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REX 3 Unit 10A Vocabulary
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Introduction to Free Fall A free-falling object is an object that is falling under the sole influence of gravity. Any object that is being acted upon only by the force of gravity is said to be in a state of free fall. There are two important motion characteristics that are true of free-falling objects: • Free-falling objects do not encounter air resistance. • All free-falling objects (on Earth) accelerate downwards at a rate of 9.8 m/s/s (often approximated as 10 m/s/s for back-of-the-envelope calculations) Because free-falling objects are accelerating downwards at a rate of 9.8 m/s/s, a ticker tape trace or dot diagram of its motion would depict an acceleration. The dot diagram at the right depicts the acceleration of a free-falling object. The position of the object at regular time intervals - say, every 0.1 second - is shown. The fact that the distance that the object travels every interval of time is increasing is a sure sign that the ball is speeding up as it falls downward. Recall from an earlier lesson, that if an object travels downward and speeds up, then its acceleration is downward. Free-fall acceleration is often witnessed in a physics classroom by means of an ever-popular strobe light demonstration. The room is darkened and a jug full of water is connected by a tube to a medicine dropper. The dropper drips water and the strobe illuminate the falling droplets at a regular rate - say once every 0.2 seconds. Instead of seeing a stream of water free-falling from the medicine dropper, several consecutive drops with increasing separation distance are seen. The pattern of drops resembles the dot diagram shown in the graphic at the right. The Acceleration of Gravity It was learned in the previous part of this lesson that a free-falling object is an object that is falling under the sole influence of gravity. A free-falling object has an acceleration of 9.8 m/s/s, downward (on Earth). This numerical value for the acceleration of a free-falling object is such an important value that it is given a special name. It is known as the acceleration of gravity - the acceleration for any object moving under the sole influence of gravity. A matter of fact, this quantity known as the acceleration of gravity is such an important quantity that physicists have a special symbol to denote it - the symbol g. The numerical value for the acceleration of gravity is most accurately known as 9.8 m/s2. There are slight variations in this numerical value (to the second decimal place) that are dependent primarily upon on altitude. We will occasionally use the approximated value of 10 m/s2 in order to reduce the complexity of the many mathematical tasks that we will perform with this number. By so doing, we will be able to better focus on the conceptual nature of physics without too much of a sacrifice in numerical accuracy. g = 9.8 m/s2, downward Look It Up! Even on the surface of the Earth, there are local variations in the value of the acceleration of gravity (g). These variations are due to latitude, altitude and the local geological structure of the region. Recall from an earlier lesson that acceleration is the rate at which an object changes its velocity. It is the ratio of velocity change to time between any two points in an object's path. To accelerate at 9.8 m/s2 means to change the velocity by 9.8 m/s each second. If the velocity and time for a free-falling object being dropped from a position of rest were tabulated, then one would note the following pattern. Time (s) Velocity (m/s) 0 0 1 - 9.8 2 - 19.6 3 - 29.4 4 - 39.2 5 - 49.0 . Observe that the velocity-time data above reveal that the object's velocity is changing by 9.8 m/s each consecutive second. That is, the free-falling object has an acceleration of approximately 9.8 m/s2. Another way to represent this acceleration of 9.8 m/s2 is to add numbers to our dot diagram that we saw earlier in this lesson. The velocity of the ball is seen to increase as depicted in the diagram at the right. (NOTE: The diagram is not drawn to scale - in two seconds, the object would drop considerably further than the distance from shoulder to toes.) Representing Free Fall by Graphs • Early in Lesson 1 it was mentioned that there are a variety of means of describing the motion of objects. One such means of describing the motion of objects is through the use of graphs - position versus time and velocity vs. time graphs. In this part of Lesson 5, the motion of a free-falling motion will be represented using these two basic types of graphs. Representing Free Fall by Position-Time Graphs A position versus time graph for a free-falling object is shown below. Observe that the line on the graph curves. As learned earlier, a curved line on a position versus time graph signifies an accelerated motion. Since a free-falling object is undergoing an acceleration (g = 9.8 m/s/s), it would be expected that its position-time graph would be curved. A further look at the position-time graph reveals that the object starts with a small velocity (slow) and finishes with a large velocity (fast). Since the slope of any position vs. time graph is the velocity of the object (as learned in Lesson 3), the small initial slope indicates a small initial velocity and the large final slope indicates a large final velocity. Finally, the negative slope of the line indicates a negative (i.e., downward) velocity. Representing Free Fall by Velocity-Time Graphs A velocity versus time graph for a free-falling object is shown below. Observe that the line on the graph is a straight, diagonal line. As learned earlier, a diagonal line on a velocity versus time graph signifies an accelerated motion. Since a free-falling object is undergoing an acceleration (g = 9,8 m/s/s, downward), it would be expected that its velocity-time graph would be diagonal. A further look at the velocity-time graph reveals that the object starts with a zero velocity (as read from the graph) and finishes with a large, negative velocity; that is, the object is moving in the negative direction and speeding up. An object that is moving in the negative direction and speeding up is said to have a negative acceleration (if necessary, review the vector nature of acceleration). Since the slope of any velocity versus time graph is the acceleration of the object (as learned in Lesson 4), the constant, negative slope indicates a constant, negative acceleration. This analysis of the slope on the graph is consistent with the motion of a free-falling object - an object moving with a constant acceleration of 9.8 m/s/s in the downward direction. The Kinematic Equations The goal of this first unit has been to investigate the variety of means by which the motion of objects can be described. The variety of representations that we have investigated includes verbal representations, pictorial representations, numerical representations, and graphical representations (position-time graphs and velocity-time graphs). In Lesson 6, we will investigate the use of equations to describe and represent the motion of objects. These equations are known as kinematic equations. There are a variety of quantities associated with the motion of objects - displacement (and distance), velocity (and speed), acceleration, and time. Knowledge of each of these quantities provides descriptive information about an object's motion. For example, if a car is known to move with a constant velocity of 22.0 m/s, North for 12.0 seconds for a northward displacement of 264 meters, then the motion of the car is fully described. And if a second car is known to accelerate from a rest position with an eastward acceleration of 3.0 m/s2 for a time of 8.0 seconds, providing a final velocity of 24 m/s, East and an eastward displacement of 96 meters, then the motion of this car is fully described. These two statements provide a complete description of the motion of an object. However, such completeness is not always known. It is often the case that only a few parameters of an object's motion are known, while the rest are unknown. For example as you approach the stoplight, you might know that your car has a velocity of 22 m/s, East and is capable of a skidding acceleration of 8.0 m/s2, West. However you do not know the displacement that your car would experience if you were to slam on your brakes and skid to a stop; and you do not know the time required to skid to a stop. In such an instance as this, the unknown parameters can be determined using physics principles and mathematical equations (the kinematic equations). The BIG 4 The kinematic equations are a set of four equations that can be utilized to predict unknown information about an object's motion if other information is known. The equations can be utilized for any motion that can be described as being either a constant velocity motion (an acceleration of 0 m/s/s) or a constant acceleration motion. They can never be used over any time period during which the acceleration is changing. Each of the kinematic equations include four variables. If the values of three of the four variables are known, then the value of the fourth variable can be calculated. In this manner, the kinematic equations provide a useful means of predicting information about an object's motion if other information is known. For example, if the acceleration value and the initial and final velocity values of a skidding car is known, then the displacement of the car and the time can be predicted using the kinematic equations. Lesson 6 of this unit will focus upon the use of the kinematic equations to predict the numerical values of unknown quantities for an object's motion. The four kinematic equations that describe an object's motion are: There are a variety of symbols used in the above equations. Each symbol has its own specific meaning. The symbol d stands for the displacement of the object. The symbol t stands for the time for which the object moved. The symbol a stands for the acceleration of the object. And the symbol v stands for the velocity of the object; a subscript of i after the v (as in vi) indicates that the velocity value is the initial velocity value and a subscript of f (as in vf) indicates that the velocity value is the final velocity value. Each of these four equations appropriately describes the mathematical relationship between the parameters of an object's motion. As such, they can be used to predict unknown information about an object's motion if other information is known. In the next part of Lesson 6 we will investigate the process of doing this. Kinematic Equations and Problem-Solving The four kinematic equations that describe the mathematical relationship between the parameters that describe an object's motion were introduced in the previous part of Lesson 6. The four kinematic equations are: In the above equations, the symbol d stands for the displacement of the object. The symbol t stands for the time for which the object moved. The symbol a stand for the acceleration of the object. And the symbol v stands for the instantaneous velocity of the object; a subscript of i after the v (as in vi) indicates that the velocity value is the initial velocity value and a subscript of f (as in vf) indicates that the velocity value is the final velocity value. Problem-Solving Strategy In this part of Lesson 6 we will investigate the process of using the equations to determine unknown information about an object's motion. The process involves the use of a problem-solving strategy that will be used throughout the course. The strategy involves the following steps: 1. Construct an informative diagram of the physical situation. 2. Identify and list the given information in variable form. 3. Identify and list the unknown information in variable form. 4. Identify and list the equation that will be used to determine unknown information from known information. 5. Substitute known values into the equation and use appropriate algebraic steps to solve for the unknown information. 6. Check your answer to ensure that it is reasonable and mathematically correct. The use of this problem-solving strategy in the solution of the following problem is modeled in Examples A and B below. Example Problem A . Ima Hurryin is approaching a stoplight moving with a velocity of +30.0 m/s. The light turns yellow, and Ima applies the brakes and skids to a stop. If Ima's acceleration is -8.00 m/s2, then determine the displacement of the car during the skidding process. (Note that the direction of the velocity and the acceleration vectors are denoted by a + and a - sign.) The solution to this problem begins by the construction of an informative diagram of the physical situation. This is shown below. The second step involves the identification and listing of known information in variable form. Note that the vf value can be inferred to be 0 m/s since Ima's car comes to a stop. The initial velocity (vi) of the car is +30.0 m/s since this is the velocity at the beginning of the motion (the skidding motion). And the acceleration (a) of the car is given as - 8.00 m/s2. (Always pay careful attention to the + and - signs for the given quantities.) The next step of the strategy involves the listing of the unknown (or desired) information in variable form. In this case, the problem requests information about the displacement of the car. So d is the unknown quantity. The results of the first three steps are shown in the table below. Diagram: Given: Find: vi = +30.0 m/s vf = 0 m/s a = - 8.00 m/s2 d = ?? The next step of the strategy involves identifying a kinematic equation that would allow you to determine the unknown quantity. There are four kinematic equations to choose from. In general, you will always choose the equation that contains the three known and the one unknown variable. In this specific case, the three known variables and the one unknown variable are vf, vi, a, and d. Thus, you will look for an equation that has these four variables listed in it. An inspection of the four equations above reveals that the equation on the top right contains all four variables. vf2 = vi2 + 2 • a • d Once the equation is identified and written down, the next step of the strategy involves substituting known values into the equation and using proper algebraic steps to solve for the unknown information. This step is shown below. (0 m/s)2 = (30.0 m/s)2 + 2 • (-8.00 m/s2) • d 0 m2/s2 = 900 m2/s2 + (-16.0 m/s2) • d (16.0 m/s2) • d = 900 m2/s2 - 0 m2/s2 (16.0 m/s2)*d = 900 m2/s2 d = (900 m2/s2)/ (16.0 m/s2) d = (900 m2/s2)/ (16.0 m/s2) d = 56.3 m The solution above reveals that the car will skid a distance of 56.3 meters. (Note that this value is rounded to the third digit.) The last step of the problem-solving strategy involves checking the answer to assure that it is both reasonable and accurate. The value seems reasonable enough. It takes a car a considerable distance to skid from 30.0 m/s (approximately 65 mi/hr) to a stop. The calculated distance is approximately one-half a football field, making this a very reasonable skidding distance. Checking for accuracy involves substituting the calculated value back into the equation for displacement and insuring that the left side of the equation is equal to the right side of the equation. Indeed it is! Example Problem B Ben Rushin is waiting at a stoplight. When it finally turns green, Ben accelerated from rest at a rate of a 6.00 m/s2 for a time of 4.10 seconds. Determine the displacement of Ben's car during this time period. Once more, the solution to this problem begins by the construction of an informative diagram of the physical situation. This is shown below. The second step of the strategy involves the identification and listing of known information in variable form. Note that the vi value can be inferred to be 0 m/s since Ben's car is initially at rest. The acceleration (a) of the car is 6.00 m/s2. And the time (t) is given as 4.10 s. The next step of the strategy involves the listing of the unknown (or desired) information in variable form. In this case, the problem requests information about the displacement of the car. So d is the unknown information. The results of the first three steps are shown in the table below. Diagram: Given: Find: vi = 0 m/s t = 4.10 s a = 6.00 m/s2 d = ?? The next step of the strategy involves identifying a kinematic equation that would allow you to determine the unknown quantity. There are four kinematic equations to choose from. Again, you will always search for an equation that contains the three known variables and the one unknown variable. In this specific case, the three known variables and the one unknown variable are t, vi, a, and d. An inspection of the four equations above reveals that the equation on the top left contains all four variables. d = vi • t + ½ • a • t2 Once the equation is identified and written down, the next step of the strategy involves substituting known values into the equation and using proper algebraic steps to solve for the unknown information. This step is shown below. d = (0 m/s) • (4.1 s) + ½ • (6.00 m/s2) • (4.10 s)2 d = (0 m) + ½ • (6.00 m/s2) • (16.81 s2) d = 0 m + 50.43 m d = 50.4 m The solution above reveals that the car will travel a distance of 50.4 meters. (Note that this value is rounded to the third digit.) The last step of the problem-solving strategy involves checking the answer to assure that it is both reasonable and accurate. The value seems reasonable enough. A car with an acceleration of 6.00 m/s/s will reach a speed of approximately 24 m/s (approximately 50 mi/hr) in 4.10 s. The distance over which such a car would be displaced during this time period would be approximately one-half a football field, making this a very reasonable distance. Checking for accuracy involves substituting the calculated value back into the equation for displacement and insuring that the left side of the equation is equal to the right side of the equation. Indeed, it is! The two example problems above illustrate how the kinematic equations can be combined with a simple problem-solving strategy to predict unknown motion parameters for a moving object. Provided that three motion parameters are known, any of the remaining values can be determined. In the next part of Lesson 6, we will see how this strategy can be applied to free fall situations. Or if interested, you can try some practice problems and check your answer against the given solutions. Kinematic Equations and Free Fall As mentioned in Lesson 5, a free-falling object is an object that is falling under the sole influence of gravity. That is to say that any object that is moving and being acted upon only be the force of gravity is said to be "in a state of free fall." Such an object will experience a downward acceleration of 9.8 m/s/s. Whether the object is falling downward or rising upward towards its peak, if it is under the sole influence of gravity, then its acceleration value is 9.8 m/s/s. Like any moving object, the motion of an object in free fall can be described by four kinematic equations. The kinematic equations that describe any object's motion are: The symbols in the above equation have a specific meaning: the symbol d stands for the displacement; the symbol t stands for the time; the symbol a stands for the acceleration of the object; the symbol vi stands for the initial velocity value; and the symbol vf stands for the final velocity. Applying Free Fall Concepts to Problem-Solving There are a few conceptual characteristics of free fall motion that will be of value when using the equations to analyze free fall motion. These concepts are described as follows: • An object in free fall experiences an acceleration of -9.8 m/s/s. (The - sign indicates a downward acceleration.) Whether explicitly stated or not, the value of the acceleration in the kinematic equations is -9.8 m/s/s for any freely falling object. • If an object is merely dropped (as opposed to being thrown) from an elevated height, then the initial velocity of the object is 0 m/s. • If an object is projected upwards in a perfectly vertical direction, then it will slow down as it rises upward. The instant at which it reaches the peak of its trajectory, its velocity is 0 m/s. This value can be used as one of the motion parameters in the kinematic equations; for example, the final velocity (vf) after traveling to the peak would be assigned a value of 0 m/s. • If an object is projected upwards in a perfectly vertical direction, then the velocity at which it is projected is equal in magnitude and opposite in sign to the velocity that it has when it returns to the same height. That is, a ball projected vertically with an upward velocity of +30 m/s will have a downward velocity of -30 m/s when it returns to the same height. These four principles and the four kinematic equations can be combined to solve problems involving the motion of free-falling objects. The two examples below illustrate application of free fall principles to kinematic problem-solving. In each example, the problem solving strategy that was introduced earlier in this lesson will be utilized. Example Problem A Luke Autbeloe drops a pile of roof shingles from the top of a roof located 8.52 meters above the ground. Determine the time required for the shingles to reach the ground. The solution to this problem begins by the construction of an informative diagram of the physical situation. This is shown below. The second step involves the identification and listing of known information in variable form. You might note that in the statement of the problem, there is only one piece of numerical information explicitly stated: 8.52 meters. The displacement (d) of the shingles is -8.52 m. (The - sign indicates that the displacement is downward). The remaining information must be extracted from the problem statement based upon your understanding of the above principles. For example, the vi value can be inferred to be 0 m/s since the shingles are dropped (released from rest; see note above). And the acceleration (a) of the shingles can be inferred to be -9.8 m/s2 since the shingles are free-falling (see note above). (Always pay careful attention to the + and - signs for the given quantities.) The next step of the solution involves the listing of the unknown (or desired) information in variable form. In this case, the problem requests information about the time of fall. So t is the unknown quantity. The results of the first three steps are shown in the table below. Diagram: Given: Find: vi = 0.0 m/s d = -8.52 m a = - 9.8 m/s2 t = ?? The next step involves identifying a kinematic equation that allows you to determine the unknown quantity. There are four kinematic equations to choose from. In general, you will always choose the equation that contains the three known and the one unknown variable. In this specific case, the three known variables and the one unknown variable are d, vi, a, and t. Thus, you will look for an equation that has these four variables listed in it. An inspection of the four equations above reveals that the equation on the top left contains all four variables. d = vi • t + ½ • a • t2 Once the equation is identified and written down, the next step involves substituting known values into the equation and using proper algebraic steps to solve for the unknown information. This step is shown below. -8.52 m = (0 m/s) • (t) + ½ • (-9.8 m/s2) • (t)2 -8.52 m = (0 m) *(t) + (-4.9 m/s2) • (t)2 -8.52 m = (-4.9 m/s2) • (t)2 (-8.52 m)/(-4.9 m/s2) = t2 1.739 s2 = t2 t = 1.32 s The solution above reveals that the shingles will fall for a time of 1.32 seconds before hitting the ground. (Note that this value is rounded to the third digit.) The last step of the problem-solving strategy involves checking the answer to assure that it is both reasonable and accurate. The value seems reasonable enough. The shingles are falling a distance of approximately 10 yards (1 meter is pretty close to 1 yard); it seems that an answer between 1 and 2 seconds would be highly reasonable. The calculated time easily falls within this range of reasonability. Checking for accuracy involves substituting the calculated value back into the equation for time and insuring that the left side of the equation is equal to the right side of the equation. Indeed it is! Example Problem B Rex Things throws his mother's crystal vase vertically upwards with an initial velocity of 26.2 m/s. Determine the height to which the vase will rise above its initial height. Once more, the solution to this problem begins by the construction of an informative diagram of the physical situation. This is shown below. The second step involves the identification and listing of known information in variable form. You might note that in the statement of the problem, there is only one piece of numerical information explicitly stated: 26.2 m/s. The initial velocity (vi) of the vase is +26.2 m/s. (The + sign indicates that the initial velocity is an upwards velocity). The remaining information must be extracted from the problem statement based upon your understanding of the above principles. Note that the vf value can be inferred to be 0 m/s since the final state of the vase is the peak of its trajectory (see note above). The acceleration (a) of the vase is -9.8 m/s2 (see note above). The next step involves the listing of the unknown (or desired) information in variable form. In this case, the problem requests information about the displacement of the vase (the height to which it rises above its starting height). So d is the unknown information. The results of the first three steps are shown in the table below. Diagram: Given: Find: vi = 26.2 m/s vf = 0 m/s a = -9.8 m/s2 d = ?? The next step involves identifying a kinematic equation that would allow you to determine the unknown quantity. There are four kinematic equations to choose from. Again, you will always search for an equation that contains the three known variables and the one unknown variable. In this specific case, the three known variables and the one unknown variable are vi, vf, a, and d. An inspection of the four equations above reveals that the equation on the top right contains all four variables. vf2 = vi2 + 2 • a • d Once the equation is identified and written down, the next step involves substituting known values into the equation and using proper algebraic steps to solve for the unknown information. This step is shown below. (0 m/s)2 = (26.2 m/s)2 + 2 •(-9.8m/s2) •d 0 m2/s2 = 686.44 m2/s2 + (-19.6 m/s2) •d (-19.6 m/s2) • d = 0 m2/s2 -686.44 m2/s2 (-19.6 m/s2) • d = -686.44 m2/s2 d = (-686.44 m2/s2)/ (-19.6 m/s2) d = 35.0 m The solution above reveals that the vase will travel upwards for a displacement of 35.0 meters before reaching its peak. (Note that this value is rounded to the third digit.) The last step of the problem-solving strategy involves checking the answer to assure that it is both reasonable and accurate. The value seems reasonable enough. The vase is thrown with a speed of approximately 50 mi/hr (merely approximate 1 m/s to be equivalent to 2 mi/hr). Such a throw will never make it further than one football field in height (approximately 100 m), yet will surely make it past the 10-yard line (approximately 10 meters). The calculated answer certainly falls within this range of reasonability. Checking for accuracy involves substituting the calculated value back into the equation for displacement and insuring that the left side of the equation is equal to the right side of the equation. Indeed, it is! Kinematic equations provide a useful means of determining the value of an unknown motion parameter if three motion parameters are known. In the case of a free-fall motion, the acceleration is often known. And in many cases, another motion parameter can be inferred through a solid knowledge of some basic kinematic principles.The Compound Book Quiz - 6th Grade Reading Comprehension 1. What is the name of the main character in "The Compound"? A. Eli Yanakakis B. Marcus Yanakakis C. Luke Yanakakis D. Rex Yanakakis 2. Where do Eli and his family go to escape? A. A mountain cabin B. An underground compound C. A secret island D. A military base 3. Who built the compound? A. The government B. Eli's grandfather C. Eli's father D. A mysterious organization 4. How long was the family supposed to stay in the compound? A. 6 months B. 15 years C. 10 years D. 1 year 5. Which family member supposedly didn't make it into the compound? A. Eli's mother B. Eli's twin brother Eddy C. Eli's younger sister D. Eli's father 6. What does Eli's father tell the family happened outside? A. An earthquake B. A terrorist attack C. A nuclear war D. A deadly virus 7. What unusual food source does Eli's father plan to use in the compound? A. Artificial meat B. Insects C. Supplements only D. A special protein formula 8. How does Eli spend most of his time in the compound? A. Reading books B. Playing video games C. Exercising D. Isolating himself 9. What discovery does Eli make that makes him suspicious? A. Hidden cameras B. A working internet connection C. Secret passages D. Extra food supplies 10. What is the name of Eli's youngest sister born in the compound? A. Lexie B. Quinn C. Terese D. Lucy 11. What hobby does Eli take up to cope with his time underground? A. Painting B. Writing C. Playing music D. Cooking 12. What does Eli find in his father's private office? A. Family photos B. A radio C. A computer D. Survival guides 13. What makes Eli realize his father might be lying? A. Found newspapers B. Working internet C. Phone signals D. Television broadcasts 14. Who helps Eli discover the truth about the compound? A. His mother B. His sister C. The supplements supplier D. His grandmother 15. What is revealed about Eddy? A. He died in an accident B. He's living with relatives C. He's alive outside D. He never existed 16. What does Eli's father do when confronted with the truth? A. Confesses immediately B. Threatens the family C. Tries to escape D. Denies everything 17. How do they eventually escape the compound? A. Through air vents B. Using explosives C. With outside help D. Through emergency exit 18. What was Eli's father's true motivation for building the compound? A. Protection from war B. A psychological experiment C. Financial gain D. Government orders 19. How many years have they actually been in the compound when they escape? A. 6 years B. 3 years C. 15 years D. 9 years 20. What happens to Eli's relationship with his twin after the escape? A. They become close again B. They remain estranged C. They never meet D. They become rivals ANSWER KEY A B C B B C D D B A C C B B C B C B A ATHE SOAR SYSTEM A solar system is a group of planets and other celestial bodies that revolve around a star. A solar nebula- a vast cloud of gas and dust, mostly hydrogen and helium. How the Solar System Form • COLLAPSE AND SPINNING DISK FORMATION - Gravity pulls material inward. The cloud flattens into a spinning disk due to conservation of angular momentum. • PROTOSTAR FORMATION- (BIRTH OF THE SUN). Material collects at the center, and begun to heat up. When it reaches to 10 million KELVIN, nuclear fusion begins. thus, SUN is born. • PLANETESIMALS AND PROTOPLANETS. Dust and gas in the disk stick together via static and gravitational forces. These form planetesimals, which grow into protoplanets collision and accretion. • PLANET FORMATION. Inner disk: too hot for gas rocky planets form Mercury, Venus, Earth, Mars. • PLANET FORMATION. Outer disk: gas and ice giants. Jupiter, Saturn, Uranus, Neptune • LEFTOVER DEBRIS. Remaining materials forms moon, asteroids, comets and dwarf planets. DIFFERENT HYPOTHESIS IN THE FORMATION OF SOLAR SYSTEM. 1. NEBULAR HYPOTHESIS- The Solar system formed from a rotating cloud of Gas and Dust (solar nebula). As it rotates conservation of angular momentum caused the cloud to flatten into a disk. the Sun formed at the center (DISK) while planets formed from the surrounding materials through acceleration. thus, it explains the coplanar and nearly circular orbit of the planets all planets orbits around the sun on the same flat, disk shaped plane. Proposed by Immanuel Kant in 1755 and Modified by Pierre Simon Laplace in 1756. PROTOPLANET HYPOTHESIS. The Solar system formed from a rotating cloud of Gas and Dust (solar nebula). As it rotates conservation of angular momentum caused the cloud to flatten into a disk. 2. Protoplanet hypothesis. Builds on the nebular model but focuses more on the role of planetesimals which then form into full planets. PROCESS: - Small solid particles stick together through collisions. As collisions takes place, it grows into kilometer-sized planetesimals. Gravitational interactions lead to the formation of planets. Lead to formation of steroids belts and varying planet sizes 3. Encounter hypothesis. States that the sun encountered a rogue star. The encounter led to the removal of hot gas from both stars due to their gravitational interaction. The hot gas then accumulated and formed the planets. The materials from the less dense rogue star formed the other planets, while that from the sun formed the inner planets. 4. TIDAL HYPOTHESIS. (also called the Tidal Theory) is an early scientific idea about how the solar system might have formed. Proposed by James Jeans and Harold Jeffreys. A massive star passed very close to the early Sun. The hot gas then accumulated and formed the planets. The materials from the less dense rogue star formed the other planets, while that from the sun formed the inner planets. Streams of hot gas were drawn out from the Sun in elongated shape. These streams eventually condensed and cooled, forming planets, moons, and other bodies in the solar system. 5. Not accepted theory. Later studies showed the streams of hot gas would disperse too quickly into space instead of condensing into planets. The theory also couldn’t explain the specific orbital patterns and compositions we see today. Modern science favors the Nebular Hypothesis, which explains solar system formation through the collapse of a rotating gas cloud. Earth as the only habitable planet 1. Right Distance from the Sun (The Goldilocks Zone). Not too hot, not too cold — just right for liquid water to exist. 2. Atmosphere with Oxygen. Earth has a mix of gases, especially oxygen, which most living things need to survive. 3. Liquid Water. Earth has oceans, rivers, and rain — water is essential for all life. 4. Magnetic Field. Earth’s magnetic field protects us from harmful solar radiation. 5. Stable Climate. The atmosphere and natural cycles keep temperatures and weather mostly stable over time. 6. Rich Resources. Earth has soil for growing food, minerals, and energy sources that support life and technology. Solar explorations 1. AUGUST 6, 2014. First space craft to orbit a comet (ROSETTA PROBE). Captures the comet photograph. -Comets have coma and tail as it approaches to the sun. 2. JULY 14, 2015. NASA’s New Horizons spacecraft made history by becoming the first spacecraft to fly by Pluto, giving us our first close-up look at the dwarf planet. First time visiting Pluto. Before this, Pluto was just a blurry dot in telescope images. Revealed a surprising world New Horizons showed mountains of ice, smooth plains, and a heart-shaped region called Tombaugh Regio. Changed what we knew. Scientists thought Pluto would be dull and frozen — instead, it turned out to be geologically active and incredibly complex. 3. SEPTEMBER 8, 2016. NASA launched OSIRIS-REx, the first U.S. mission to collect a sample from an asteroid and return it to Earth. Changed what we knew. Scientists thought Pluto would be dull and frozen — instead, it turned out to be geologically active and incredibly complex. OSIRIS-REx stands for: Origins, Spectral Interpretation, Resource Identification, Security–Regolith Explorer It was sent to study the asteroid Bennu, a near-Earth asteroid about 500 meters wide. Mission Goals: Collect a sample of surface material from Bennu Study the asteroid’s omposition, structure, and history. Mission Goals: Help scientists understand the origins of the solar system. Learn more about asteroids that could impact Earth. 4. August 12, 2018: Launch of NASA’s Parker Solar Probe, the first spacecraft to "touch" the Sun by flying through its outer atmosphere, called the corona. Mission Goal: To study the Sun up close and help scientists understand: How the solar wind (a stream of charged particles) is formed. Why the Sun’s corona is hotter than its surface. What causes solar storms that can affect Earth’s satellites and power grids. 5. November 26, 2018: NASA’s Insight Lander Touches Down on Mars. Its mission was focused on studying the interior of the Red Planet (crust, mantle, and core of the planet). Why the Sun’s corona is hotter than its surface. What causes solar storms that can affect Earth’s satellites and power grids 6. November 26, 2018: NASA’s Insight Lander Touches Down on Mars. Its mission was focused on studying the interior of the Red Planet (crust, mantle, and core of the planet) 7. JULY 30, 2020 PERSEVERANCE PROBE. Perseverance rover as part of the Mars 2020 mission aboard an Atlas V-541 rocket This marked a major step in Mars exploration. 8. DECEMBER 25, 2021-JAMES WEBB SPACE TELESCOPE. Investigate exoplanets’ atmospheres for signs of habitability. Observe the first galaxies formed after the Big Bang. Study the formation of stars and planetary systems. Look deeper into the infrared universe than ever before. RESULTS OF EXPLORATION • Evidence of Ancient Life-friendly Environment. • Sedimentary rocks formed in water-rich environments. • Signs of clay and carbonate minerals, which can preserve biosignatures (traces of past life). • Evidence of Ancient Life-friendly Environment. • Sedimentary rocks formed in water-rich environments. • Signs of clay and carbonate minerals, which can preserve biosignatures (traces of past life). • Evidence of Ancient Life-friendly Environment. • Sedimentary rocks formed in water-rich environments. • Signs of clay and carbonate minerals, which can preserve biosignatures (traces of past life).REX Unit 3B Vocabularu QuizREX unit 10A VocabularyREX Level 10 Unit 9 vocabulary Quizrex spectaculum dat I-IIaREX UNIT 2A - LOVE: A CHEMICAL REACTION