Hsc Physics Notes – Space Essay

Reasons for variations of g – Experimental Errors: ; Time: o Human reaction time error of approximately 0. As o Stopwatch limit of reading of + or – 0. Ass 4. Calculate result o False Judgment of the maximum angle of displacement and intern the end of a period ; Length: o Ruler limit of reading of + or – 0. Mm o Incorrect estimation of centre of mass ; General: o Our resistance which opposed the falling of the picket fence which the pendulum hung was not fixed Computer Assisted Technology Experiment: o Point from Aim: To determine acceleration due to gravity using computer assisted technology.

Apparatus: Computer, data studio software, picket fence, light gate As an object falls freely it accelerates due to the applied net force, gravity. If air resistance is neglected and the speed of the object is measured over a number of short, consecutive intervals as it falls the speed can be used to determine gravity. A picket fence, a clear plastic strip with uniformly spaced opaque sections will be dropped through a Photostat. The opaque sections block the Photostat beam and as the time between each block becomes shorter the speed increases.

The software on the computer has knowledge of the exact measurements of the picket fence bands is able to calculate the speed between each band and hence the change in velocity or acceleration due to gravity. Method: 1 . Hold the picket fence vertically at one end with thumb and forefinger so that it is just above the Photostat beam. 2. Click the record button on the computer software and drop the picket fence. 3. Click the stop button can record the value of 4. Repeat steps 1-3 at least 5 times to obtain 5 experimental values of g.

I Trial: I Acceleration due to gravity (g) (ms-2) Experimental value for g = 9. Moms-2 ; Air resistance ; Human error – picket fence may have been dropped at an angle Reasons for variations of g – Actual Variations: Height/Altitude: ; Increase in altitude ( Decrease in acceleration due to gravity ; Evident through the expression g = GM, as M and G are constant therefore g a 1/re Variation with Geographical location and altitude The actual value of the acceleration due to gravity, g, that will apply in a given situation will depend upon geographical location and variation with altitude.

If the body is the same then mass is constant, hence the weight force would be dependent on the planet and its gravitational field. In this instance the relationship W a g may be used. Find the weight of a keg man standing on the surface of the Earth (g = 9. Ms-2) and the Moon (1. Ms-2). Hearth Won = 70 x 9. 8 = NON Context 2 – Many factors have to be taken into account to achieve a successful rocket launch, 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 Trajectory – The path followed by a projectile during flight The trajectory of an object in projectile motion on the Earth is parabolic. This is derived through the vector addition of the vertical and horizontal components of motion that produce a resulting direction and magnitude.

Vertical Component: ; Gravity acts vertically downwards on the projectile towards the centre of the earth ; Downwards net force causes an acceleration in this direction ; As the projectile accelerates the vertical displacement per unit time increases ;sat ova=u+at so = UT + hath ova -? – u + as so = h (u + Horizontal Component: No net force – gravity can have no effect as the motions are perpendicular to each other ; Uniform and constant according to the principle of Inertia ; No acceleration/change in velocity in the absence of air resistance ; Equal intervals of distance at equal intervals of time ; Calculation: s = Vt Describe Galileo analysis of projectile motion Galileo analyses in his 1638 publication “Dialogues of the Two New Sciences” that the trajectory of any object undergoing projectile motion within the Earth’s gravitational field is parabolic. He discovered that there were two motions acting on a projectile tooth completely independent of each other but when added provided the total motion of the object. He realized that the vertical component would undergo uniform acceleration and change (excluding air resistance) whilst the horizontal would remain constant, hence resulting in the parabolic shape.

Explain the concept of escape velocity in terms of the: ; Gravitational constant ; Mass and radius of the planet Escape velocity is the minimum velocity that allows an object to escape from a point in the gravitational field of a planet (or other body) to infinity without further energy input (scalar quantity). 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. For this reason h move = (Game)/re and by canceling the formula of escape velocity may be derived. [pick] This formula demonstrates that velocity is dependent upon the gravitational constant, the mass and the radius of the planet. It does not depend on any intrinsic property of the projectile. As v = RASH of the escape velocity of the rocket it will be able to escape the gravitational field.

Escape velocity increases as the mass of the planet increases and decreases as the radius of the planet increases Outline Newton’s concept of escape velocity Newton conducted a thought experiment in which he envisaged a cannon firing a projectile horizontally from a peak on the earth’s surface. Ignoring the factor of air resistance the projectile would follow a parabolic trajectory and return to earth. However, if the speed of the projectile was increased it would take progressively longer to reach the ground. By increasing the speed enough the projectile will never circle will become an ellipse and the trajectory hyperbolic escaping the earth’s revelation field. The velocity corresponding to the time when this first occurs is then the escape velocity.

Identify why the term ‘g forces’ is used to explain the forces acting on an astronaut during launch G-forces refers to the force experienced by an astronaut in terms of the Earth’s gravitational field strength at the earths surface. GIG is equal to the force experienced by an astronaut on the surface of the earth. On a rocket accelerating upwards at 9. Ms-2, then the astronaut experiences a net force equal to ass. When an astronaut is in farewell they experience ass. The term g forces is used because it is easy to relate to and eases calculations as to the forces which the human body can withstand during launch. Calculating g forces: G force = acceleration due to gravity + object’s acceleration 9. 8 Fine = ma and Fine = if Example: Free-fall t Fine = ma = m x (-9. ) = -MGM t Fine = FEE = Normal – Weight Equating N -W MGM N =- MGM + MGM N = O (MGM) = GO environment Discuss the effect of the Earth’s orbital motion and its rotational motion on the launch of a rocket The orbital motion and rotational motion of the Earth both have related effects, the onto orbit. This effect arises when a rocket is launched its velocity is not Just provided by the rocket but also the velocity of the earth. Rotational Motion: The earth rotates on its axis in an anticlockwise direction as viewed from the North Pole with a velocity of 0. Simms-1. Rockets launched in an easterly direction at the equator (where the orbital velocity is the greatest) hence have an initial velocity of 0. Simms-1. This means that the rocket requires a lower velocity to achieve orbit (ii. Parking orbit in preparation for Journeying further into space). This in turn reduces munch costs and enables less fuel and a greater payload to be carried.

Orbital Motion: The earth rotates around the sun in an anticlockwise direction as viewed from the North Pole with an average speed of 10,Khmer-1. If trying to reach an outer planet such as Mars when the rocket is in parking orbit, third-stage rockets are fired to accelerate the rocket in the direction of the earth’s orbit providing it with a greater velocity than the earth, this is known as a Human Transfer Orbit. This enables it to reach higher orbits and outer planets as well as reduce launch costs and increase payload. Conversely the rocket may also be launched in the opposite direction to the earth’s orbit enabling it to reach inner planets.

However, the rotational and orbital motion of the earth creates difficulties in launching in the direction against the motion. It would require a greater velocity and in turn incur greater launch costs. Analyses the changing acceleration of a rocket during launch in terms of the: of Conservation of Momentum ; Forces experienced by astronauts Law of Conservation of Momentum: ; Law Rocket propulsion is derived from a force pair (as described in Newton’s third law). The initial momentum of the rocket and its fuel is zero. This sum must be preserved (the Law of Conservation of Momentum). When mass is ejected from the rocket with a momentum P the rocket will gain an equal momentum P in the opposite direction.

When the force of the rocket on the gases is greater than the weight the rocket will accelerate upwards. If the thrust produced by the engines remains constant, the rate mass (fuel) (F = ma), a decrease in g at higher altitudes resulting in a lesser weight and a decrease in air density and hence air resistance. Forces experienced by astronauts: The two forces that act upon an astronaut during launch are the upward thrust (T) and the downward weight (W or MGM). As described above, if the mass of the rocket decreases during flight and the thrust remains constant, the acceleration of the rocket (and astronauts) increases. Thus the g forces experienced by the astronaut increase so that it is greater than one.

They may reach up to approximately ass during lift-off. As each stage of the rocket falls away the astronauts experience momentary weightlessness. The g forces then rise again but not to the same maximum and eventually are not effected by the earth’s gravitational field. Analyses the forces involved in uniform circular motion for a range of objects, including satellites orbiting the Earth Uniform circular motion is circular motion with a uniform orbital speed. The key force in uniform circular motion is centripetal force. This is a force that always acts in a direction towards the centre of the circle in uniform circular motion and perpendicular to the tangential velocity.

Centripetal force = Resultant (Net) Force in UCM- [Pick]. Circular Motion I Conical Pendulum I Car on a Circular Track I Source of Centripetal Force I Tension in the string I Friction between the trees and the road I Satellite Orbiting the Earth I Gravitational attraction between the Earth and the satellite I Compare qualitatively low earth and gee-stationary orbits Low Earth orbits are generally 250 – smoke above the earth’s surface as below this range there is too much atmospheric drag and above it the Van Allen Radiation belt begins to appear. It has an orbital period of about 90 minutes at km above with an orbital velocity of 27900kmh-1.

Weather and remote sensing satellites are both in low Earth orbits as they are able to provide high quality visuals of the earth. In comparison gee-stationary orbits are a particular type of geosynchronous orbit which means they have a period equal to the earth’s rotational period (about 24 hours). Geostationary orbits lie on the equatorial plane and maintain the same position with respect to the rotating Earth at all times. It has a much slower orbital velocity of about 11000kmh-1 and is at a much higher altitude of approximately 36000km. This type of orbit is used for communication, satellite TV as well as data transmission. Define the term orbital velocity and the quantitative and qualitative relationship