# Rocket Launches and Orbital Motion

## Describe the trajectory of an object undergoing projectile motion within the Earth’s gravitational field in terms of horizontal and vertical components

The trajectory of an object in projectile motion on Earth is a parabola. The motion of an object can be derived through analysing the horizontal and vertical components of its motion and then adding the vectors to produce the resulting direction and magnitude of the object’s velocity (the object’s net velocity vector). In standard projectile motion on Earth, the horizontal component is constant, and is equal to the original horizontal component at the point of release. The vertical component is constantly changing, being affected by the gravitational field. The change occurs directly towards the centre of the field, and in the Earth’s case, acts in this direction at 9.8m/s for every second in flight. At any given time, the vertical component is equal to the initial vertical component at the time of release, minus 9.8 times the time elapsed, where a negative value is downward motion.

Remember- An object in projectile motion travels in a parabola with a constant x-component and a changing y-component.

## Describe Galileo’s analysis of projectile motion

You’ll need to memorise what Galileo said and how he devised his vector analysis. This is a history lesson, but it also tests whether you understand how the component system works so make sure you explain that too.

Galileo was the first to analyse projectile motion mathematically and have his work documented. Instead of considering the motion of the object as a whole, he divided the motion into a horizontal and a vertical component, which when added provide the total motion of the object. Galileo realised that during projectile motion, only the vertical component would change (excluding air resistance) while the horizontal component would remain constant. He also realised that the motion of projectiles is parabolic in nature due to the uniform acceleration vertically with constant horizontal motion.

Remember- Galileo was the first to break a projectile’s motion into components.

## Explain the concept of escape velocity in terms of the gravitational constant and the mass and radius of the planet

This dotpoint is essentially memorising the formula, and explaining the concept of what escape velocity is.

Escape velocity is the velocity required at a planet’s surface to completely leave its gravitational field without further energy input. This means that it must have the same amount of kinetic energy as the absolute value of the gravitational potential energy it has at the point of takeoff. Assuming takeoff from the planet’s surface, this means $\frac{1}{2}mv^2 = \frac{Gmm_p}{r_p}$ where $m_p$ refers to the mass of the planet. Cancelling, $v^2 = \frac{2Gm_p}{r_p}$. This formula links escape velocity to the gravitational constant and the mass and radius of the planet. If at the surface of the planet $v^2$ is equal to the RHS, then the rocket will be able to escape the gravitational field. Thus the $v$ value at this point is the escape velocity. Escape velocity increases as the mass of the planet increases, and decreases as the radius of the planet increases.

Remember- Escape velocity is the velocity needed at the surface to exit the gravitational field. More mass and a smaller radius make it bigger.

## Outline Newton’s concept of escape velocity

Make sure you can properly explain this, it has caused people trouble before. Memorise it.

Newton envisaged a cannon firing a projectile horizontally from the Earth’s surface. Ignoring air resistance, the projectile would prescribe a parabola, eventually falling back to Earth. However, as the speed of the projectile is increased, the projectile will take progressively longer to hit the ground, because although gravity is pulling towards the centre of the field, the Earth’s surface is falling away from the projectile at the same time due to its horizontal motion. Increase the speed enough, and the projectile will never hit the ground, instead travelling in a circle around the Earth. As the velocity increases even more, the circle becomes an ellipse, and if the speed is increased enough, the trajectory becomes hyperbolic. At this point, the projectile has enough velocity to leave the gravitational field. The velocity corresponding to the time when this first occurs is then the escape velocity.

Remember- Newton used a horizontal cannon to visualise orbits and escape velocity.

## Identify why the term “g forces” is used to explain the forces acting on an astronaut during launch

This dotpoint is comparatively easy, but when considering G-forces take care to add the forces correctly. It may be easiest to visualise yourself in the scenario to get an idea as to how the forces interact.

‘G-Forces’ refers to the force experienced by an astronaut in terms of the Earth’s gravitational field strength at the Earth’s surface. 1G is equal to the force experienced by an astronaut on the surface of the Earth: w = mg where g = 9.8. If a rocket is accelerating upwards at 9.8m/s2 , then the astronaut experiences a net force equal to 2Gs (twice the force they would experience due to Earth’s gravity). If an astronaut is in freefall, they experience 0Gs. The term g forces is used because it is easy to relate to, and because it is eases calculations as to the forces which the human body can withstand during launch.

Remember- G-force measures acceleration in terms of Earth’s gravity.

## Perform a first-hand investigation, gather information and analyse data to calculate initial and final velocity, maximum height reached, range and time of flight of a projectile for a range of situations by using simulations, data loggers and computer analysis

In this experiment, we placed a grid against a wall and then threw a ball in a parabola in front of the grid. A video camera recorded the experiment so that we could see the ball travelling in front of the grid. Using the grid, we were able to calculate the position of the ball. Times were calculated based on each video frame representing $\frac{1}{25}$th of a second. By analysing the movement of the ball between frames, we were able to use the standard motion equations in the X and Y axes to calculate the initial and final velocities, as well as the maximum height reached and the range of the projectiles, in this case, a tennis ball. There would have been errors caused by the ball not travelling in a straight line (i.e. It did not travel only vertically and horizontally, but laterally too) resulting in erroneous readings, and it is likely that the camera did not record frames at exactly $\frac{1}{25}$th of a second intervals, producing further errors.

Remember- Grid on the wall, tennis ball, video camera, analyse changes between frames.

## Analyse the changing acceleration of a rocket during launch in terms of the Law of Conservation of Momentum and the forces experienced by astronauts

The key part of this dotpoint is analysis in terms of Conservation of Momentum. To say that the thrust is constant and the weight decreases, so acceleration increases by F = ma is incomplete. Make sure you deal with Conservation of Momentum as well.

The Law of Conservation of Momentum states that in a closed system, the sum of the momenta before a change is equal to the sum of momenta after the change. In a rocket, the change is the release of exhaust gas. The momentum of the exhaust gas is the same as the rocket’s momentum, with a reversed direction, so that when added, they amount to 0. P = mv . This equation links velocity to mass and momentum. Because the sum of the momentum of the exhaust gas and the rocket is zero, |mexhaust × vexhaust| = |mrocket × vrocket| (taking absolute values because one side of this equation will be negative, since the rocket and the exhaust travel in opposite directions). As the rocket travels into space, it burns fuel and so its mass decreases. But because the momentum of the exhaust is constant, this means the rocket’s velocity must rise in order to balance the equation. This means that when the burn is completed, the rocket is travelling faster than if the rocket had maintained a constant mass (because vrocket is now larger as mrocket decreased while procket and pexhaust remained constant). This in turn implies that the acceleration of the rocket has increased during the burn in order to fulfil conservation of energy. This can be seen through F = ma, where F is the thrust of the rocket motor. Because the rocket motor provides constant thrust, F is a constant. As the rocket burns fuel, its mass decreases, and so for ma to remain constant the rocket’s acceleration must increase. This means that as the rocket takes off, its acceleration becomes progressively higher as it burns its fuel and becomes lighter. For the astronauts, this means an increasing force. So as the rocket lifts off, its thrust needs to be progressively reduced to protect its occupants.

Remember- As a rocket burns fuel, it accelerates faster.

## Discuss the effect of the Earth’s orbital motion and its rotational motion on the launch of a rocket

A question on this area will need a comprehensive answer, so make sure that you address the positives and negatives of both Earth’s rotation and its orbital motion.

The orbital motion of the Earth and the rotational motion of the Earth both have related effects, the orbital motion affecting interplanetary travel and rotational motion affecting satellites orbiting the Earth. The effect arises because when a rocket is launched, its velocity is not simply that provided by the rocket motor, but also the velocity it has because of the Earth’s movement through space.

In terms of orbital motion, space probes launched in the same direction as the Earth’s orbit carry its orbital velocity, again reducing fuel requirements, resulting in greater payloads or cheaper missions.

For rotation, the Earth rotates constantly in an anticlockwise direction as viewed from above the North Pole. Rockets launched in an easterly direction therefore carry extra momentum with them, giving them around an additional 0.5km/s towards their velocity. This means that to achieve orbit, the rocket only needs to accelerate 7.5km/s, with the additional 0.5km/s resulting from the motion of the Earth. This means that less fuel is required, and/or a greater payload can be carried.

On the other hand, the orbital and rotational motion makes it hard to launch rockets in a direction against the motion. For example, to launch a rocket in a westerly direction into orbit would take an acceleration of 8.5km/s, significantly greater. Likewise, to launch a space probe against the motion of the Earth would result in far greater fuel requirements to achieve the same trajectory.

Remember- If you launch a satellite in the direction of the Earth’s orbit or rotation, it effectively has more velocity, saving fuel.

## Analyse the forces involved in uniform circular motion for a range of objects including satellites orbiting the Earth

The forces involved for uniform circular motion for satellites are the same as uniform circular motion in any situation. There will always be tangential velocity, and there will always be a centripetal force that causes the object to travel in a circular path. The only difference is the source of the forces. For an examination of the virtual force centrifugal force, see the Extra Content section at the end of the Guide.

Uniform circular motion refers to the motion of objects that prescribe a perfect circle as they move. The key force in uniform circular motion is centripetal force. Centripetal force is a centre-seeking force that always acts in a direction towards the centre of the circle in uniform circular motion. The
formula for centripetal force is $F = \frac{mv^2}{r}$ The forces for uniform circular motion may be sourced differently, but all are centripetal in nature and all follow this formula. This is true of satellites in orbit around the Earth, cars as they turn, and a charged particle in a magnetic field.

Remember- Uniform circular motion always requires centripetal force, which can come from a variety of sources.

## Compare qualitatively low Earth and geo-stationary orbits

A low Earth orbit is one that is approximately 300km from the Earth’s surface, although technically it refers to any satellite below 1500km in altitude. LEOs (Low Earth Orbit satellites) have an orbital period of around 90 minutes, with an orbital velocity of about 8km/s. Geostationary satellites remain above a fixed position on the Earth, because their orbital period is exactly 24 hours. They are far higher up than LEOs, at around 36000km in altitude, and have a lower orbital velocity (around 3km/s). A geo-stationary orbit is a special type of geo-synchronous orbit. A geo-synchronous orbit refers to any orbit with a period of 24 hours. However, not all geo-synchronous orbits are geo- stationary, because geo-stationary orbits must be equatorial, travelling directly above the equator. A polar orbit may be geo-synchronous, but it cannot be geo-stationary.

So essentially, compared to the low Earth orbit, a geostationary orbit is higher up, has a longer orbital period and a lower orbital velocity.

Remember- A low Earth orbit is low and fast with a short period, and geo-stationary is high and slow with a 24-hour period.

## Identify data sources, gather, analyse and present information on the contri- bution of one of the following to the development of space exploration: Tsiolkovsky, Oberth, Goddard, Esnault-Pelterie, O’Neill or von Braun

Konstantin Tsiolkovsky (1857-1935), a Russian scientist, while not contributing directly to space travel during his lifetime, devised many new ideas that were almost prophetic and extremely important in space travel. The key ideas he had were firstly the principles behind rocket propulsion, secondly the use of liquid fuels, and finally multi-stage rockets. Tsiolkovsky showed how Newton’s 3rd law and how conservation of momentum can be applied to rockets. This principle underlies the functioning of all rockets, and is vital to understanding their operation. Secondly, Tsiolkovsky proposed using liquid hydrogen and liquid oxygen as rocket fuels so that the thrust produced by a rocket could be varied. These same fuels were implemented in the Saturn V rocket that powered the Apollo missions to the moon, and the use of liquid fuels has proved vital in manned spaceflight because they allow g-forces experienced by astronauts to be controlled, unlike in solid fuel engines. Also, liquid fuels are used in satellites and space probes, where intermittent firing of rockets is desired rather than a continuous burn as provided by a solid rocket motor. Finally, Tsiolkovsky visualised a 20-stage rocket train that dropped stages as each stage ran out of fuel, to cut weight and improve efficiency. Although 20 stages was rather extreme, the multistage rocket proved vital in high-energy launches for manned space missions such as Apollo as well as missions with large payloads. So while Tsiolkovsky didn’t directly impact space exploration during his lifetime, he devised many ideas that are vital to spaceflight today.

Remember- Tsiolkovsky devised concepts well before they could be practically implemented.

## Define the term orbital velocity and the quantitative and qualitative relation- ship between orbital velocity, the gravitational constant, mass of the central body, mass of the satellite and the radius of the orbit using Kepler’s Law of Periods

Although the dotpoint mentions the relationship between orbital velocity and the mass of the satellite, the mass of the satellite is irrelevant. Looking at the 2 equations provided here, the only 2 variables are the mass of the central body and the orbital radius. This means that there is no relationship between the mass of the satellite and orbital velocity, providing the satellite is significantly lighter than the central body (as otherwise more complicated effects would come into play).

Orbital velocity is simply the speed at which a satellite is travelling, calculated by dividing the distance it travels in its orbit (which is the circumference of the circle in a circular orbit) by its orbital period. Orbital velocity is linked to the gravitational constant, the mass of the central body and the radius of the orbit according to the formulae $\frac{r^3}{T_2} = \frac{Gm_c}{4 \pi^2}$ and $v = \frac{2 \pi r}{T}$. Essentially, orbital velocity increases when the mass of the central body increases, and decreases when the radius of the orbit is increased. The mass of the satellite has no bearing on the orbital velocity, as it cancels out when calculating orbital velocity.

## Account for the orbital decay of satellites in low Earth orbit

LEOs continually lose orbital speed and require periodic rocket boosts in order to stay in orbit, preventing them from crashing. The reason LEOs lose velocity is because the Earth’s atmosphere extends far into space. The boundary between the atmosphere and the vacuum of space isn’t clearly defined, and there are still air particles high above the Earth’s surface. As LEOs collide with these particles they slowly lose orbital velocity through friction, resulting in orbital decay. Orbital decay is where a satellite loses orbital velocity and therefore moves into a lower orbit closer to the Earth’s surface. If orbital decay continues, the satellite will eventually crash.

Remember- LEOs crash because they collide with air particles.

## Discuss issues associated with safe re-entry into the Earth’s atmo- sphere and landing on the Earth’s surface (including “Identify that there is an op- timum angle for safe re-entry for a manned spacecraft into the Earth’s atmosphere and the consequences of failing to achieve this angle”)

Re-entry is a complex procedure due to the high velocities and temperatures encountered, as well as the fine balance of trajectory required to land safely. To land a space vehicle, the vehicle must firstly slow down, and secondly travel back down through the atmosphere. These are done simultaneously

with atmospheric drag slowing the vehicle as it descends. The high velocity of the vehicle results in a great deal of friction, which heats the vehicle to up to 3000◦C depending on airflow. This necessitates highly temperature resistant shielding, usually ceramic or carbon based, that can withstand the

temperatures and protect the rest of the vehicle as it descends. Modern designs also feature blunt noses and have the spacecraft descend belly-first, which ensures the majority of the vehicle is shielded. Without appropriate shielding, the vehicle will be unable to return, as recently seen in the 2004 Columbia space shuttle accident in which its heat shielding was compromised. Secondly, the angle of re-entry is critical. If the angle is too steep, the descent rate will be too fast, and the vehicle will encounter the higher density atmosphere closer to the Earth’s surface while it retains too much of its velocity. Higher density air provides more drag, which therefore decelerates the vehicle faster and leads to higher temperatures. This will result in at the very minimum excess g-forces for the crew, and at worst, the extra heating could destroy the entire vehicle. On the other hand, if the angle is too shallow, the spacecraft will retain too much of its velocity and exit the atmosphere by effectively skimming it, returning to space. The vehicle must have an angle between 5.2 and 7.2 degrees to make a safe re-entry. During re-entry, the high temperature of the spacecraft results in the air around it becoming ionised. This results in an ionisation blackout, with the ionised air blocking radio communication with the ground during re-entry. Although not a direct hazard, it can cause complications in the event of a safety issue arising during re-entry which could endanger the spacecraft. Finally, in order to land, the descent rate must be slowed dramatically. In the Apollo missions and with non-reusable space probes, parachutes are used to slow the descent to make a gentle landing. The space shuttle uses wings to generate lift, enabling it to glide to a gentle landing.

Remember- To re-enter, you need strong heat shielding and an approach with a specific angle of descent.