SATHEE: Escape And Orbital Velocity

Escape And Orbital Velocity

Escape and Orbital Velocity

If we want to define escape and orbital velocities in simple terms, then we can say that orbital velocity is the speed needed for an object to orbit around another object, while escape velocity is the minimum speed needed for an object to escape the gravitational pull of another mass. Below, we will look at the concepts of escape and orbital velocities and their relationship in greater detail.

What is Escape Velocity?
Orbital Velocity
Relationship Between Escape Velocity and Orbital Velocity
Motion of Satellites Around Earth

What is Escape Velocity?(#what-is-escape-velocity?)

Escape Velocity is the minimum speed needed for an object to escape the gravitational pull of a planet or other body.

The range of a projectile, as studied in kinematics, is proportional to the square of the initial velocity, i.e. $${{R}_{max}}\propto {{u}^{2}} \Rightarrow {{R}_{max}}=\frac{{{u}^{2}}}{2g}$$, which implies that the particle will fly away from the gravitational influence of the earth for a given initial velocity.

The minimum amount of velocity required for a particle to escape the gravitational sphere of influence of a planet is known as escape velocity (ve). When this velocity is provided to a body, it theoretically goes to infinity.

The Law of Conservation of Energy applies to the gravitational force, as it is a conservative force. This law can be applied to a particle that is given a minimum velocity to reach infinity.

$$(\begin{array}{l}{U_i} + {K_i} = {U_f} + {K_f}\end{array})$$

At infinity, the particles experience no interaction, so we can deduce that the final potential energy and kinetic energy of the particle are both zero. This is based on the fact that, as stated in the Motion in 1D chapter, the final velocity of a body becomes zero after reaching its maximum height.

Then,

$\begin{array}{l} U_{i} + K_{i} = 0 and we know that, U_{i} = \frac{- G M m}{R}, K_{i} = \frac{1}{2} m {v_{e}}^{2} \end{array}$

We get,

\begin{array}{l}m{{v}_{e}}^{2} = \frac{2GMm}{R}\Rightarrow {{v}_{e}} = \sqrt{\frac{2GM}{R}}\end{array}

That implies:

\begin{array}{l}{v}_{e}=\sqrt{\frac{2GM}{R}}……………(1)\end{array}

It is evident from the formula that escape velocity is independent of the test mass (m).

If the source mass is Earth, then the escape velocity has a value of 11.2 km/s.

If v = v_e, the body escapes the planet’s gravitational sphere of influence. If $$0 \leq v < v_e$$, the body either falls back onto the earth or continues to orbit around the planet within the sphere of influence of the planet.

Orbital Velocity

The orbital velocity (v0) of a test mass orbiting around a source mass in a circular path of radius ‘r’ with the centre of the source mass as the centre of the circular path is provided by the centripetal force of the gravitational force, which is always an attracting force directed towards the centre of the source mass.

\begin{array}{l} \frac{GMm}{r^2} = \frac{mv_o^2}{r} \end{array}

\begin{array}{l}{v}_{o}^{2}=\frac{GM}{r}\end{array}

$v_o = \sqrt{\frac{GM}{r}}$

If the test mass is at small distances from the source mass $$r \approx R$$(radius of the source mass) he said

He said, “Then.”

$$v_{o}=\sqrt{\frac{GM}{r}} \qquad \qquad \qquad (2)$$

The above formula suggests that the orbital velocity is independent of the test mass (the mass which is orbiting).

Relationship Between Escape Velocity and Orbital Velocity

The relationship between escape velocity and orbital velocity can be mathematically represented by:

$\sqrt 2 V ₀ = V ₑ$

Meanwhile, if we divide Eq.(1) and Eq.(2), we get,

\begin{array}{l}\frac{\sqrt{2GM}}{\sqrt{GM}}=\frac{{{v}_{e}}}{{{v}_{o}}}\end{array}

$$(\frac{{{v}_{e}}}{{{v}_{o}}}=\sqrt{2})$$

It shows that escape velocity is $$\sqrt{2}$$ times greater than orbital velocity.

Certain conditions need to be taken into consideration. The most important one is that the escape velocity should be at least $$\sqrt{2}$$ times larger than the orbital velocity to be free.

When the velocities are the same, the object will remain in constant orbit at the same elevation.

If the escape velocity is less than the orbital velocity, then the orbit will decrease, leading to the object crashing.

If the object’s velocity is greater than the escape velocity, it will be free in the orbit and will likely float into space.

Motion of Satellites Around the Earth

Let us examine the kinematics of a satellite revolving around the earth in a circular orbit of radius r, with the centre of the earth as its centre.

The orbital velocity is the velocity with which a satellite orbits around the Earth, where the test mass is the satellite and the source mass is the Earth.

$\sqrt{\frac{G M}{r}} = v_{o}$

Time Period of a Satellite

The time period of a satellite is the amount of time it takes to complete one revolution around the Earth. This can be calculated by dividing the total distance travelled by the satellite around the Earth by its orbital velocity.

\begin{array}{l}T = \frac{Total~distance~travelled~by~the~satellite}{Orbital~velocity} \end{array}

\begin{array}{l}T=\frac{2\pi r}{{{v}_{O}}} \end{array}

\begin{array}{l}T=2\pi\sqrt{\frac{r^3}{GM}}\end{array}

$$T=\frac{2\pi}{\sqrt{GM}}r^{3/2}$$

Squaring $$T^2 = \frac{4\pi^2}{GM}\left(r\right)^3$$

which is Kepler’s Third Law

The constant of proportionality in the above equation depends only on the source mass and not on the test mass.

Kinetic Energy of a Satellite

The kinetic energy of a satellite in pure rotational motion is given by:

\begin{array}{l}K = \frac{1}{2} mr^2\omega^2\end{array}

The angular velocity of a satellite,ω, is related to its time period, T, by the formula $ω = \frac{2 π}{T}$ .

Substituting $[K = \frac{1}{2} m r^{2} {(\frac{2 π}{T})}^{2}]$

From Kepler’s 3rd law, we know the time period of a satellite. By substituting this value into the formula above, we can calculate the distance of the satellite.

$$K = \frac{GMm}{2r}$$

Kinetic energy can never be negative for any force.

Check Out: Kepler’s Laws

Potential Energy of a Satellite

Potential energy is the energy possessed by a body in a particular position. When the position of the body changes, potential energy changes as well. In order to study potential energy, two bodies are required; a source mass (M) that provides the gravitational force and a test mass (m) that experiences the gravitational force of the source mass. In this case, the satellite is the test mass and the Earth is the source mass. The potential energy possessed by the satellite at a distance ‘r’ from the centre of the Earth is given by:

$(\frac{- G M m}{r} = U)$

Total Energy of a Satellite

The total energy of the satellite is the sum of kinetic and potential energies, as only the mechanical motion of the satellite is considered in its orbit around the Earth.

Total energy of the satellite = Kinetic energy of the satellite + Potential energy of the satellite

\begin{array}{l}K=E-U\end{array}

$$E=\frac{GMm}{2r}-\frac{GMm}{r}$$

\begin{array}{l}\Rightarrow E=-\frac{GMm}{2r}\end{array}

From the above equation,

Total energy of the satellite = -(kinetic energy of the satellite)

\begin{array}{l}\text{Potential energy of the satellite} = 2 \times \text{Total energy of the satellite}\end{array}

Virial Theorem

According to the Virial Theorem, if the potential energy is proportional to the $n^{t h}$ power of the position (r), then the average kinetic energy is equal to $\frac{n}{2}$ times the potential energy at that position.

\begin{array}{l}K=\frac{n}{2} \left( U \right) \ \text{if}\ U\propto {{r}^{n}}\end{array}

Binding Energy of a Satellite

The binding energy of a satellite is the amount of energy required for it to escape the gravitational pull of the Earth. This energy is numerically equal to the negative of the total energy possessed by the satellite.

\begin{array}{l}\text{Binding energy of a satellite}=\frac{GMm}{2r}\end{array}

The binding energy of a system of two bodies is the amount of minimum energy required to separate them by infinite distance, or it is simply the amount of energy required to make the potential energy of the system equal to zero, which is numerically equal to the kinetic energy of the body under motion.

Angular Momentum of a Satellite

The angular momentum of a satellite of mass ’m’ orbiting on an orbital path of radius ‘r’ with an angular speed ω is given by $L = m r^{2} ω$ .

We know that $ω = \frac{2 π}{T}$ , so by substituting this into the equation, we get

$L = \frac{m \cdot r^{2} \cdot 2 π}{T}$

\begin{array}{l}\text{Using}\ T=\frac{2\pi }{\sqrt{GM}}{{\left( r \right)}^{\frac{3}{2}}},\ \text{we get};\end{array}

\begin{array}{l}L = \sqrt{GM_{r}m}\end{array}

It is evident from the equation that the angular momentum of a satellite is dependent on the mass of the satellite, the mass of the Earth, and the radius of the satellite’s orbit.

Frequently Asked Questions on Escape and Orbital Velocity

What is Escape Velocity?

Escape velocity is the minimum speed needed for an object to escape the gravitational pull of a massive body, such as a planet or star. It is equal to the square root of two times the gravitational constant times the mass of the massive body divided by the radius of the body.

The escape velocity is the minimum speed required to project a body vertically upwards from the surface of the earth so that it never returns to the surface of the earth.

Orbital velocity is the speed at which a satellite orbits around a larger body, such as a planet or moon.

The velocity needed to place a satellite into its orbit around the Earth is referred to as orbital velocity.

Hydrogen is lighter than oxygen, so it is able to escape the gravitational pull of the Earth more easily than oxygen.

Hydrogen molecules possess a much higher thermal speed than oxygen molecules, thus allowing them to acquire escape velocity more easily. Therefore, hydrogen escapes from the Earth faster than oxygen.

What is the equation for escape velocity?

$\sqrt 2 g R = v_{e}$

Here, R is the radius of the Earth.

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