Waste less time on Facebook — follow Brilliant.
×
Gravitational Physics

Bound orbits

Most of the two thousand or so man-made satellites orbiting Earth range in altitude from about $$\SI{100}{\kilo\meter}$$ to over $$\SI{35\times10^3}{\kilo\meter}.$$ Some are in nearly circular orbits and travel at a constant speed, while others spend long periods far from the surface, then make quick passes near the top of the atmosphere. With so many satellites in orbit, it is surprising that we rarely hear about collisions due to satellites straying from their predicted trajectories.

In this quiz and the next, we will map out the trajectories of satellites and other orbiting bodies using the the conservation laws for energy $$E$$ and angular momentum $$\ell.$$ It turns out that these two parameters, along with the masses of Earth and the satellite, are enough to determine the elliptical shape of the orbit. This remarkable result is called Kepler's first law

Let's start by taking a closer look at a common class of orbits for man-made satellites and one we're already familiar with: circular orbits.

Let's imagine a satellite with mass $$m$$ in a circular orbit, radius $$R_\text{circ},$$ around Earth.

Calculate the satellite's angular momentum $$\ell = mvR_\text{circ}$$ in the circular orbit.

Details

• Express $$\ell$$ in terms of the mass of the satellite is $$m,$$ the mass of Earth $$M_\text{Earth},$$ the radius of the circular orbit $$R_\text{circ},$$ and the gravitational constant $$G.$$

As we saw in the last quiz, energy is central to determining an object's trajectory in a gravitational field.

Find the total energy $$E_\textrm{circ} = K + U$$ of the satellite in circular orbit.

Details

• Gravitational potential energy at a radius $$r$$ is $U=-GmM_\text{Earth}/r.$
• Express $$\ell$$ in terms of the mass of the satellite is $$m,$$ the mass of Earth $$M_\text{Earth},$$ the radius of the circular orbit $$R_\text{circ},$$ and the gravitational constant $$G.$$

Many satellites are in circular orbits (or near-circular orbits). Each satellite with different angular momentum $$\ell$$ has a unique orbital radius $$R_\text{circ}$$ and energy $$E_\text{circ},$$ as you saw in the last two questions.

However, some bound orbits are non-circular. Circular orbits are one special case of all possible bound orbits in which the radial coordinate is constant, $$\dot{r}=0.$$ Non-circular orbits relax this condition, and the satellite's distance from Earth varies along its orbit.

Like the case of circular orbits, conservation of $$\ell$$ and $$E$$ determine the shape and size of the orbit. We will see this is a consequence of the total-energy relationship: \begin{align} E &= \frac12 m \dot{r}^2 + \frac{\ell^2}{2mr^2} -\frac{GmM_\textrm{Earth}}{r} \\ &= \frac12 m \dot{r}^2+\phi_\ell(r). \end{align} $$\phi_\ell$$ is the effective potential of a body with angular momentum $$\ell$$ derived in the last quiz. Much of the essential information needed to predict satellite orbits can be derived directly from this relationship.

One central feature of the effective potential $\phi_\ell(r)=\frac{\ell^2}{2mr^2} - \frac{GmM_\textrm{Earth}}{r}$ is the energy minimum at $$r=R_\text{min}.$$ Because the function is concave-up at the minimum, this stationary point is stable.

What is the energy $$E_\text{min}$$ of an orbiting body at this stable equilibrium point?

Details & Hint

• Recall that the condition for a potential energy minimum is $$\phi_\ell^\prime=0.$$
• Express your answer in terms of $$E_\text{circ},$$ the total energy of the satellite in a circular orbit with angular momentum $$\ell.$$
• Eliminate $$R_\text{circ}$$ between the two previous results for $$\ell$$ and $$E_\text{circ}.$$

The previous question suggests that there is a spectrum of non-circular, bound orbits available to the our satellite with energies greater than $$E_\text{circ}.$$

Just as a pendulum oscillates about its potential energy minimum, an orbiting satellite's radial position oscillates about $$R_\text{circ}$$ when its total energy $$E \gt E_\text{circ}.$$
With this picture in mind, in the remainder of this quiz we will probe the physics of non-circular orbits, firing a rocket to boost our satellite in the radial direction.

Let's light it up in $$3 \dots 2 \dots 1 \dots.$$

Imagine our satellite now has total energy $$E \gt E_\text{circ}$$ and can move through a range of values $$r_p \lt r \lt r_a$$ within the effective potential. We would like to calculate the endpoints $$r_a$$ and $$r_p$$ of this range, the so-called apsides of the orbit.

When our satellite reaches its maximal or minimal orbital radius, what is $$\dot{r}$$?

Using the condition you found in the previous question, and the expression for total energy $E = \frac12 m \dot{r}^2 + \frac{\ell^2}{2mr^2} -\frac{GmM_\textrm{Earth}}{r},$ solve for the range of allowed $$r$$ values.

$\begin{array}{c|c} \mathbf{A}& \frac{R_\text{circ}}{1+\sqrt{\frac{-\Delta E}{E_\text{circ}}}}\leq r \leq \frac{R_\text{circ}}{1-\sqrt{\frac{-\Delta E}{E_\text{circ}}}} \\ \hline \mathbf{B}& \frac{R_\text{circ}}{1+\sqrt{\frac{-2\Delta E}{E_\text{circ}}}}\leq r \leq \frac{R_\text{circ}}{1-\sqrt{\frac{-2\Delta E}{E_\text{circ}}}} \\ \hline \mathbf{C}& \frac{R_\text{circ}}{1+\sqrt{\frac{-\Delta E}{2E_\text{circ}}}}\leq r \leq \frac{R_\text{circ}}{1-\sqrt{\frac{-\Delta E}{2E_\text{circ}}}} \\ \end{array}$

Details & Hint

• Express the bounds in terms of the parameters of the circular orbit when the satellite has angular momentum $$\ell,$$ $$R_\text{circ}$$ and $$E_\text{circ},$$ and the energy difference $\Delta E=E-E_\text{circ}.$
• Try introducing a new variable $$u=1/r$$ in the total energy; a quadratic polynomial can have two real roots.

We have found that the satellite's orbit is bounded from above and below. For objects orbiting Earth, the closest approach is called the perigee and the maximum is called the apogee.

Suppose a satellite is in a circular orbit at altitude $$\SI{15000}{\kilo\meter}$$ (measured from Earth's center). To enter a new orbit, it fires a rocket to gain speed $$\dot r=v_t/2,$$ where $$v_t$$ is the satellite's speed in the circular orbit before the burn.

What is the apogee of the new orbit (in $$\si{\kilo\meter}$$)?

Details & Assumptions

• The impulse delivered to the satellite is only in the radial direction, so its angular momentum is, for our purposes, the same before and after the burn.
• The burn is quick compared to the period of the orbit.

What is the satellite's new perigee in the previous question?

Details

• The radius of its original circular orbit is $$\SI{15000}{\kilo\meter}.$$
• Express your answer in $$\si{\kilo\meter}.$$

By boosting an Earth satellite out of a circular orbit, we have established some basic ideas in orbital mechanics.

• Circular orbits are just a special type of bound orbit for which $$\dot{r}=0.$$
• In general, an orbiting body undergoes an oscillation around the minimum $$R_\text{circ}$$ of the effective potential $$\phi_\ell$$ determined by its angular momentum.
• The extremal points on an orbit depend on the difference between the total energy $$E$$ and the energy minimum $$E_\text{circ}$$ of $$\phi_\ell$$ $$\Delta E=E-E_\text{circ}$$:

$r_\text{apsides}=\frac{R_\text{circ}}{1 \pm \sqrt{\frac{-\Delta E}{E_\text{circ}}}}.$

We will use these results to pin down the exact orbital geometry in the next quiz.

×