Physics first Secondary 2020 1st Session Gravitational Field The gravitational field: ● The universal gravitational law states that the gravitational attraction between two bodies is inversely proportional to square the distance between their centers. Thus, the gravitational force decreases gradually as the distance increases till the distance between the centers of the two bodies reaches a point at which the attraction ceases. ● Within this distance, there is a region in which the gravitational force appear, this region is called the gravitational field. Deduction of the gravitational field intensity (g): ● By assuming that a body of mass 1kg is placed in the Earth’s gravitational field and at a distance r from the center of the Earth, then the attraction force of the Earth to the body: F = mg = g……………1 And by applying the universal gravitational law: F = G = …………2 From 1 and 2: g = Where : (M) is the mass of the Earth that equals (5.98×1024kg). If the body is at: The Earth’s surface. A height (h) above A depth (h) below the the Earth’s surface. Earth’s surface. Then g = g = g = Where: (R) is the radius of the Earth that equals 1 MSE Physics first Secondary 2020 ● From the previous we notice that the gravitational field intensity at a certain point equals numerically the acceleration due to gravity of the Earth at this point. The gravitational field intensity: It is the gravitational force acting on a mass of 1 kg at a certain point. To compare the acceleration due gravity for two planets: = Note: The Earth is not perfectly spherical but it is oblate spheroid (a sphere that is squashed at its poles and swollen at the equator) and that is due to the centripetal force that originates from the rotation of the Earth around itself. The factors that affect gravitational field intensity (g) at a point: Mass of planet: Distance from the center of the The gravitational field intensity is planet: directly proportional to the mass The gravitational field intensity of the planet when the distance is inversely proportional to between the point and the center square the distance from the of the planet is center of the planet. constant. ∆g Slope = 1 = GM ∆g ∆ Slope = = r ∆ g = 퐆퐌 퐫ퟐ Example 1: A planet has a mass twice that of Earth and a diameter twice that of Earth. Find the ratio of the acceleration due gravity of the two planets. 2 MSE Physics first Secondary 2020 Answer: g Mp = 2 Me R1 = 2 Re = ? g g = = = g Example2: A satellite of mass 104kg orbits the Earth at a height of 600km from its surface, calculate: a) The acceleration due to the gravity of the Earth that affects the satellite on its orbit. b) The weight of the satellite in its orbit (Knowing that: R = 6378km, M = 5.98×1024kg, G = 6.67×10-11n.m2/kg2) Answer: m = 104kg h = 600km R = 6378km M = 5.98×1024kg G = 6.67×10-11N.m2/kg2 g = ? w = ? 11 a) g = = = = 8.19 m/s2 b) w = mg = 8.19 × 104N HW 1. The acceleration due to Earth’s gravity is…………… a) a general universal constant b) changeable according the height from the Earth’s surface c) different through the seasons of the year d) changeable depending on the distance between Earth and Sun 3 MSE Physics first Secondary 2020 2. A planet of mass 5.98×1024kg and its radius = 6378km, if G = 6.67×10-11 N.m2/kg2, so the intensity of the planet’s gravitational field at a point that lies at a distance 36000km from its surface equals…………..N/kg. a) 22.2×104 b) 22.2×102 c) 22.2×10-2 d) 22.2×10-4 3. On the scale of a building, gravitational field is……….. a) increasing b) decreasing c) uniform d) varying 4. Which of the following affect the strength of the gravitational field on the surface of a planet?........... (more than one answer is correct) a) The mass of the object at the surface b) The mass of the planet c) The radius of the planet d) The presence of air at the surface of the planet 5. Assign the acceleration of the Moon the symbol am and the mass of the Moon Mm. If a satellite with a mass Ms was placed in the same orbit of the Moon at the same speed, what is the gravitational field strength at the satellite’s orbit?............... a) am b) (Ms/Mm) am c) (Mm/Ms) am d) (Ms+Mm) am/Ms 6. If the acceleration due to gravity of the Earth at the orbit of a satellite that orbits the Earth is 2.5 m/s2, then the distance between the satellite and the Earth’s surface (h) equals…………. (Where: R is the radius of the Earth and the acceleration due to gravity at the Earth’s surface = 10 m/s2) a) 2 R b) R c) d) 7. If R is the radius of the Earth, so the height at which the weight of a body becomes its weight on the surface of the Earth is…………… a) 2 R b) R c) d) 4 MSE Physics first Secondary 2020 8. At a place, value of (g) is less by 1% than its value on the surface of the Earth (radius of Earth (R) = 6400 km). The place is………… a) 64km below the surface of the Earth b) 64km above the surface of the Earth c) 30km below the surface of the Earth d) 32km above the surface of the Earth 9. An object has a mass of 50kg and a weight of 500N when it is resting on the surface of the Earth. If it is moved to a height equal to three times the Earth’s radius, what is the object’s new weight?..................... a) 300 N b) 31.25 N c) 130 N d) 60.5 N 10. Two asteroids in the space are close enough to each other. Each has a mass of 6.69×1015 kg. If they are 100000 m apart, what is the gravitational acceleration that they experience?................... a) 3.89×1010 m/s2 b) 5.12×104 m/s2 c) 2.99×1011 m/s2 d) 4.46×10-5 m/s2 11. Two asteroids, one with a mass of 7.12×1018 kg and the other with a mass of 5.33×108 kg and the distance between them is 10×1010 m. What is the acceleration of the smaller asteroid?................. a) 3.79×105m/s2 b) 9.41×10-7m/s2 c) 4.75×10-14m/s2 d) 6.11×10-24m/s2 12. When climbing from the sea level the top of mount Everest, a hiker change elevation by 8848 m. By what percentage will the gravitational field of the Earth change during the climb?................ (where: the Earth’s mass is 6×1024 kg and its radius is 6.4×106 m) a) It will increase by approximately 0.3 % b) It will decrease by approximately 0.3 % c) It will increase by approximately 12 % d) it will decrease by approximately 12 % 5 MSE Physics first Secondary 2020 2nd Session Satellites ● Man has dreamed for centuries to explore the space and to achieve that dream, he continued developing probes and rockets that launch space ships to orbit the Earth or reach further to another planet like Mars. That dream has come true on the 4th of October 1957 when the first satellite (Sputnik) has sent into space. ● That was followed by sending other satellites and stepping onto the Moon. Still space exploration continues successfully. The idea of launching a satellite: ● A satellite in its orbit is considered as an object that falls freely towards the Earth’s surface (because it is under the effect of Earth’s gravity). In spite of this it never reaches the Earth’s surface. Isaac Newton explained this where he imagines that when projecting a cannon projectile from the top of a mountain in a horizontal direction (neglecting the air resistance): ● The projectile moves through a certain horizontal distance before falling on the ground. By increasing the velocity of projection, the horizontal distance moved by the projectile before falling on the ground increases. ● At a certain projecting velocity, where the curvature of the projectile path equalize with the curvature of the Earth’s surface, the projectile orbits the Earth in a roughly circular path and it becomes a follower of the Earth and named a satellite and this velocity is called the orbital velocity of the satellite. The orbital velocity of a satellite: It is the velocity that makes the satellite orbit the Earth in a roughly circular path where it is still at a constant distance from the Earth’s surface. Deducing the orbital velocity of a satellite (v) ● Assume that a satellite of mass (m) moves with a velocity (v) in an orbit of radius (r) around the Earth of mass (M) as in figure: 6 MSE Physics first Secondary 2020 − The force of attraction between the Earth and the satellite is given by the relation: F = G − The force of attraction between the Earth and the satellite is normal to the direction of the motion of the satellite so it makes the satellite move in a circular path: F = So the attraction force between the Earth and the satellite is the same centripetal force that acts on the satellite. ∴ G = , v2 = ∴ v = √ Where (G) is the gravitational constant and if the height that the satellite was launched to is (h) and the radius of the Earth is (R): r = R + h ∴ v = √ The factors that affect orbital velocity of a satellite: The planet mass: The orbit radius: Orbital velocity of a satellite is Orbit velocity of directly proportional to the square a satellite is root of the mass of the planet at inversely constant radius of proportional to the the orbit.