceleration. However, this is practically not possible due to the high air resistance on the Earth's surface.

Deriving the second cosmic velocity using calculus

In theory to escape the gravitational field of the Earth, a missile has to move infinitely far from the Earth. The required energy can be calculated:

\begin{aligned} \Delta W &= G \cdot m_\rm{E} \cdot m_\rm{R} \cdot \left( \dfrac{1}{r_1} - \dfrac{1}{r_2} \right) \\ \\ &= G \cdot m_\rm{E} \cdot m_\rm{R} \cdot \left( \dfrac{1}{r_\rm{E}} - \dfrac{1}{\infty} \right) \\ \\ &= G \cdot m_\rm{E} \cdot m_\rm{R} \cdot \dfrac{1}{r_\rm{E}} \\ \end{aligned}

This work must be available in the kinetic energy of the missile.

\begin{aligned} E_\rm{kin} &= \Delta W \\ \\ \dfrac{\cancel m_\rm{R}}{2} \cdot (v_2)^2 &= G \cdot m_\rm{E} \cdot \cancel m_\rm{R} \cdot \dfrac{1}{r_\rm{E}} \\ \\ (v_2)^2 &= 2 \cdot G \cdot m_\rm{E} \cdot \dfrac{1}{r_\rm{E}} \\ \\ v_2 &= \sqrt{ 2 \,\, G \dfrac{m_\rm{E}}{r_\rm{E}} } \\ \\ \end{aligned}

3. Cosmic velocity

$$ v_3 = \sqrt{(v_\mathrm{2S})^2 + (v_\mathrm{2E})^2} = 16,7 \, \rm{\dfrac{km}{s}} $$ A missile theoretically requires at least the third cosmic velocity
to escape the gravitational field of the sun.

An example would be a missile launched with the third cosmic velocity from the Earth that flies without further acceleration away from the earth and then also away from the Sun. However, this is practically not possible due to the high air resistance on the Earth's surface.

Deriving the third cosmic velocity using calculus

First, calculate the velocity, which is needed to escape the gravitational field of the Sun from a stationary Earth. For this you put the mass of the Sun and the Earth-Sun distance into the formula for the second cosmic velocity.

$$ v_{2S} = \sqrt{2 \, G \, \dfrac{m_\rm{S}}{r_\rm{E, S}}} = \sqrt{2 \, G \, \rm \dfrac{\SI{2e30}{kg}}{\SI{149.6e9}{m}}} = 42,2 \,\, \rm \dfrac{km}{s} $$

As the earth already orbits around the Sun with \( v = 29,8 \,\, \rm \frac{km}{s} \), the required velocity is reduced to:

$$ v_{2S} = 42,2 \,\, \rm \dfrac{km}{s} - 29,8 \,\, \rm \dfrac{km}{s} = 12,4 \,\, \rm \dfrac{km}{s} $$

When starting from the Earth's surface we must add to this velocity the escape velocity of the Earth squared. The result is the third cosmic velocity:

$$ v_3 = \sqrt{(v_{2S})^2 + (v_{2E})^2} = \sqrt{(12,4 \,\, \rm \tfrac{km}{s})^2 + (11,2 \, \rm{\tfrac{km}{s}})^2} = 16,7 \, \rm{\dfrac{km}{s}} $$

Sources