TOPIC ONE: SPACE (THEORY)
1. The Earth has a gravitational field that exerts a force on objects both on it and around it.
Define weight as the force on an object due to a gravitational field
- Weight is defined as the force on a mass due to the gravitational field of a large celestial body, such as the Earth.
- It is a force and therefore a vector quantity.
- Newton’s Second Law states and hence:
Where:
W = weight (N)
m = mass (kg)
g = acceleration due to gravity (m/s^2)
Explain that a change in gravitational potential energy is related to work done
- To have a change in gravitational potential of an object within a gravitational field, work is done. This means that if you move an object in a gravitational field, work is done on that object in order for it to move.
- If an object is moved against the gravitational field, positive work is done and if the object is moved with the gravitational field, negative work is done.
- For objects within a gravitational field, gravitational potential energy is given by the formula:
Where:
Ep = gravitational potential energy (J)
m = mass (kg),
g = acceleration due to gravity (m/s^2),
h = height (m).
Define gravitational potential energy as the work done to move an object from a very large distance away to a point in a gravitational field
- Gravitational potential energy is defined as the work done to move an object from a very large distance away to a point within a gravitational field.
- Due to the inverse relationship, (Constant/ r), Ep will only become zero when r is at a very large distance. Thus if its initial potential energy is zero then its final potential energy must be negative.
- Therefore gravitational potential energy is given by the formula:
Where:
Ep = gravitational potential energy (J)
G = 6.67 x10^-11
M = mass of central body (kg)
m = mass of object (kg)
r = distance separating from the centre of each mass (m)
- The change in potential energy of a mass m is defined as the work done to move the mass m from its initial distance Ri to the centre of a body of mass M to its final position Rf. This is expressed as:
Where:
Ep = change in gravitational potential energy (J)
G = 6.67 x10^-11
M = mass of central body (kg)
m = mass of object (kg)
Ri = initial distance separating from the centre of each mass (m)
Rf = final distance separating from the centre of each mass (m)
2. Many factors have to be taken into account to maintain a stable orbit and return to earth.
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 any projectile is parabolic (ignoring air resistance).
- The velocity of the projectile continuously changes throughout its trajectory.
- The motion of a projectile can be split into two independent components, the horizontal (x) and the vertical (y) component.
- The horizontal (x) component is not subject to the acceleration due to gravity and therefore the velocity stays constant throughout its motion.
- The vertical (y) component is subject to the acceleration due to gravity and therefore the velocity changes throughout its motion.
- Adding these two vectors together, the initial or launch velocity of the projectile can be calculated via the use of Pythagoras theorem, and the launch angle can be calculated via the use of trigonometry.
Describe Galileo’s analysis of projectile motion
- The motion of a projectile can be split into two independent components, the horizontal (x) and the vertical (y) component.
The horizontal (x) component is not subject to the acceleration due to gravity and therefore the velocity stays constant throughout its motion.
The vertical (y) component is subject to the acceleration due to gravity and therefore the velocity changes throughout its motion.
- Galileo was responsible for stating that the trajectory of a projectile was parabolic. (ignoring air resistance)
- Galileo’s analysis of projectile motion led him to consider reference frames. This is what all measurements are referred to.
- Galileo supported the use of mathematics and experimentation in explaining natural phenomena.
Explain the concept of escape velocity in terms of the:
i. Universal gravitational constant
ii. Mass and radius of the planet
- By considering the kinetic and gravitational potential energy of a projectile, it can be shown mathematically that the escape velocity is only dependant upon the universal gravitational constant, the mass and the radius of the planet.
- Mathematical proof:
(At a distance of infinity both Ek and Ep = 0, similarly due to the principal of conservation of momentum Ek and Ep must also equal to zero at the planet’s surface. )
Where:
V = escape velocity (m/s)
G = 6.67 x10^-11
M = mass of central body (kg)
r = radius of the orbit from the centre of the central body (m)
- Therefore the escape velocity is only dependant upon the universal gravitational constant, the mass and the radius of the planet.
- It can be seen from the equation that escape velocity is not affected by the mass of the object meaning that the same escape velocity applies for all masses. Also the greater the mass of the planet the greater the escape velocity required while the greater the radius the smaller escape velocity required.
Outline Newton’s concept of escape velocity.
- In a thought experiment, Newton imagined that projectiles being launched horizontally from a very high above the surface above of the Earth will fall down to earth following a parabolic path.
- As the projectiles are given higher and higher velocities, they reach further and further around the earth’s curved surface.
- If the projectile is launched at just the right velocity, it will follow the curve of the Earth’s surface and orbit the earth.
- However if the speed of the projectile is equal or greater than the escape velocity, the projectile will fly off into space and never return to earth.
Identify why the term g-forces is used to explain the forces acting on an astronaut during launch
- During a launch an astronaut is subject to many forces such as thrust. The term ‘g’ force is a term used to express a person’s apparent weight (weight experienced at that instant) over the person’s true weight (mass x acceleration due to gravity).
- The astronaut feels an apparent weight of mg+ma and the true weight is 9.8m.
- Since g force is equal to apparent weight over true weight, therefore
G-force
Where:
a = acceleration of rocket (m/s^2)
g = acceleration due to gravity (m/s^2)
- By using this formula, the term g-forces can be used to explain the forces acting on an astronaut during a launch.
- Further a force of 5g is equivalent to 5 times the acceleration due to gravity.
Discuss the effect on the earth’s orbital motion and it’s its rotational motion on the launch of a rocket
- The Earth is a moving platform with two different motions which can be utilized in a rocket launch to gain a boost in velocity and therefore conserve fuel.
- The first is the orbital motion, Earth revolves around the Sun at 107,000km/h relative to the Sun and the second is its rotational motion, Earth rotates once on its axis per day so that a point on the Equator has a rotational velocity of approximately 1,700km/h relative to the Sun.
- A rocket heading into orbit is launched to the east at the equator to receive a velocity boost from the Earth’s rotational motion.
- The flight of a rocket heading into space is timed so that it can head out in the direction of the Earth’s rotational and orbital motion to receive a velocity boost; this is an inconvenience as if the rocket misses the opportunity it may have to wait for a long time before the rocket can be launched again.
Analyse the changing acceleration of a rocket launch in terms of:
i. Law of conservation of momentum
- The Law of conservation of momentum states that the total momentum of any group of objects remains the same unless outside forces act on the objects.
- The Law of conservation of momentum applies during a launch. Rocket engines generate thrust by burning fuel and expelling the resulting gases. Conservation of momentum means that as the gases move in one way, the rocket moves the other. (Momentum before the launch is zero; hence the momentum after the launch is also zero. The gases carry the momentum in one direction, and so the rocket carries an equal momentum in the opposite direction up). This can be expressed mathematically as:
Thus, the rocket is forcing a large volume of gases backwards behind it which in turn propels the rocket forward according to Newton’s 3rd Law.
- As fuel is consumed and the gases expelled, the mass of the system decreases. Therefore if thrust is constant as according to the law of conservation of momentum, the acceleration will increase as mass decreases.
- Note:
Impulse = (force) x (time) = Ft
Momentum (P) = (mass) x (velocity) = mv
Ft = ΔP
Ft = mv – mu
Ft = m (v - u)
F = m (v - u) / t
Analyse the changing acceleration of a rocket launch in terms of:
ii. Forces experienced by an astronaut
- Two forces act upon an astronaut during launch. The upwards thrust (T) as well as the downward weight (mg). A simple expression for acceleration of a rocket that is launched directly upwards: can be derived from Newton’s 2nd law:
Where:
a = acceleration of rocket (m/s^2)
T= thrust (N)
m = mass of rocker (kg)
g = acceleration due to gravity (m/s^2)
- As the rocket rises, the acceleration increases which in turn increases the forces experienced by the astronauts. This is because of the following factors:
1. As fuel is consumed and the gases expelled, the mass of the system decreases. According to the equation, acceleration is proportional to the thrust and inversely proportional to the mass, as the mass decreases, the acceleration increases.
2. The acceleration due to gravity decreases when altitude increases.
3. The air resistance decreases as air becomes thinner at higher altitude.
- Variations in the forces and g-forces experienced by astronauts can be accounted from the sequential shutdowns during a multistage rocket. The above factors can also account for the variations in the forces experienced by the astronauts.
Analyse the forces involved in uniform circular motion for a range of objects including satellites orbiting the earth
.
- When an object is undergoing uniform circular motion:
It is moving at constant speed
The magnitude of its velocity remains the same
The direction of its velocity is always changing, so it is continuously accelerating.
- There are two types of circular motions: angular (or orbital) and tangential (or linear)
- Angular velocity is the rate at which the angle is changing.
It is expressed as [w=θ/t] where:
w is the angular velocity in radian per second (rad-1)
θ: is the angle travelled in radians (rad)
t: is time taken in seconds (s)
One revolution/period is equal to 2pie radians, thus w = 2pie/t and the time is t = 2pie/ w.
- Tangential velocity: is the rate at which distance is changing.
Tangential velocity = change in distance / change in time
= circumference of circle/ period
That is v -= 2pier/t s since w = 2pie/ t in this case therefore
Tangential velocity can be expressed as [v=wr] where:
v= linear velocity (ms-1)
w= angular velocity (rads-1)
r = radius of circle (m)
- Centripetal acceleration is when an object moves in uniform circular motion, the magnitude of the velocity remains constant but its direction is always changing and so the object is therefore continuously accelerating.
The direction of this acceleration is always towards the centre because this is the direction in which the force must be applied to keep the object moving in a circle. The vectors are always perpendicular at every point.
It is expressed as[Ac= v^2/r] or [Ac=w^2r] where:
Ac= centripetal acceleration (ms-2)
v= velocity (ms-1)
r= radius (m)
- Centripetal force is the force needed to keep an object moving in uniform circular motion. Like centripetal acceleration, centripetal force is also directed towards the centre of the circle. An easy way to remember this is centripetal means “centre seeking”.
It is expressed as F = ma = mw^2r = mv^2/r
- Contextual info:
A radian is defined as the able subtended at the centre of a circle by an arc which is equal in length to its radius.
The size of the angle in radians is the ratio of the length of the arc, l, to the length
of the radius, r, of the circle.
It is expressed as θ = l / r
Theatre radian/ 2pie = theatre degrees / 360 degrees
Therefore theatre radian= 2pie x (theatre degrees / 360 degrees)
Theatre degrees= 360 degrees x (Theatre radian/ 2pie)
- Examples on above theory
1.Convert the following into radians per second:
i) 250rpm
ii) 25000rpm
Ai) knowing that one revolution is equal to 2 pie radians thus, 250rpm = 250 x 2pie therefore 250rps = (250 x 2pie)/60 = 26.18 rad/s
Aii) knowing that one revolution is equal to 2 pie radians thus, 25000rpm = 25000 x 2pie x 60 therefore 25000rps = (25000 x 2pie)/60 = 2618 rad/s