Physics Ch5
What keeps a satellite up in its orbit around the earth?
A satellite remains in orbit due to the combination of gravitational force on the satellite directed towards the center of the orbit and the tangential speed of the satellite. First, the proper tangential speed had to be established by some other force than the gravitational force. Then, if the satellite has the proper combination of speed and radius such that the force required for circular motion is equal to the force of gravity on the satellite, then the satellite will maintain circular motion.
Why do airplanes bank when they turn? How would you compute the banking angle given its speed in radius of the turn?
Airplanes bank when they turn because in order to turn there must be a force that will be exerted towards the center of a circle. By tilting the wings, do you lift force on the wings has a non-vertical component which points toward the center of the curve, providing the Centripetal force. The banking angle can be computed from the free body diagram. The sum of vertical forces must be zero for the plane to execute a level turn, and so Fliftcos0=mg. The horizontal component of the lifting force must provide the centripetal force to move the airplane in a circle. Fliftsin0=mv2/r->mgsin0/cos0=mv2/r->tan0=v2/rg.
Will an object weigh more at the equator or at the poles? What two effects are at work? Do they oppose each other?
And object weighs more at the polls, due to two effects which complement (not oppose) each other. First of all, the Earth Who is slightly flattened at the poles and expanded at the equator, relative to a perfect sphere. Thus The mass at the poles is slightly closer to the center, and so experiences a slightly larger gravitational force. Secondly, objects at the equator have a centripetal acceleration due to the rotation of the earth that objects at the poles do not have. to provide that centripetal acceleration, the apparent weight (The radially outward normal force of the earth on an object) isslightly less than the gravitational pull inward. So the two effects both make the weight of an object at the equator less than at the poles.
The earth moves faster in its orbit around the sun in January than in July. Is the earth closer to the sun in January, or in July? Explain.
By Kepler's second law, the earth moves faster around the sun when it is nearest the sun. Kepler's second law says that an imaginary line drawn from the sun to the earth sweeps out equal areas in equal times. So when the earth is close to the sun, it must move faster to sweep out a given area then when the earth is far from the sun. Thus the earth is closer to the sun in January.
Explain how a runner experiences "free fall" or "apparent weightlessness" between steps.
When the runner has both feet off the ground, the only force on the runner is gravity-there is no normal force from the ground on the runner. This lack of normal force is interpreted as "free fall" and "apparent weightlessness".
Would it require less speed to launch a satellite towards the east or toward the west? Consider the earths rotation direction
In order to orbit, a satellite must reach an orbital speed relative to the center of the earth. Since the satellite is already moving eastward when launch due to the rotation speed at the surface of the earth, it requires less additional speed to launch it east to obtain the final orbital speed.
The mass of Pluto was not known until it was discovered to have a moon. Explain how this discovery enabled an estimate of Pluto's mass.
Let the mass of Pluto be M, the mass of the moon be m, the radius of the Means orbit be R, and the period of the moon's orbit be T. Then newtons second law for the moon orbiting Pluto will be F=GmM/R2. If that means orbit is a circle, then the form of the force must be centripetal, and so F=mv2/R. Equate that these two expressions for the force on the moon, and substitute the relationship for the circular orbit that v=2piR/T. M=4pi2R3/GT2. Bus a value for the mass of Pluto can be calculated knowing the period in radius of the moon's orbit.
A girl is whirling a ball on a string around her head in a horizontal plane. She wants to let go at precisely the right time so that the ball will hit a target on the other side of the yard. When should she let go of the string?
She should let go of the string when the ball is at a position where the tangent to the circle at the balls location, when extended, passes through the targets position. That tangent line indicates the direction of the velocity at that instant, and if the centripetal force is removed, then the ball will follow that line horizontally.
When will your apparent weight be the greatest, as measured by a scale in a moving elevator: when the elevator accelerates downward, accelerates upward, is in freefall, moves upward at a constant speed? In which case would your weight be the least? When would it be the same as when you are on the ground?
The apparent weight (the normal force) would be largest when the elevator is accelerating upward. From the freebody diagram, with up as positive, we have Fn-mg=ma->Fn=ma+mg. With a positive acceleration, the normal force is greater than your weight. The apparent weight would be the least when in freefall, because there the apparent weight is zero, since a=-g. When the elevator is moving with a constant speed, your apparent weight would be the same as it is on the ground, since a=0 and so Fn=mg.
Does an Apple exert a gravitational force on the earth? If so, how large a force? Consider an apple (a) attached to a tree, and (b) falling.
The apple does exert a gravitational force on the earth. By Newton's third law, the force on the earth due to the apple is the magnitude as the force on the apple due to the earth - the weight of the apple. The force is also independent of the state of motion of the apple. So for both a hanging apple and a falling apple, the force on the earth due to the apple is equal to the weight of the apple.
Is the centripetal acceleration of mars in its orbit around the sun larger or smaller than the centripetal acceleration of the earth?
The centripetal acceleration of Mars is smaller than the earth. The acceleration of each planet can be found by dividing force on each planet by the planet's mass. The resulting acceleration is inversely proportional to the square of the distance of the planet from the sun. Since Mars is further from the sun than earth, the acceleration of mars will be smaller.
The sun's gravitational pull on the earth is much larger than the moon's. Yet the moon's is mainly responsible for the tides. Explain. (hint: consider the difference in gravitational pull from one side of the earth to the other.)
The difference in force on the two sides of the earth form from the gravitational pull of either the sun or the moon is the primary cause of the tides. That difference in the force comes about from the fact that the two sides of the earth are a different distance away from the pulling body. Relative to the side, the difference in distance (Earth diameter) of the two sides from the sun, relative to the average distance to the sun, is given by 2Rearth/Rearth to sun=8.5x10-5. 2Rearth/Rearth=3.3x10-2. Since the relative changein distance is much greater for the earth - moon combination, we see that the moon is primary cause of the earth's tides.
If the earth's mass were double what it is, in what ways would the Moon's orbit be different?
The gravitational force on the moon given by GMearthMmoon/R2, where R is the radius of the moon's orbit. This is a radial force, and so can be expressed as MmoonVmoon2/R. This can be changed using the relationship Vmoon=2piR/T, where T is the orbital period of the moon, to 4pi2MmoonR/T2. If we equate to expressions for the forest, we get the following: GMearthMmoon/R2=4pi2MmoonR/T2->R3/T2=GMearth/4pi2. Thus the mass of the earth determines the ratio R'3/T'2. If the mass of the earth were doubled, then the ratio R'3/T'2 would double, and so R'3/T'2=R3/T2, where the primes indicated the "after doubling" conditions. For example, the radius might stay the same, and the period decrease by a factor of square root of two, which means the speed increased by a factor of square root of two. Or the period might stay the same, and the radius increase by a factor of 2^1/3, which means the speed increased by the same factor of 2^1/3. Or if both R and T were to double, keeping the speed constant, then R3/T2 would double. There are an infinite number of other combinations that would also satisfy the doubling of R3/T2.
Which pulls harder gravitationally, the earth on the moon, or the moon on the earth? Which accelerates more?
The gravitational pullis the same in each case, by newton's third law. The magnitude of that pull is given by F=GMearthMmoon/r2earth-moon. To find the acceleration of each body, the gravitational pulling force is divided by the mass of the body. Since the moon has the smaller mass, it will have the larger acceleration.
The gravitational force on the moon due to the earth is only about half of the force on the moon due to the sun. Why isn't the moon pulled away from the earth?
The moon is not pulled away from the earth because both the moon and the earth are experiencing the same radio acceleration due to the sun. They both have the same period around the sun because they are both, on average, the same distance away from the sun, and so they travel around the sun together.
Astronauts who spend long periods outer space could be adversely affected by weightlessness. One way to simulate gravity is to shape the spaceships like a cylindrical shell that rotates, with the astronauts walking on the inside surface. Explain how this simulate gravity. Consider (a) how objects fall, (b) the force we feel on our feet, and (c) any other aspects of gravity you can think of.
The passengers are standing on the floor. (a) if a passenger held an object beside their waist and then released it, the object would move in a straight line, tangential to the circle in which the passenger's waist was moving when the object was released. In the figure, we see that the released object would hit the rotating she'll, and so fall on the floor, but behind the person. The passenger might try to explain the motion by inventing some kind of "retarding" force on dropped objects, when there really is no such force. (b) the floor exerts a centripetal force on the feet, pushing them toward the center. This force has the same direction that a passenger would experience on earth, and so it seems to the passenger that gravity must be pulling them "down". Actually, the passengers are pushing down on the floor, because the floor is pushing up from them. (c) the "normal" way of playing catch, for example, would have to change. Since the artificial gravity is not uniform, passengers would have to re-learn how to throw something across the room to each other. There would not be projectile motion as we experience get it on earth. Also, if the cylinder were small, there might be a noticeable difference in the acceleration of our head vs. our feet. Since the head is closer to the center of the circle then the feet, and both the head and the feet of the same period of rotation, the centripetal acceleration is smaller for the head. This might cause dizziness or a lightheaded feeling.
How many "accelerators" do you have in your car? There are at least three controls in the car which can be used to cause the car to accelerate. What are they? What acceleration do they produce?
The three major accelerators are the accelerator pedal, the brake pedal, and the steering wheel. The accelerator pedal can be used to increase speed or to decrease speed in combination with friction. The brake pedal can be used to decrease speed by depressing it. The steering wheel is used to change direction, which is also an acceleration.
Why do bicycle riders lean inward when rounding a curve at high speed?
When a bicycle rider leans inward, the bike tire pushes down the ground at an angle. The road surface then pushes back on the tire both vertically (to provide the normal force which counteracts gravity) and horizontally toward the center of the curve (to provide the centripetal frictional force, enabling them to turn).
A child on a sled comes flying over the crest of a small hill. His sled does not leave the ground, but he feels the normal force between his chest and the sled decrease as he goes over the hill. Explain this decrease using Newton's second law.
When the child is on a level surface, the normal force between his chest and the sled is equal to the child's weight, and thus he has no vertical acceleration. When he goes over the hill, the normal force on him will be reduced. Since the child is moving on a curved path, there must be a net centripetal force towards the center of the path, and so the normal force does not completely support the weight. Fr=mg-Fn=mv2/r->Fn=mg-mv2/r. Normal force is reduced from mg by the centripetal force.
