Escape Velocity

Imagine throwing a ball upwards. It goes up for a while and then comes back down due to Earth's gravity. If you throw it harder, it goes higher, but it still comes back. Now, what if you could throw it so hard that it never comes back? That speed is called escape velocity.

Escape velocity is the minimum speed an object must have to overcome the gravitational pull of a celestial body and escape into outer space, never to return. For Earth, this speed is approximately 11.2 kilometers per second (km/s).

Derivation of Escape Velocity

To derive the formula for escape velocity, we use the principle of conservation of energy. When an object is at the surface of a celestial body (like Earth), it has kinetic energy (due to its motion) and potential energy (due to its position in the gravitational field). To escape, the object needs to reach an infinite distance from the body with zero kinetic energy.

Let:

• $m$ be the mass of the object
• $M$ be the mass of the celestial body (e.g., Earth)
• $R$ be the radius of the celestial body
• $G$ be the universal gravitational constant
• $v_e$ be the escape velocity

The initial energy of the object at the surface is the sum of its kinetic energy and gravitational potential energy: $$ E_{initial} = \frac{1}{2}mv_e^2 - \frac{GMm}{R} $$ The final energy of the object when it has escaped to an infinite distance is: $$ E_{final} = 0 \text{ (kinetic energy) } + 0 \text{ (potential energy at infinity) } = 0 $$

According to the conservation of energy, $E_{initial} = E_{final}$. $$ \frac{1}{2}mv_e^2 - \frac{GMm}{R} = 0 $$ $$ \frac{1}{2}mv_e^2 = \frac{GMm}{R} $$ The mass of the object ($m$) cancels out: $$ v_e^2 = \frac{2GM}{R} $$ Therefore, the escape velocity is: $$ v_e = \sqrt{\frac{2GM}{R}} $$

We also know that the acceleration due to gravity at the surface of the Earth is $g = \frac{GM}{R^2}$. We can rewrite the escape velocity formula in terms of $g$: $$ v_e = \sqrt{\frac{2(GM)}{R}} = \sqrt{\frac{2(gR^2)}{R}} = \sqrt{2gR} $$

For Earth, $G \approx 6.674 \times 10^{-11} \, \text{N m}^2/\text{kg}^2$, $M \approx 5.972 \times 10^{24} \, \text{kg}$, and $R \approx 6.371 \times 10^6 \, \text{m}$. Using $v_e = \sqrt{\frac{2GM}{R}}$: $$ v_e = \sqrt{\frac{2 \times (6.674 \times 10^{-11}) \times (5.972 \times 10^{24})}{6.371 \times 10^6}} \approx \sqrt{1.25 \times 10^8} \approx 11180 \, \text{m/s} \approx 11.2 \, \text{km/s} $$ Using $v_e = \sqrt{2gR}$ with $g \approx 9.8 \, \text{m/s}^2$: $$ v_e = \sqrt{2 \times 9.8 \times 6.371 \times 10^6} \approx \sqrt{1.249 \times 10^8} \approx 11176 \, \text{m/s} \approx 11.2 \, \text{km/s} $$

It's important to note that escape velocity is a speed, not a velocity in the sense of direction. While the minimum speed is required, the direction of launch is generally upwards, perpendicular to the surface.

Motion of a Satellite

A satellite is an object that orbits around a larger celestial body. This can be a natural satellite like the Moon orbiting Earth, or an artificial satellite like the International Space Station (ISS) orbiting Earth. The motion of a satellite is governed by gravity.

For a satellite to maintain a stable orbit, the gravitational force pulling it towards the central body must provide the necessary centripetal force for its circular motion.

Orbital Velocity of a Satellite

Consider a satellite of mass $m$ revolving around a celestial body of mass $M$ in a circular orbit of radius $r$. The radius $r$ is measured from the center of the celestial body. If the satellite is orbiting close to the surface of a planet of radius $R$, then $r \approx R$.

The gravitational force between the celestial body and the satellite is: $$ F_g = \frac{GMm}{r^2} $$ This gravitational force acts as the centripetal force required to keep the satellite in its circular orbit. The centripetal force is given by: $$ F_c = \frac{mv^2}{r} $$ where $v$ is the orbital velocity of the satellite.

Equating the gravitational force and the centripetal force: $$ F_g = F_c $$ $$ \frac{GMm}{r^2} = \frac{mv^2}{r} $$ The mass of the satellite ($m$) cancels out: $$ \frac{GM}{r^2} = \frac{v^2}{r} $$ Solving for $v^2$: $$ v^2 = \frac{GM}{r} $$ Therefore, the orbital velocity is: $$ v = \sqrt{\frac{GM}{r}} $$

If the satellite is orbiting close to the surface of the Earth (radius $R$), then $r \approx R$. The orbital velocity ($v_0$) is: $$ v_0 = \sqrt{\frac{GM}{R}} $$ We can also express this in terms of $g$: $$ v_0 = \sqrt{\frac{GM}{R}} = \sqrt{\frac{gR^2}{R}} = \sqrt{gR} $$

Notice the relationship between escape velocity and orbital velocity: $$ v_e = \sqrt{2} \times v_0 $$ This means the escape velocity is $\sqrt{2}$ times the orbital velocity for an object at the same distance from the center of the celestial body.

For a satellite orbiting Earth just above its surface ($r \approx R \approx 6.371 \times 10^6 \, \text{m}$), with $g \approx 9.8 \, \text{m/s}^2$: $$ v_0 = \sqrt{9.8 \times 6.371 \times 10^6} \approx \sqrt{6.244 \times 10^7} \approx 7900 \, \text{m/s} \approx 7.9 \, \text{km/s} $$ This is indeed less than the escape velocity of 11.2 km/s.

Time Period of a Satellite

The time period ($T$) of a satellite is the time it takes to complete one full revolution around the celestial body. For a circular orbit, the distance covered in one period is the circumference of the orbit, which is $2\pi r$.

We know that speed = distance / time. So, orbital velocity ($v$) = $\frac{2\pi r}{T}$. $$ v = \frac{2\pi r}{T} $$ We can rearrange this to solve for $T$: $$ T = \frac{2\pi r}{v} $$

Now, substitute the expression for orbital velocity $v = \sqrt{\frac{GM}{r}}$ into this equation: $$ T = \frac{2\pi r}{\sqrt{\frac{GM}{r}}} $$ $$ T = 2\pi r \sqrt{\frac{r}{GM}} $$ $$ T = 2\pi \sqrt{\frac{r^3}{GM}} $$

This formula shows that the time period of a satellite depends on the radius of its orbit and the mass of the central body. Squaring both sides gives Kepler's Third Law for circular orbits: $$ T^2 = \frac{4\pi^2}{GM} r^3 $$ This means $T^2 \propto r^3$, which is a fundamental law of planetary motion.

If the satellite is orbiting close to the surface of the Earth, $r \approx R$. The time period is: $$ T_0 = 2\pi \sqrt{\frac{R^3}{GM}} $$ Using $GM = gR^2$: $$ T_0 = 2\pi \sqrt{\frac{R^3}{gR^2}} = 2\pi \sqrt{\frac{R}{g}} $$

Let's calculate the time period for a satellite orbiting just above the Earth's surface: $$ T_0 = 2\pi \sqrt{\frac{6.371 \times 10^6}{9.8}} \approx 2\pi \sqrt{6.501 \times 10^5} \approx 2\pi \times 806.3 \approx 5065 \, \text{seconds} $$ This is approximately 84.4 minutes. So, a satellite in a low Earth orbit completes one revolution in about 84 minutes.

Energy of a Satellite

A satellite in orbit possesses two forms of energy: kinetic energy and gravitational potential energy.

Kinetic Energy ($K$)

The kinetic energy of the satellite of mass $m$ with orbital velocity $v$ is: $$ K = \frac{1}{2}mv^2 $$ Substituting $v^2 = \frac{GM}{r}$: $$ K = \frac{1}{2}m\left(\frac{GM}{r}\right) = \frac{GMm}{2r} $$

Gravitational Potential Energy ($U$)

The gravitational potential energy of the satellite at a distance $r$ from the center of the celestial body is defined as: $$ U = -\frac{GMm}{r} $$ The negative sign indicates that the potential energy is zero at infinity and decreases as the satellite gets closer to the central body.

Total Mechanical Energy ($E$)

The total mechanical energy of the satellite is the sum of its kinetic and potential energies: $$ E = K + U $$ $$ E = \frac{GMm}{2r} + \left(-\frac{GMm}{r}\right) $$ $$ E = \frac{GMm}{2r} - \frac{2GMm}{2r} $$ $$ E = -\frac{GMm}{2r} $$

The total energy is negative, which signifies that the satellite is bound to the central body by gravity and is in a stable orbit. To escape the gravitational pull, this negative energy must be overcome. The energy required to escape is equal to the magnitude of the total energy, i.e., $\frac{GMm}{2r}$.

Notice that the kinetic energy is positive and equal in magnitude to half of the potential energy, but with the opposite sign. Also, the total energy is equal in magnitude to the kinetic energy, but negative. $$ K = \frac{GMm}{2r} $$ $$ U = -2K $$ $$ E = -K = \frac{1}{2}U $$

If a satellite moves to a higher orbit (larger $r$), its total energy $E$ increases (becomes less negative). This means energy must be supplied to the satellite to move it to a higher orbit. Conversely, if a satellite loses energy, it spirals inwards towards the central body.

For a satellite to be in a stable orbit, the gravitational force must exactly balance the centripetal force required for that orbit. If the speed is too low, gravity will pull it down. If the speed is too high, it will move to a higher orbit or escape.

The energy required to move a satellite from an orbit of radius $r_1$ to an orbit of radius $r_2$ is $\Delta E = E_2 - E_1 = \left(-\frac{GMm}{2r_2}\right) - \left(-\frac{GMm}{2r_1}\right) = \frac{GMm}{2} \left(\frac{1}{r_1} - \frac{1}{r_2}\right)$.