# Falling in Orbit…

“Orbital motion” is a wonderful concept in the theory of gravitation. Almost all celestial objects have their own rotation, and revolution around something. As an example, the eight planets move around the Sun, the sun moves around the galactic center, so do all the other stars, probably determined by the galaxy rotation. Any massive object in outer space can provide a good orbit for a satellite.

## Newton’s cannonballs

At his time, Newton was probably the genius. His laws of motion, combined with his theory of gravity can predict the motion of celestial objects traversing through the night sky (i.e) provided the current position and velocity of the body, it’s possible to determine its acceleration. But, Newton has to demonstrate this kind of motion. He was aware of Galileo’s observations like the feather & coin experiment, that Jupiter has moons (which supported the belief that Earth is at the center) and further experiments on projectiles. His illustration of orbital motion is called the Newton’s cannonballs

Newton imagines a cannon at the top of a mountain or something. According to his first law (inertia), in the absence of other forces, if a cannonball is fired, it moves in a straight line. When gravity is included, the trajectory of the projectile is largely affected.

At rest, the cannonball simply falls down (to point O). With increasing magnitudes of imparted horizontal velocities, the ball follows different trajectories to points A, B and C. This velocity is the orbital velocity, which is the minimum velocity required to orbit (approximately 7.3 km/s, in case of Earth). If the object’s velocity equals this velocity, it follows a circular path.. Here’s a simulation of the experiment to show how different velocities affect the path of the ball.

In this way, Newton showed that a satellite in orbit is continuously trying to fall down (into Earth). But, it simply misses the target every time

## Conics as orbits…

Two kinds of velocities have to be accounted for plotting an orbit, the escape velocity $v_e$ and orbital velocity $v_o$. Let $M$ and $m$ be masses of planet and satellite, $R$ be the distance from center of planet to the satellite. In case of a circular orbit, the projectile experiences an acceleration of $a=\frac{v^2}r$ directed towards the center of planet. Equating centripetal and gravitational forces, one can obtain

$\frac{GMm}{r^2}=\frac{mv_o ^2}r$

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

The elegance of orbits comes from the fact that an orbit uses conic sections as trajectories (i.e) the path can be circular, elliptical, parabolic or hyperbolic. Because, the force is inversely related to the distance squared. In reality, one can’t really expect a circular orbit (satisfying the above condition), as it’s very difficult to maintain. Any perturbations in the value of the above expression may lead to other conics (as shown in the picture).

The ratio $v_o/v_e$ is simply the eccentricity of the conic. So, the cases are…

• As $v_o>v_e$, the projectile follows a hyperbolic trajectory and escapes from the planet into outer space.
• As $v_o=v_e$, the path is a parabola. But still, the object leaves the planet.
• If the velocity is more or less the orbital velocity (i.e) $v\propto v_o$, the orbit is elliptical.

## In terms of total energy…

The different orbits can also be expressed in terms of the sum of potential and kinetic energies of the object (total energy). Any body in the gravitational field has gravitational potential energy to some extent. Being the energy associated by virtue of an object’s position, it’s value becomes zero only at infinity. Field is really the most abstract (but, fruitful) concept in Physics. For now, let’s say that it’s just a number allocated to positions in space. Different points have different values in the field.

For example, take Earth which has a gravitational field. An object at the surface (say, you) has low potential energy relative to an object at some height (say someone at top of a mountain). Because, you’ve done some amount of work in placing the object over there. Now, let’s proceed with the common convention that the gravitational potential energy is zero at infinite separation and so, the object’s potential energy is always negative, whereas it’s kinetic energy is never negative.

So, there are three cases…

• If $K.E+P.E>0$ (positive), the object has enough kinetic energy to escape the planet’s gravitational influence and whizz towards infinity. Hence, the path is hyperbolic.
• If $K.E+P.E=0$, it’s parabolic, where the object has the threshold energy to reach infinity.
• If $K.E+P.E<0$ (negative), the object is bound to the planet. It’s elliptic (or theoretically, circular too). There’s also the worst case of falling into the planet too, but it’s not necessary here.

It’s worth noting that bound orbits (elliptical & circular) have negative total energy, and open orbits (parabolic & hyperbolic) have zero or positive total energy.

## Funny math…

A fast runner (say, Usain Bolt) can have maximum speeds up to $\mathrm{40\ km/hr}$ ($\mathrm{v_o=11.11\ m/s}$). Well, Is there any asteroid on which he can run and end up orbiting?

Several assumptions has to be made… First, the object is spherical, as most other celestial objects. It’s a rock dwarf and hence, it has a mean density of around $=\mathrm{2\ g/cm^3}$ ($\mathrm{\rho=2000\ kg/m^3}$). Bolt has a mass of around $\mathrm{100\ kg}$ and he’s gonna have a circular orbit.

The centripetal force equals the gravitational force,

$\frac{GMm}{r^2}=\frac{mv_o ^2}r$

$M=\frac{4}3\pi r^3\rho$

Plugging into the expression, we obtain…

$GM=rv_o ^2$

$G(\frac{4}3\pi r^3\rho)=rv_o ^2$

$\implies r=v_o\sqrt{\frac{3}{4G\pi \rho}}$

This gives the radius of asteroid to be $\mathrm{14.86\ km}$. Note that this includes the height from the surface too. It’s very negligible and let’s approximate it to $\mathrm{14\ km}$. There’s a whole list of asteroids that can fit our specifications. Or, let’s take 142 Polana (looked like a nice name).

This is totally meant for fun and is in no way applicable in reality. The wonderful thing to note here is that the gravity of such an object is very small, so small that Bolt can barely walk (or even stand there). When he starts to walk, he’ll be out into space…

Inspired by this post at Physics Stack Exchange…