Physics 101 Homework Set 8 Exercises: ---------- Chapter 9: 2: The gravitational force is proportional to the mass (ie: if you double the mass, you double the gravitational force (=weight)). However, the acceleration equals the force divided by the mass (a = F/m) so if you double the mass, you halve the acceleration. Therefore the two effects cancel and heavy and light bodies fall at the same rate.. 10: If the space pod is inbetween the Earth and the moon, then the Earth's gravity and the moon's gravity pull in opposite directions. For the forces to cancel, they must have the same magnitude. If the space pod was exactly halfway between, then the Earth's gravity would exert more force than the moon's gravity because the Earth's mass is greater than the moon's mass (and the distances are the same). Therefore, for the force of gravity from both bodies to have the same magnitude, the space pod must be closer to the moon. 11: 1 N. Newton's third law. If the force of the Earth on the apple is 1 N, then the force of the apple on the Earth is also 1 N. 14: If the Earth's radius increased, then the distance from you to the center of the Earth also increased. Therefore, since the gravitational force is proportional to 1/r^2, the force, and hence your weight, will decrease. If the Earth shrinks (assuming that you are still on the Earth's surface), then the distance from you to the center of the Earth decreases. Since F is proportional to 1/r^2, the force, and hence your weight, increases. [Note that if you can somehow keep your old distance from the Earth while it shrinks, your weight will not change.] 26: If there are no tidal forces, then gravity tugs on the entire object equally. However, when there are tidal forces, gravity pulls harder on the part of the object closest to the planet (or moon or sun). Thus, tidal forces from the moon try to pull the Earth apart. Tidal forces from Saturn probably pulled moons apart [or kept them from forming in the first place] to form the planet's rings. Chapter 10: ----------- 12: Vertical and horizontal motion are independent of each other. Therefore your vertical motion will be the same, no matter how fast (or slow) you are moving horizontally. Therefore your hang time depends only on your vertical component of velocity when you jump. 18: The speed of a falling object does not depend on its mass. We explained this in Chapter 9, exercise 2. A satellite in orbit is falling around the Earth. It is also moving only under the influence of gravity (ie: gravity is the only force acting on it). Therefore its speed does not depend on its mass. 38: The orbiting space shuttle is already falling. It is moving only under the influence of gravity (ie: gravity is the only force acting on it). If you are already in free fall, then you cannot fall any faster than that. If you drop a wrench, it falls with you. This is why astronauts in the space shuttle appear weightless. The shuttle is in free fall and the astronauts inside the shuttle are in free fall. You can achieve the same effect on Earth either a) by jumping [let go of an object while you are in the middle of a jump and it will fall with you] or b) by riding the NASA 'Vomet Comet', an airplane that flies free fall trajectories so that astronaut trainees can experience a minute of weightlessness. Chapter 10 extra exercise: Suppose that you drop an object from an airplane travelling at constant velocity and further suppose that you can ignore air resistance. a) What will be its falling path (ie: its trajectory) as observed by someone at rest on the ground but off to the side where they have a clear view of the plane and object? (Draw a picture to show your answer.) b) What will be the falling path as observed by you looking downward from the airplane? c) Where will the object strike the ground relative to you in the airplane? a) As seen from an observer on the ground, its trajectory will be a parabola and will look just like the trajectory shown in figure 10.3c. b) As seen from the plane, it will appear to fall straight down. (Assuming no air resistance.) c) It will land directly beneath the airplane (NOT directly beneath the point where it was dropped, but directly beneath where the airplane is when it hits). Problems: --------- Chapter 9: ---------- 2: Jupiter is 300 times more massive than Earth but its gravitational force is only three times greater. F = G m1 m2 / r^2. m1 is the same (say 1 kg), G is the same, m2 is 300 times bigger, but F is only 3 times bigger. The only variable left is r. We can do this one of two ways. a) symbolic way: The force is 100 times weaker than we expect. If the radius is 10 times larger, then the force will be 100 times weaker. Thus, Jupiter's radius is 109 times that of Earth. b) number crunch way: A 1 kg mass weighs 10 N on Earth so it must weigh 30 N on Jupiter. If we solve the force equation for r, we get r = sqrt( (G m1 m2) / F ) = sqrt( (6.67*10^(-11) N m^2/kg^2) * 1 kg * 300 * 6*10^24 kg / 30 N ) = 6.3*10^7 m = 10 * r_earth 6: (See the footnote on p 165) g = G m_earth / r^2 = (6.67*10^(-11) N m^2/kg^2) * 6*10^24 kg / (6.58*10^6 m)^2 = 9.24 m/s^2 This is 100 * 9.24 / 9.8 = 94% of the gravitational acceleration on the Earth's surface. Chapter 10: ----------- 6: This problem is just like the problem we did in class where the rock fell 5 m vertically as it moved 20 m horizontally. First we figure out how long it takes to fall, then we figure out its horizontal velocity. The tennis ball must fall 1 m so it hits the ground before it goes out. Since d = 1/2 g t^2, we know d = 1 m and g = 10 m/s^2 so we need to find t. We can solve for t t = sqrt(2 d / g) = sqrt(2 * 1 m / 10 m/s^2) = 0.45 s Now we know the maximum horizontal velocity is v = distance / time = 12 m / 0.45 s = 27 m/s This is about 55 mph. We know that tennis serves can exceed 100 mph. Therefore, tennis serves must cross the net heading downward (as they clearly do). Extra credit: ------------- Chapter 9, problem 10: T_F = 4 G M R / d^3 We calculate various tidal forces here. (Note that the tidal force is actually an acceleration since the units are N/kg. Don't worry about that detail.) a) The moon on you. The distance from you to the moon is 3.8*10^8 m and the moon's mass is 7.3*10^22 kg. Your size is about 1 m so T_F = 4 * (6.67*10^(-11) N m^2/kg^2) * 7.3*10^22 kg * 1 m ___________________________________________________ (3.8 * 10^8 m)^3 = 3.5 * 10^(-13) N/kg b) A melon on you. T_F = 4 * (6.67*10^(-11) N m^2/kg^2) * 1 kg * 1 m / (1 m)^3 = 2.7 * 10^(-10) N/kg Gee, the melon exerts a tidal force on you that is 1000 times bigger than that of the moon!