## Table of Contents

## Circular Orbits: The Basics

Consider a planet of mass $m$ in a circular orbit around a star of mass $M$. This scenario is a classic example of circular motion under the influence of gravitational forces. The force of gravity acts as the centripetal force necessary to keep the planet in its orbit. Mathematically, this relationship is expressed as:

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

where $G$ is the gravitational constant, $r$ is the radius of the orbit, and $v$ is the orbital speed of the planet.

Rearranging the terms, we find an expression for the orbital speed:

$$v = \sqrt{\frac{GM}{r}} \tag{1}$$

This equation beautifully links the speed of a planet’s orbit directly to its distance from the sun. Furthermore, the time period of the orbit ($T$), which is the time it takes to complete one full orbit, is connected to the orbital radius and speed through the relationship:

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

This demonstrates that the orbital period squared is proportional to the orbital radius cubed, a reflection of Kepler’s third law of planetary motion.

The above relation can be derived via linking the speed of the orbit ($v = \omega r$ from Circular Motion: Period & Frequency) to the circumference ($2 \pi r$) and the time period of the orbit, $T$.

$$v = \omega r = \frac{2 \pi r}{T}$$

$$v^{2} = \frac{4 \pi^{2} r^{2}}{T^{2}}$$

From Equation 1 above, we know that $v^{2} = \frac{GM}{r}$. Hence,

$$\begin{aligned} \frac{4 \pi^{2} r^{2}}{T^{2}} &= \frac{GM}{r} \\ \frac{r^{3}}{T^{2}} &= \frac{GM}{4 \pi^{2}} \\ T^{2} &= \left( \frac{4 \pi^{2}}{GM} \right) r^{3} \end{aligned}$$

## The Shift to Elliptical Orbits

While circular orbits provide a simplified model, the reality is that most celestial orbits are elliptical. An ellipse has two focal points, and in the context of planetary orbits, one of these points is occupied by the sun. The major axis of the ellipse represents the longest diameter, and the semi-major axis is half of this length. The elliptical nature of orbits introduces variations in speed; a planet or comet moves fastest at its closest approach to the sun (perihelion) and slowest at its furthest point (aphelion).

The fundamental equation linking the time period of an orbit to its semi-major axis in elliptical orbits is given by:

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

where ($r_{sma}$) is the semi-major axis of the ellipse. This equation underscores that the properties governing circular orbits extend to elliptical ones, with the semi-major axis playing a crucial role analogous to the radius in circular orbits.

## Dynamics of Elliptical Orbits

In elliptical orbits, a celestial body’s velocity is not constant. At the perihelion, the body moves at its highest speed due to the stronger gravitational pull from the sun. As it moves away, its speed decreases, reaching a minimum at the aphelion. This variation in speed is a result of the conservation of energy, with kinetic and potential energy converting into each other as the body moves along its orbit.

## Satellite Orbits Around Earth

Orbiting Earth, satellites follow similar principles. Depending on their purpose, satellites may occupy low Earth orbits for detailed observation or higher geosynchronous orbits for continuous communication coverage. A geosynchronous orbit is particularly fascinating as it matches Earth’s rotational period, allowing the satellite to remain fixed relative to a point on Earth’s surface.

## Worked Examples

### Example 1: Calculating Orbital Speed

A planet has a circular orbit around its star with a radius of $1.5 \times 10^{11}$ meters. The mass of the star is $2 \times 10^{30}$ kg. Calculate the orbital speed of the planet.

**Click here to show/hide answer**

The orbital speed $v$ can be found using the formula $v = \sqrt{\frac{GM}{r}}$, where $G$ is the gravitational constant $6.67 \times 10^{-11} \, \text{Nm}^2/\text{kg}^2$, $M$ is the mass of the star, and $r$ is the radius of the orbit.

$$\begin{aligned} v &= \sqrt{\frac{(6.67 \times 10^{-11})(2 \times 10^{30})}{1.5 \times 10^{11}}} \\ &= \sqrt{\frac{13.34 \times 10^{19}}{1.5 \times 10^{11}}} \\ &= \sqrt{8.8933 \times 10^{8}} \\ &\approx 29805 \, \text{m/s} \end{aligned}$$

So, the orbital speed of the planet is approximately 29,805 m/s.

### Example 2: Time Period of an Elliptical Orbit

A comet orbits the Sun in an elliptical path with a semi-major axis of $4 \times 10^{12}$ meters. Calculate the time period of the comet’s orbit.

**Click here to show/hide answer**

The time period $T$ of an orbit is given by $T^2 = \frac{4\pi^2}{GM} (r_{sma})^3$, where ($r_{sma}$) is the semi-major axis.

Let’s plug in the values: $G = 6.67 \times 10^{-11} \, \text{Nm}^2/\text{kg}^2$, $M = 1.99 \times 10^{30} \, \text{kg}$, and $r_{sma} = 4 \times 10^{12} \, \text{m}$.

$$\begin{aligned} T^2 &= \frac{4\pi^2}{(6.67 \times 10^{-11})(1.99 \times 10^{30})} (4 \times 10^{12})^3 \\ T^2 &= \frac{39.4784}{1.32733 \times 10^{20}} \times 64 \times 10^{36} \\ T^2 & \approx 1.902 \times 10^{17} \\ T & \approx 1.38 \times 10^{8} \, \text{s} \end{aligned}$$

To convert seconds into years: $\frac{1.38 \times 10^{8}}{3.156 \times 10^{7}} \approx 4.37 \, \text{years}$.

The time period of the comet’s orbit is approximately 4.37 years.

### Example 3: Height of a Geosynchronous Satellite

Calculate the height above Earth’s surface a satellite needs to be placed to maintain a geosynchronous orbit. Assume Earth’s mass is $5.97 \times 10^{24}$ kg and its radius is $6.371 \times 10^{6}$ meters.

**Click here to show/hide answer**

A geosynchronous orbit means the orbital period $T$ matches Earth’s rotational period, which is $T = 24 \times 3600$ seconds. The formula to find the orbital radius $r$ for geosynchronous orbit is derived from $T^2 = \frac{4\pi^2}{GM}r^3$.

Solve for $r$:

$$r^3 = \frac{GMT^2}{4\pi^2}$$

Plug in the values:

$$\begin{aligned} r^3 &= \frac{(6.67 \times 10^{-11})(5.97 \times 10^{24})(24 \times 3600)^2}{4\pi^2} \\ r & \approx 4.225 \times 10^7 \, \text{m} \end{aligned}$$

The height above Earth’s surface is $r-\text{Earth’s radius}$:

$$4.225 \times 10^7 \, \text{m}-6.371 \times 10^6 \, \text{m} \approx 3.588 \times 10^7 \, \text{m}$$

The satellite needs to be approximately 35,880 km above Earth’s surface.

### Example 4 – Moon’s Mass Doubling

Imagine if the Moon doubles in mass, which of the following is likely to happen?

- Each month will be longer than 30 days.
- Nights will be a lot brighter since the Moon is nearer to the Earth
- Each day will be longer since the period of the Moon increases.
- The Moon continues to revolve around the Earth in the same orbit.

**Click here to show/hide answer**

We note that:

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

From the above equation, doubling the Moon’s mass will not affect the orbital period.

### Example 5: Geostationary Satellites

Which quantity is not necessarily the same for satellites that are in geostationary orbits around the Earth?

- Angular velocity
- Kinetic energy
- Centripetal acceleration
- Orbital period

**Click here to show/hide answer**

For satellites in geostationary orbits:

- Angular velocity must be the same for all, as they must match the Earth’s rotation to appear stationary over a point on the equator.
- Orbital period must be the same, equal to one Earth day, to maintain a geostationary position.

However, kinetic energy ($\text{KE} = \frac{1}{2} mv^{2}$) depends on the mass $m$ and the velocity $v$ of the satellite. Since the mass of satellites can vary, their kinetic energy can also vary even though they have the same velocity in a geostationary orbit.

**Correct Answer:** 2. Kinetic energy

### Example 6: Mar’s Satellites

Mars is known to possess two satellites, Phobos and Deimos. The former is at a distance of 9,500 km from the centre of Mars and the latter at a distance of 24,500 km. Find the ratio of the period of Phobos to that of Deimos in their revolutions around Mars.

**Click here to show/hide answer**

To find the ratio of the periods of Phobos and Deimos in their revolutions around Mars, we can use the following equation:

$$\begin{aligned} T^{2} &= \left( \frac{4 \pi^{2}}{GM} \right) r^{3} \\ &\text{OR} \\ T^{2} & \propto r^{3} \end{aligned}$$

Where:

- $T$ is the orbital period,
- $r$ is the radius of the orbit (distance from the center of Mars to the satellite).

Given that we are comparing the ratios of the periods of Phobos and Deimos, we can express the above equation for each moon and then take the ratio of these expressions.

For Phobos:

$$T_{\text{Phobos}}^2 \propto r_{\text{Phobos}}^3$$

For Deimos:

$$T_{\text{Deimos}}^2 \propto r_{\text{Deimos}}^3$$

Taking the ratio of the periods squared for Phobos and Deimos gives us:

$$\frac{T_{\text{Phobos}}^2}{T_{\text{Deimos}}^2} = \frac{r_{\text{Phobos}}^3}{r_{\text{Deimos}}^3}$$

Given distances:

- $ r_{\text{Phobos}} = 9,500 \, \text{km} = 9.5 \times 10^6 \, \text{m}$
- $ r_{\text{Deimos}} = 24,500 \, \text{km} = 24.5 \times 10^6 \, \text{m}$

Substituting these values into the ratio:

$$\frac{T_{\text{Phobos}}^2}{T_{\text{Deimos}}^2} = \left( \frac{9.5 \times 10^6}{24.5 \times 10^6} \right)^3$$

Simplifying this expression:

$$\begin{aligned} \frac{T_{\text{Phobos}}^2}{T_{\text{Deimos}}^2} &= \left( \frac{9.5}{24.5} \right)^3 \\ & \approx 0.00583 \end{aligned}$$

Since we are interested in the ratio of the periods $T_{\text{Phobos}}$ to $T_{\text{Deimos}}$, we take the square root of both sides (since we initially compared the squares of the periods):

$$\frac{T_{\text{Phobos}}}{T_{\text{Deimos}}} = \sqrt{0.00583} \approx 0.241$$

Therefore, the ratio of the period of Phobos to that of Deimos, based on their distances from Mars, is approximately $0.241$, indicating that Phobos completes its orbit in roughly 24.1% of the time it takes Deimos to complete one orbit around Mars.

### Example 7: Moon & LEO Satellites

a) The Moon is orbiting at a certain height above the Earth. The Earth exerts a force of $1.99 \times 10^{20}$ N on the Moon.

Given: mass of the Earth = $5.98 \times 10^{24}$ kg

mass of the Moon = $7.35 \times 10^{22}$ kg

radius of the Earth = $6.38 \times 10^{6}$ m

i) Find the height above the Earth’s surface where the Moon is orbiting.

ii) Using Newton’s Second Law of motion, calculate the linear velocity of the Moon.

iii) Calculate the centripetal acceleration of the Moon.

b) Most artificial satellites are placed in Low Earth Orbits (LEO), at a height typically around 200 km – 1200 km above the Earth’s surface. A 700 kg satellite is in the Low Earth Orbit at a height of 1200 km above the Earth’s surface.

i) What is the change in gravitational potential energy of the satellite before it is launched and when it is in orbit?

ii) Calculate the period of the satellite. Express your answer in hours.

iii) Satellites in LEO are used in telecommunications. State one difference between the LEO and geostationary satellites, and an advantage of using LEO satellites in telecommunications.

**Click here to show/hide answer**

a) i)

First, we’ll find the distance $r$ between the centers of the Earth and the Moon using Newton’s Law of Universal Gravitation and then subtract the Earth’s radius to find the height above the Earth’s surface.

$$F = G \frac{m_1 m_2}{r^2}$$

Solving for $r$, we get:

$$r = \sqrt{ \frac{G m_1 m_2}{F} }$$

Substituting $G = 6.674 \times 10^{-11} \, \text{N m}^2/\text{kg}^2$, $m_1 = 5.98 \times 10^{24} \, \text{kg}$, $m_2 = 7.35 \times 10^{22} \, \text{kg}$, and $F = 1.99 \times 10^{20} \, \text{N}$:

$$\begin{aligned} r &= \sqrt{ \frac{6.674 \times 10^{-11} \times 5.98 \times 10^{24} \times 7.35 \times 10^{22}}{1.99 \times 10^{20}} } \\ &= 3.839 \times 10^{8} \text{ m} \end{aligned}$$

Then, the height above the Earth’s surface $h$ is $r-\text{radius of the Earth}$.

$h = 3.839 \times 10^{8}-6.38 \times 10^{6} = 3.78 \times 10^{8} \text{ m}$

ii) Using Newton’s Second Law of motion, calculate the linear velocity of the Moon.

$$\begin{aligned} F &= m_{\text{Moon}} a \\ a &= \frac{v^2}{r} \\ F &= m_{\text{Moon}} \frac{v^2}{r} \\ v &= \sqrt{\frac{F r}{m_{\text{Moon}}}} \\ v &= \sqrt{\frac{1.99 \times 10^{20} \times 3.839 \times 10^{8}}{7.35 \times 10^{22}}} \\ v &= 1.02 \times 10^{3} \text{ ms}^{-1} \end{aligned}$$

iii) Calculate the centripetal acceleration of the Moon.

$$\begin{aligned} a_c &= \frac{v^2}{r} \\ &= \frac{(1.02 \times 10^{3})^{2}}{3.839 \times 10^{8}} \\ &= 2.71 \times 10^{-3} \text{ms}^{-2} \end{aligned}$$

b) (i)

The gravitational potential energy (GPE) of an object relative to Earth can be calculated using the formula:

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

where:

- $U$ is the gravitational potential energy,
- $G$ is the gravitational constant ($6.674 \times 10^{-11} \, \text{N m}^2/\text{kg}^2$),
- $M$ is the mass of the Earth ($5.98 \times 10^{24} \, \text{kg}$),
- $m$ is the mass of the satellite ($700 \, \text{kg}$),
- $r$ is the distance from the center of the Earth to the object. For an object on the Earth’s surface, $r = \text{radius of the Earth}$, and for an object in orbit, $r = \text{radius of the Earth} + \text{height of the orbit}$.

The change in gravitational potential energy ($\Delta U$) as the satellite goes from the surface to orbit is the difference in $U$ at these two positions.

$$\Delta U = 6.93 \times 10^{9} \text{ J}$$

ii)

The period $T$ of an orbiting satellite can be found using Kepler’s third law for circular orbits, which in terms of given quantities is:

$$T = 2\pi \sqrt{\frac{r^3}{G M}}$$

We’ll calculate this using $r = \text{radius of the Earth} + 1200 \, \text{km}$.

$$T = 1.82 \text{ hours}$$

iii)

**Difference:**

**Orbital Height and Position:**LEO satellites orbit at a height typically between 200 km and 1200 km above the Earth’s surface, making several orbits around the Earth each day. In contrast, geostationary satellites orbit at a height of approximately 35,786 km directly above the equator, remaining stationary relative to a point on the Earth. This means LEO satellites move relative to the Earth’s surface, while geostationary satellites do not.

**Advantage of Using LEO Satellites in Telecommunications:**

**Lower Latency:**Due to their closer proximity to the Earth, LEO satellites can provide lower latency communication compared to geostationary satellites. This is crucial for applications requiring real-time communication, such as video calls, online gaming, and certain financial transactions. Lower latency leads to faster data transmission times, improving the overall speed and responsiveness of communication networks.