Physics 1st Ch 06 Gravitation and Gravity

Physics - First Paper Chapter 6: Gravitation and Gravity Selected Questions and Answers (HSC English Version) Chapter 6: Gravitation and Gravity B. Creative Questions 1. M R Earth Artificial satellite 700 km v M →Mass of the earth; R →Radius of the earth. [M = 6 × 1024 kg and R = 6.4 × 106 m] (a) What is gravitational intensity? (b) Can the kinetic energy of a body be negative? - explain. (c) Calculate the horizontal velocity of the satellite. Solution : M = 6 × 1024 kg, R = 6.4 × 106 m, h = 700 × 103 m. Orbital radius: r = R + h = (6.4 + 0.7) × 106 = 7.1 × 106 m. Orbital velocity: v = p GM/r = p (6.67 × 10−11 × 6 × 1024)/(7.1 × 106). v = r 4.002 × 1014 7.1 × 106 = p 5.637 × 107 v ≈7508 m/s ≈7.51 km/s (d) What change of time period will appear if the satellite is placed 1000 km above the earth’s surface? Analyse it. Answer: (a) Gravitational field intensity (strength) at a point is the gravitational force per unit mass placed at that point:⃗E =⃗F/m = −GM/r2 ˆr. Its SI unit is N/kg = m/s2. It is a vector directed toward the source mass. (b) No. Kinetic energy KE = 1 2mv2. Since mass m > 0 always and v2 ≥0 (a square is non- negative), KE ≥0 always. KE is zero only when the body is at

(d) For h = 1000 km: r′ = (6.4 + 1.0) × 106 = 7.4 × 106 m. From Kepler’s third law: T ∝r3/2. T ′ T = r′ r 3/2 = 7.4 7.1 3/2 = (1.0423)1.5 = 1.0639 T at h = 700 km: T = 2πr/v = 2π × 7.1 × 106/7508 ≈5940 s ≈99 min. T ′ = 1.0639 × 99 ≈105 min. Conclusion: Raising the satellite from 700 km to 1000 km increases the orbital radius by ≈4.2% and the time period from ≈99 min to ≈105 min (an increase of ≈6 min). By Kepler’s third law, a larger orbit has a longer period. 2. If a stone is thrown upward it comes back to the earth. But if a body is thrown at a particular or more velocity, then the body does not come back to the earth. The magnitude of this velocity is different for moon, mars, neptune. (a) What is escape velocity? (b) Why gravitational force is conservative force? (c) If the radius of the earth R = 6.4 × 106 m and g = 9.8 m/s2, then find the escape velocity of a body from the earth. (d) Why escape velocity from the earth, moon, mars, neptune and other planets are different? Give justification. Answer: (a) Escape velocity is the minimum initial speed with which a body must be projected vertically upward from the surface of a celestial body to overcome its gravitational field

(d) Give justification by mathematical analysis whether the weight of the astronaut will increase or decrease on the moon. Answer: (a) Centre of gravity is the point at which the entire weight of a body is considered to act. For a uniform gravitational field, the centre of gravity coincides with the centre of mass. It is the point where the resultant of all gravitational forces on the body acts. (b) When a body is thrown upward, it decelerates due to Earth’s gravitational attraction (F = mg, downward). If the initial velocity v < ve (escape velocity), the body cannot overcome the gravitational potential well. Its kinetic energy is converted to gravitational PE as it rises; it slows to zero velocity at maximum height and then falls back under gravity. Only if v ≥ve = 11.2 km/s can it escape. (c) (d) Astronaut’s weight on Earth: WE = 630 N. Mass: m = WE/gE = 630/9.8 ≈64.3 kg. gm = gE × 16/81 = 9.8 × 16/81 ≈1.936 m/s2. Weight on Moon: Wm = mgm = 64.3 × 1.936 ≈124.5 N. Conclusion: The astronaut’s weight decreases from 630 N to approximately 124.5 N on the Moon (about 1 5 of Earth weight), because the Moon’s gravitational acceleration is only ≈1.94 m/s2 vs Earth’s 9.8 m/s2. Mass remains